Introduction To Computational Finance and Financial Econometrics
Introduction To Computational Finance and Financial Econometrics
Introduction To Computational Finance and Financial Econometrics
Financial Econometrics
Chapter 1 Asset Return Calculations
Eric Zivot
Department of Economics, University of Washington
December 31, 1998
Updated: January 7, 2002
F Vn = $V · (1 + R)n .
If interest is paid m time per year then the future value after n years is
µ ¶m·n
R
F Vnm = $V · 1 + .
m
1
R
m
is often referred to as the periodic interest rate. As m, the frequency of
compounding, increases the rate becomes continuously compounded and it
can be shown that future value becomes
µ ¶m·n
R
F Vnc lim $V · 1 +
= m→∞ = $V · eR·n ,
m
Example 2 If the simple annual percentage rate is 10% then the value of
$1000 at the end of one year (n = 1) for different values of m is given in the
table below.
2
Example 3 To determine the simple annual rate with quarterly payments
that produces an effective annual rate of 12%, we solve
µ ¶
R 4
1.12 = 1+ =⇒
³
4 ´
R = (1.12)1/4 − 1 · 4
= 0.0287 · 4
= 0.1148
Rc = 2 · ln(1.05) = 0.09758.
3
2.1 Simple Returns
Let Pt denote the price in month t of an asset that pays no dividends and
let Pt−1 denote the price in month t − 11 . Then the one month simple net
return on an investment in the asset between months t − 1 and t is defined
as
Pt − Pt−1
Rt = = %∆Pt . (4)
Pt−1
Pt −Pt−1 Pt
Writing Pt−1
= Pt−1
− 1, we can define the simple gross return as
Pt
1 + Rt =
. (5)
Pt−1
Notice that the one month gross return has the interpretation of the future
value of $1 invested in the asset for one month. Unless otherwise stated,
when we refer to returns we mean net returns.
(mention that simple returns cannot be less than 1 (100%) since prices
cannot be negative)
Example 5 Consider a one month investment in Microsoft stock. Suppose
you buy the stock in month t − 1 at Pt−1 = $85 and sell the stock the next
month for Pt = $90. Further assume that Microsoft does not pay a dividend
between months t − 1 and t. The one month simple net and gross returns are
then
$90 − $85 $90
Rt = = − 1 = 1.0588 − 1 = 0.0588,
$85 $85
1 + Rt = 1.0588.
The one month investment in Microsoft yielded a 5.88% per month return.
Alternatively, $1 invested in Microsoft stock in month t − 1 grew to $1.0588
in month t.
4
Pt Pt Pt−1
Since Pt−2
= Pt−1
· Pt−2
the two-month return can be rewritten as
Pt Pt−1
Rt (2) = · −1
Pt−1 Pt−2
= (1 + Rt )(1 + Rt−1 ) − 1.
Example 6 Continuing with the previous example, suppose that the price of
Microsoft stock in month t − 2 is $80 and no dividend is paid between months
t − 2 and t. The two month net return is
$90 − $80 $90
Rt (2) = = − 1 = 1.1250 − 1 = 0.1250,
$80 $80
or 12.50% per two months. The two one month returns are
$85 − $80
Rt−1 = = 1.0625 − 1 = 0.0625
$80
$90 − 85
Rt = = 1.0588 − 1 = 0.0588,
$85
and the geometric average of the two one month gross returns is
5
2.3 Annualizing returns
Very often returns over different horizons are annualized, i.e. converted to
an annual return, to facilitate comparisons with other investments. The an-
nualization process depends on the holding period of the investment and an
implicit assumption about compounding. We illustrate with several exam-
ples.
To start, if our investment horizon is one year then the annual gross and
net returns are just
Pt
1 + RA = 1 + Rt (12) = = (1 + Rt )(1 + Rt−1 ) · · · (1 + Rt−11 ),
Pt−12
Pt
RA = − 1 = (1 + Rt )(1 + Rt−1 ) · · · (1 + Rt−11 ) − 1.
Pt−12
In this case, no compounding is required to create an annual return.
Next, consider a one month investment in an asset with return Rt . What
is the annualized return on this investment? If we assume that we receive
the same return R = Rt every month for the year then the gross 12 month
or gross annual return is
1 + RA = 1 + Rt (12) = (1 + R)12 .
Notice that the annual gross return is defined as the monthly return com-
pounded for 12 months. The net annual return is then
RA = (1 + R)12 − 1.
Example 7 In the first example, the one month return, Rt , on Microsoft
stock was 5.88%. If we assume that we can get this return for 12 months then
the annualized return is
RA = (1.0588)12 − 1 = 1.9850 − 1 = 0.9850
or 98.50% per year. Pretty good!
6
Here the annual gross return is defined as the two month return compounded
for 6 months.
Example 8 In the second example, the two month return, Rt (2), on Mi-
crosoft stock was 12.5%. If we assume that we can get this two month return
for the next 6 two month periods then the annualized return is
RA = (1.1250)6 − 1 = 2.0273 − 1 = 1.0273
or 102.73% per year.
7
3 Continuously Compounded Returns
3.1 One Period Returns
where ln(·) is the natural log function2 . To see why rt is called the con-
tinuously compounded return, take the exponential of both sides of (6) to
give
Pt
ert = 1 + Rt = .
Pt−1
Rearranging we get
Pt = Pt−1 ert ,
so that rt is the continuously compounded growth rate in prices between
months t − 1 and t. This is to be contrasted with Rt which is the simple
growth rate in prices between
³ ´ months t − 1 and t without any compounding.
Furthermore, since ln xy = ln(x) − ln(y) it follows that
à !
Pt
rt = ln
Pt−1
= ln(Pt ) − ln(Pt−1 )
= pt − pt−1
Example 10 Using the price and return data from example 1, the continu-
ously compounded monthly return on Microsoft stock can be computed in two
ways:
rt = ln(1.0588) = 0.0571
2
The continuously compounded return is always defined since asset prices, Pt , are
always non-negative. Properties of logarithms and exponentials are discussed in the ap-
pendix to this chapter.
8
or
rt = ln(90) − ln(85) = 4.4998 − 4.4427 = 0.0571.
Notice that rt is slightly smaller than Rt . Why?
Rt = ert − 1
Rt = e.0571 − 1 = 0.0588.
9
so that rt (2) is the continuously compounded growth rate of prices between
months t − 2 and t. Using PPt−2 t
= PPt−1
t
· PPt−1
t−2
and the fact that ln(x · y) =
ln(x) + ln(y) it follows that
à !
Pt Pt−1
rt (2) = ln ·
Pt−1 Pt−2
à ! à !
Pt Pt−1
= ln + ln
Pt−1 Pt−2
= rt + rt−1 .
Hence the continuously compounded two month return is just the sum of the
two continuously compounded one month returns. Recall that with simple
returns the two month return is of a multiplicative form (geometric average).
The second way uses the sum of the two continuously compounded one month
returns. Here rt = ln(90) − ln(85) = 0.0571 and rt−1 = ln(85) − ln(80) =
0.0607 so that
rt (2) = 0.0571 + 0.0607 = 0.1178.
Notice that rt (2) = 0.1178 < Rt (2) = 0.1250.
10
3.3 Annualizing Continuously Compounded Returns
Just as we annualized simple monthly returns, we can also annualize contin-
uously compounded monthly returns.
To start, if our investment horizon is one year then the annual continu-
ously compounded return is simply the sum of the twelve monthly continu-
ously compounded returns
rA = rt (12) = rt + rt−1 + · · · + rt−11
11
X
= rt−j .
j=0
Notice that
11
X
12 · rm = rt−j
j=0
4 Further Reading
This chapter describes basic asset return calculations with an emphasis on
equity calculations. Campbell, Lo and MacKinlay provide a nice treatment
of continuously compounded returns. A useful summary of a broad range
of return calculations is given in Watsham and Parramore (1998). A com-
prehensive treatment of fixed income return calculations is given in Stigum
(1981) and the official source of fixed income calculations is “The Pink Book”.
11
5 Appendix: Properties of exponentials and
logarithms
The computation of continuously compounded returns requires the use of
natural logarithms. The natural logarithm function, ln(·), is the inverse of
the exponential function, e(·) = exp(·), where e1 = 2.718. That is, ln(x) is
defined such that x = ln(ex ). Figure xxx plots ex and ln(x). Notice that ex
is always positive and increasing in x. ln(x) is monotonically increasing in x
and is only defined for x > 0. Also note that ln(1) = 0 and ln(−∞) = 0. The
exponential and natural logarithm functions have the following properties
6. ex ey = ex+y
7. ex e−y = ex−y
8. (ex )y = exy
9. eln(x) = x
d x
10. dx
e = ex
d f (x) d
11. dx
e = ef (x) dx f (x) (chain-rule)
6 Problems
Exercise 6.1 Excel exercises
12
download data from Yahoo. Read the data into Excel and make sure to re-
order the data so that time runs forward. Do your analysis on the monthly
closing price data (which should be adjusted for dividends and stock splits).
Name the spreadsheet tab with the data “data”.
1. Make a time plot (line plot in Excel) of the monthly price data over the
period (end of December 1996 through (end of) December 2001. Please
put informative titles and labels on the graph. Place this graph in a
separate tab (spreadsheet) from the data. Name this tab “graphs”.
Comment on what you see (eg. price trends, etc). If you invested
$1,000 at the end of December 1996 what would your investment be
worth at the end of December 2001? What is the annual rate of return
over this five year period assuming annual compounding?
2. Make a time plot of the natural logarithm of monthly price data over
the period December 1986 through December 2000 and place it in the
“graph” tab. Comment on what you see and compare with the plot of
the raw price data. Why is a plot of the log of prices informative?
3. Using the monthly price data over the period December 1996 through
December 2001 in the “data” tab, compute simple (no compounding)
monthly returns (Microsoft does not pay a dividend). When computing
returns, use the convention that Pt is the end of month closing price.
Make a time plot of the monthly returns, place it in the “graphs” tab
and comment. Keep in mind that the returns are percent per month
and that the annual return on a US T-bill is about 5%.
4. Using the simple monthly returns in the “data” tab, compute simple
annual returns for the years 1996 through 2001. Make a time plot of the
annual returns, put them in the “graphs” tab and comment. Note: You
may compute annual returns using overlapping data or non-overlapping
data. With overlapping data you get a series of annual returns for every
month (sounds weird, I know). That is, the first month annual return
is from the end of December, 1996 to the end of December, 1997. Then
second month annual return is from the end of January, 1997 to the
end of January, 1998 etc. With non-overlapping data you get a series of
5 annual returns for the 5 year period 1996-2001. That is, the annual
return for 1997 is computed from the end of December 1996 through
13
the end of December 1997. The second annual return is computed from
the end of December 1997 through the end of December 1998 etc.
5. Using the monthly price data over the period December 1996 through
December 2001, compute continuously compounded monthly returns
and place then in the “data” tab. Make a time plot of the monthly
returns, put them in the ”graphs” tab and comment. Briefly compare
the continuously compounded returns to the simple returns.
Consider the following (actual) monthly closing price data for Microsoft
stock over the period December 1999 through December 2000
1. Using the data in the table, what is the simple monthly return between
December, 1999 and January 2000? If you invested $10,000 in Microsoft
at the end of December 1999, how much would the investment be worth
at the end of January 2000?
14
2. Using the data in the table, what is the continuously compounded
monthly return between December, 1999 and January 2000? Convert
this continuously compounded return to a simple return (you should
get the same answer as in part a).
3. Assuming that the simple monthly return you computed in part (1)
is the same for 12 months, what is the annual return with monthly
compounding?
5. Using the data in the table, compute the actual simple annual return
between December 1999 and December 2000. If you invested $10,000 in
Microsoft at the end of December 1999, how much would the investment
be worth at the end of December 2000? Compare with your result in
part (3).
6. Using the data in the table, compute the actual annual continuously
compounded return between December 1999 and December 2000. Com-
pare with your result in part (4). Convert this continuously com-
pounded return to a simple return (you should get the same answer
as in part 5).
7 References
References
[1] Campbell, J., A. Lo, and C. MacKinlay (1997), The Econometrics of
Financial Markets, Princeton University Press.
[2] Handbook of U.W. Government and Federal Agency Securities and Re-
lated Money Market Instruments, “The Pink Book”, 34th ed. (1990), The
First Boston Corporation, Boston, MA.
[3] Stigum, M. (1981), Money Market Calculations: Yields, Break Evens and
Arbitrage, Dow Jones Irwin.
15
[4] Watsham, T.J. and Parramore, K. (1998), Quantitative Methods in Fi-
nance, International Thomson Business Press, London, UK.
16
Introduction to Financial Econometrics
Chapter 2 Review of Random Variables and
Probability Distributions
Eric Zivot
Department of Economics, University of Washington
January 18, 2000
This version: February 21, 2001
1 Random Variables
We start with a basic de&nition of a random variable
For example, consider the price of Microsoft stock next month. Since the price
of Microsoft stock next month is not known with certainty today, we can consider
it a random variable. The price next month must be positive and realistically it
can t get too large. Therefore the sample space is the set of positive real numbers
bounded above by some large number. It is an open question as to what is the
best characterization of the probability distribution of stock prices. The log-normal
distribution is one possibility1 .
As another example, consider a one month investment in Microsoft stock. That
is, we buy 1 share of Microsoft stock today and plan to sell it next month. Then
the return on this investment is a random variable since we do not know its value
today with certainty. In contrast to prices, returns can be positive or negative and are
bounded from below by -100%. The normal distribution is often a good approximation
to the distribution of simple monthly returns and is a better approximation to the
distribution of continuously compounded monthly returns.
As a &nal example, consider a variable X de&ned to be equal to one if the monthly
price change on Microsoft stock is positive and is equal to zero if the price change
1
If P is a positive random variable such that ln P is normally distributed the P has a log-normal
distribution. We will discuss this distribution is later chapters.
1
is zero or negative. Here the sample space is trivially the set {0, 1}. If it is equally
likely that the monthly price change is positive or negative (including zero) then the
probability that X = 1 or X = 0 is 0.5.
De&nition 2 A discrete random variable X is one that can take on a &nite number
of n different values x1 , x2 , . . . , xn or, at most, an in&nite number of different values
x1 , x2 , . . . .
De&nition 3 The pdf of a discrete random variable, denoted p(x), is a function such
that p(x) = Pr(X = x). TheP pdf must satisfy (i) p(x) ≥ 0 for all x ∈ SX ; (ii) p(x) = 0
for all x ∈
/ SX ; and (iii) x∈SX p(x) = 1.
As an example, let X denote the annual return on Microsoft stock over the next
year. We might hypothesize that the annual return will be in! uenced by the general
state of the economy. Consider &ve possible states of the economy: depression, reces-
sion, normal, mild boom and major boom. A stock analyst might forecast different
values of the return for each possible state. Hence X is a discrete random variable
that can take on &ve different values. The following table describes such a probability
distribution of the return.
Table 1
State of Economy SX = Sample Space p(x) = Pr(X = x)
Depression -0.30 0.05
Recession 0.0 0.20
Normal 0.10 0.50
Mild Boom 0.20 0.20
Major Boom 0.50 0.05
2
The probability distribution described above can be given an exact mathematical
representation known as the Bernoulli distribution. Consider two mutually exclusive
events generically called success and failure . For example, a success could be a
stock price going up or a coin landing heads and a failure could be a stock price going
down or a coin landing tails. In general, let X = 1 if success occurs and let X = 0
if failure occurs. Let Pr(X = 1) = π, where 0 < π < 1, denote the probability of
success. Clearly, Pr(X = 0) = 1 − π is the probability of failure. A mathematical
model for this set-up is
p(x) = Pr(X = x) = π x (1 − π)1−x , x = 0, 1.
When x = 0, p(0) = π 0 (1 − π)1−0 = 1 − π and when x = 1, p(1) = π 1 (1 − π)1−1 = π.
This distribution is presented graphically in Figure 2.
That is, Pr(X ∈ A) is the area under the R∞probability curve over the interval A”. The
pdf p must satisfy (i) p(x) ≥ 0; and (ii) −∞ p(x)dx = 1.
A typical bell-shaped pdf is displayed in Figure 3. In that &gure the total area
under the curve must be 1, and the value of Pr(a ≤ X ≤ b) is equal to the area of
the shaded region. For a continuous random variable, p(x) 6= Pr(X = x) but rather
gives the height of the probability curve at x. In fact, Pr(X = x) = 0 for all values of
x. That is, probabilities are not de&ned over single points; they are only de&ned over
intervals.
3
and is presented graphically in Figure 4. Notice that the area under the curve over
the interval [a, b] integrates to 1 since
Z b Z b
1 1 1 1
dx = dx = [x]ba = [b − a] = 1.
a b−a b−a a b−a b−a
Suppose, for example, a = −1 and b = 1 so that b − a = 2. Consider computing
the probability that the return will be between -50% and 50%.We solve
Z 0.5
1 1 1 1
Pr(−50% < X < 50%) = dx = [x]0.5
−0.5 = [0.5 − (−0.5)] = .
−0.5 2 2 2 2
Next, consider computing the probability that the return will fall in the interval [0, δ]
where δ is some small number less than b = 1 :
Z
1 δ 1 1
Pr(0 ≤ X ≤ δ) = dx = [x]δ0 = δ.
2 0 2 2
Hence, probabilities are de&ned on intervals but not at distinct points. As a result,
for a continuous random variable X we have
4
The standard normal distribution is graphed in Figure 5. Notice that the distribution
is symmetric about zero; i.e., the distribution has exactly the same form to the left
and right of zero.
The normal distribution has the annoying feature that the area under the normal
curve cannot be evaluated analytically. That is
Z b
1 1 2
Pr(a < X < b) = √ · e− 2 x dx
a 2π
does not have a closed form solution. The above integral must be computed by
numerical approximation. Areas under the normal curve, in one form or another, are
given in tables in almost every introductory statistics book and standard statistical
software can be used to &nd these areas. Some useful results from the normal tables
are
Pr(−1 < X < 1) ≈ 0.67,
Pr(−2 < X < 2) ≈ 0.95,
Pr(−3 < X < 3) ≈ 0.99.
Finding Areas Under the Normal Curve In the back of most introductory
statistics textbooks is a table giving information about areas under the standard
normal curve. Most spreadsheet and statistical software packages have functions for
&nding areas under the normal curve. Let X denote a standard normal random
variable. Some tables and functions give Pr(0 ≤ X < z) for various values of z > 0,
some give Pr(X ≥ z) and some give Pr(X ≤ z). Given that the total area under
the normal curve is one and the distribution is symmetric about zero the following
results hold:
• Pr(X ≤ z) = 1 − Pr(X ≥ z) and Pr(X ≥ z) = 1 − Pr(X ≤ z)
• Pr(X ≥ z) = Pr(X ≤ −z)
• Pr(X ≥ 0) = Pr(X ≤ 0) = 0.5
The following examples show how to compute various probabilities.
Example 6 Find Pr(X ≥ 2). We know that Pr(X ≥ 2) = Pr(X ≥ 0) − Pr(0 ≤ X ≤
2) = 0.5 − Pr(0 ≤ X ≤ 2). From the normal tables we have Pr(0 ≤ X ≤ 2) = 0.4772
and so Pr(X ≥ 2) = 0.5 − 0.4772 = 0.0228.
Example 7 Find Pr(X ≤ 2). We know that Pr(X ≤ 2) = 1 − Pr(X ≥ 2) and using
the result from the previous example we have Pr(X ≤ 2) = 1 − 0.0228 = 0.9772.
Example 8 Find Pr(−1 ≤ X ≤ 2). First, note that Pr(−1 ≤ X ≤ 2) = Pr(−1 ≤
X ≤ 0) + Pr(0 ≤ X ≤ 2). Using symmetry we have that Pr(−1 ≤ X ≤ 0) = Pr(0 ≤
X ≤ 1) = 0.3413 from the normal tables. Using the result from the &rst example we
get Pr(−1 ≤ X ≤ 2) = 0.3413 + 0.4772 = 0.8185.
5
1.3 The Cumulative Distribution Function
De&nition 9 The cumulative distribution function (cdf), F, of a random variable X
(discrete or continuous) is simply the probability that X ≤ x :
The cdf for the discrete distribution of Microsoft is given in Figure 6. Notice that
the cdf in this case is a discontinuous step function.
The cdf for the uniform distribution over [a, b] can be determined analytically
since Z x
1 1 x−a
F (x) = Pr(X < x) = dt = [t]xa = .
b−a a b−a b−a
Notice that for this example, we can determine the pdf of X directly from the cdf via
d 1
p(x) = F 0 (x) = F (x) = .
dx b−a
The cdf of the standard normal distribution is used so often in statistics that it
is given its own special symbol:
Z x
1 1
Φ(x) = P (X ≤ x) = √ exp(− z 2 )dz,
−∞ 2π 2
where X is a standard normal random variable. The cdf Φ(x), however, does not
have an anaytic representation like the cdf of the uniform distribution and must be
approximated using numerical techniques.
FX (qα ) = Pr(X ≤ qα ) = α
6
The median of the distribution is 50% quantile. That is, the median satis&es
The 5% quantile and the median are illustrated in Figure xxx using the CDF FX as
well as the pdf fX .
If FX is invertible then qa may be determined as
qa = FX−1 (α)
where FX−1 denotes the inverse function of FX . Hence, the 5% quantile and the median
may be determined as
Using the inverse cdf, the 5% quantile and median, for example, are given by
Example 11 Let X˜N(0, 1). The quantiles of the standard normal are determined
from
qα = Φ−1 (α)
where Φ−1 denotes the inverse of the cdf Φ. This inverse function must be approxi-
mated numerically. Using the numerical approximation to the inverse function, the
5% quantile and median are given by
7
1.5 Shape Characteristics of Probability Distributions
Very often we would like to know certain shape characteristics of a probability distri-
bution. For example, we might want to know where the distribution is centered and
how spread out the distribution is about the central value. We might want to know
if the distribution is symmetric about the center. For stock returns we might want to
know about the likelihood of observing extreme values for returns. This means that
we would like to know about the amount of probability in the extreme tails of the
distribution. In this section we discuss four shape characteristics of a pdf:
Example 12 Using the discrete distribution for the return on Microsoft stock in
Table 1, the expected return is
E[X] = (−0.3) · (0.05) + (0.0) · (0.20) + (0.1) · (0.5) + (0.2) · (0.2) + (0.5) · (0.05)
= 0.10.
E[X] = 0 · (1 − π) + 1 · π = π
That is, the expected value of a Bernoulli random variable is its probability of success.
8
Example 14 Suppose X has a uniform distribution over the interval [a, b]. Then
Z b · ¸b
1 1 1 2
E[X] = xdx = x
b−a a b−a 2 a
1 £ 2 ¤
= b − a2
2(b − a)
(b − a)(b + a) b+a
= = .
2(b − a) 2
Example 15 Suppose X has a standard normal distribution. Then it can be shown
that Z ∞
1 1 2
E[X] = x · √ e− 2 x dx = 0.
−∞ 2π
Example 16 Using the discrete distribution for the return on Microsoft stock in
Table 1 and the result that µX = 0.1, we have
var(X) = (−0.3 − 0.1)2 · (0.05) + (0.0 − 0.1)2 · (0.20) + (0.1 − 0.1)2 · (0.5)
+(0.2 − 0.1)2 · (0.2) + (0.5 − 0.1)2 · (0.05)
= 0.020.
9
Example 17 Let X be a Bernoulli random variable with success probability π. Given
that µX = π it follows that
var(X) = (0 − π)2 · (1 − π) + (1 − π)2 · π
= π2 (1 − π) + (1 − π 2 )π
= π(1 − π) [π + (1 − π)]
= π(1 − π).
10
Hence, for a general normal random variable about 95% of the time we expect to see
values within ± 2 standard deviations from its mean. Observations more than three
standard deviations from the mean are very unlikely.
Y = eX .
Example 20 Let rt = ln(Pt /Pt−1 ) denote the continuously compounded monthly re-
turn on an asset and assume that rt ~ N (µ, σ 2 ). Let Rt = Pt −P
Pt
t−1
denote the simple
monthly return. The relationship between rt and Rt is given by rt = ln(1 + Rt ) and
1 +Rt = ert . Since rt is normally distributed 1+Rt is log-normally distributed. Notice
that the distribution of 1 + Rt is only de&ned for positive values of 1 + Rt . This is
appropriate since the smallest value that Rt can take on is −1.
11
lower expected return but we also get less return variability or risk. This example
illustrates the fundamental no free lunch principle of economics and &nance: you
can t get something for nothing. In general, to get a higher return you must take on
extra risk.
1.5.7 Skewness
The skewness of a random variable X, denoted skew(X), measures the symmetry of
a distribution about its mean value using the function g(X) = (X − µX )3 /σ 3X , where
σ 3X is just SD(X) raised to the third power. For a discrete random variable X with
sample space SX
P 3
E[(X − µX )3 ] x∈SX (x − µX ) · Pr(X = x)
skew(X) = = .
σ 3X σ 3X
Example 21 Using the discrete distribution for the return on Microsoft stock in
Table 1, the results that µX = 0.1 and σ X = 0.141, we have
skew(X) = [(−0.3 − 0.1)3 · (0.05) + (0.0 − 0.1)3 · (0.20) + (0.1 − 0.1)3 · (0.5)
+(0.2 − 0.1)3 · (0.2) + (0.5 − 0.1)3 · (0.05)]/(0.141)3
= 0.0
Example 22 Suppose X has a general normal distribution with mean µX and vari-
ance σ 2X . Then it can be shown that
Z ∞
(x − µX )3 1 − 12 (x−µX )2
skew(X) = 3
· √ e 2σ
X dx = 0.
−∞ σ X 2πσ 2
This result is expected since the normal distribution is symmetric about it s mean
value µX .
12
1.5.8 Kurtosis
The kurtosis of a random variable X, denoted kurt(X), measures the thickness in the
tails of a distribution and is based on g(X) = (X − µX )4 /σ 4X . For a discrete random
variable X with sample space SX
P 2
E[(X − µX )4 ] x∈SX (x − µX ) · Pr(X = x)
kurt(X) = = ,
σ 4X σ 4X
where σ 4X is just SD(X) raised to the fourth power. Since kurtosis is based on
deviations from the mean raised to the fourth power, large deviations get lots of
weight. Hence, distributions with large kurtosis values are ones where there is the
possibility of extreme values. In contrast, if the kurtosis is small then most of the
observations are tightly clustered around the mean and there is very little probability
of observing extreme values.
Example 23 Using the discrete distribution for the return on Microsoft stock in
Table 1, the results that µX = 0.1 and σ X = 0.141, we have
kurt(X) = [(−0.3 − 0.1)4 · (0.05) + (0.0 − 0.1)4 · (0.20) + (0.1 − 0.1)4 · (0.5)
+(0.2 − 0.1)4 · (0.2) + (0.5 − 0.1)4 · (0.05)]/(0.141)4
= 6.5
For a continuous random variable X with pdf p(x)
R∞
E[(X − µX )4 ] −∞
(x − µX )4 · p(x)dx
kurt(X) = = .
σ 4X σ 4X
Example 24 Suppose X has a general normal distribution mean µX and variance
σ 2X . Then it can be shown that
Z ∞
(x − µX )4 1 1 2
kurt(X) = 4
·p e− 2 (x−µX ) dx = 3.
−∞ σX 2
2πσ X
Hence a kurtosis of 3 is a benchmark value for tail thickness of bell-shaped distribu-
tions. If a distribution has a kurtosis greater than 3 then the distribution has thicker
tails than the normal distribution and if a distribution has kurtosis less than 3 then
the distribution has thinner tails than the normal.
Sometimes the kurtosis of a random variable is described relative to the kurtosis
of a normal random variable. This relative value of kurtosis is referred to as excess
kurtosis and is de&ned as
excess kurt(X) = kurt(X) − 3
If excess the excess kurtosis of a random variable is equal to zero then the random
variable has the same kurtosis as a normal random variable. If excess kurtosis is
greater than zero, then kurtosis is larger than that for a normal; if excess kurtosis is
less than zero, then kurtosis is less than that for a normal.
13
1.6 Linear Functions of a Random Variable
Let X be a random variable either discrete or continuous with E[X] = µX , var(X) =
σ 2X and let a and b be known constants. De&ne a new random variable Y via the
linear function of X
Y = g(X) = aX + b.
Then the following results hold:
• E[Y ] = aE[X] + b or µY = aµX + b.
• var(Y ) = a2 var(X) or σ 2Y = a2 σ 2X .
The &rst result shows that expectation is a linear operation. That is,
E[aX + b] = aE[X] + b.
In the second result notice that adding a constant to X does not affect its variance
and that the effect of multiplying X by the constant a increases the variance of X by
the square of a. These results will be used often enough that it useful to go through
the derivations, at least for the case that X is a discrete random variable.
Proof. Consider the &rst result. By the de&nition of E[g(X)] with g(X) = b+aX
we have
X
E[Y ] = (ax + b) · Pr(X = x)
x∈SX
X X
= a x · Pr(X = x) + b Pr(X = x)
x∈SX x∈SX
= aE[X] + b · 1
= aµX + b
= µY .
Next consider the second result. Since µY = aµX + b we have
var(Y ) = E[(Y − µy )2 ]
= E[(aX + b − (aµX + b))2 ]
= E[(a(X − µX ) + (b − b))2 ]
= E[a2 (X − µX )2 ]
= a2 E[(X − µX )2 ] (by the linearity of E[·])
= a2 var(X)
a2 σ 2X .
Notice that our proof of the second result works for discrete and continuous random
variables.
A normal random variable has the special property that a linear function of it is
also a normal random variable. The following proposition establishes the result.
14
Proposition 25 Let X ∼ N (µX , σ 2X ) and let a and b be constants. Let Y = aX + b.
Then Y ∼ N(aµX + b, a2 σ 2X ).
The above property is special to the normal distribution and may or may not hold
for a random variable with a distribution that is not normal.
Example 26 Let X ∼ N(2, 4) and suppose we want to &nd Pr(X > 5). Since X is
not standard normal we can t use the standard normal tables to evaluate Pr(X > 5)
directly. We solve the problem by standardizing X as follows:
µ ¶
X −2 5−2
Pr (X > 5) = Pr √ > √
4 4
µ ¶
3
= Pr Z >
2
¡ ¢
where Z ∼ N (0, 1) is the standardized value of X. Pr Z > 32 can be found directly
from the standard normal tables.
X = µX + σ X Z.
15
This result is useful for modeling purposes. For example, in Chapter 3 we will consider
the Constant Expected Return (CER) model of asset returns. Let R denote the
monthly continuously compounded return on an asset and let µ = E[R] and σ 2 =
var(R). A simpli&ed version of the CER model is
R =µ+σ·ε
where ε is a random variable with mean zero and variance 1. The random variable ε
is often interpreted as representing the random news arriving in a given month that
makes the observed return differ from the expected value µ. The fact that ε has mean
zero means that new, on average, is neutral. The value of σ represents the typical
size of a news shock.
(Stuff to add: General functions of a random variable and the change of variables
formula. Example with the log-normal distribution)
• What is the probability that end of month wealth is less than $9, 000 and what
must the return on Microsoft be for this to happen?
• What is the monthly VaR on the $10, 000 investment in Microsoft stock with
5% probability? That is, what is the loss that would occur if the return on
Microsoft stock is equal to its 5% quantile, q.05 ?
To answer the &rst question, note that end of month wealth W1 is related to initial
wealth W0 and the return on Microsoft stock R via the linear function
W1 = W0 (1 + R) = W0 + W0 R
= $10, 000 + $10, 000 · R.
E[W1 ] = W0 + W0 E[R]
= $10, 000 + $10, 000(0.05) = $10, 500
16
and
var(W1 ) = (W0 )2 var(R)
= ($10, 000)2 (0.10)2 ,
SD(W1 ) = ($10, 000)(0.10) = $1, 000.
Further, since R is assumed to be normally distributed we have
W1 ~ N ($10, 500, ($1, 000)2 )
To answer the second question, we use the above normal distribution for W1 to
get
Pr(W1 < $9, 000) = 0.067
To &nd the return that produces end of month wealth of $9, 000 or a loss of $10, 000 −
$9, 000 = $1, 000 we solve
$9, 000 − $10, 000
R∗ = = −0.10.
$10, 000
In other words, if the monthly return on Microsoft is −10% or less then end of
month wealth will be $9, 000 or less. Notice that −0.10 is the 6.7% quantile of the
distribution of R :
Pr(R < −0.10) = 0.067
The third question can be answered in two equivalent ways. First, use R ~N (0.05, (0.10)2 )
and solve for the the 5% quantile of Microsoft Stock:
R R
Pr(R < q.05 ) = 0.05 ⇒ q.05 = −0.114.
That is, with 5% probability the return on Microsoft stock is −11.4% or less. Now,
if the return on Microsoft stock is −11.4% the loss in investment value is $10, 000 ·
(0.114) = $1, 144. Hence, $1, 144 is the 5% VaR over the next month on the $10, 000
R
investment in Microsoft stock. In general, if W0 represents the initial wealth and q.05
is the 5% quantile of distribution of R then the 5% VaR is
R
5% VaR = |W0 · q.05 |.
For the second method, use W1 ~N ($10, 500, ($1, 000)2 ) and solve for the 5%
quantile of end of month wealth:
W1 W1
Pr(W1 < q.05 ) = 0.05 ⇒ q.05 = $8, 856
This corresponds to a loss of investment value of $10, 000 − $8, 856 = $1, 144. Hence,
W1
if W0 represents the initial wealth and q.05 is the 5% quantile of the distribution of
W1 then the 5% VaR is
W1
5% VaR = W0 − q.05 .
(insert VaR calculations based on continuously compounded returns)
17
1.8 Log-Normal Distribution and Jensen s Inequality
(discuss Jensen s inequality: E[g(X)] < g(E[X]) for a convex function. Use this
to illustrate the difference between E[W0 exp(R)] and W0 exp(E[R]) where R is a
continuously compounded return.) Note, this is where the log-normal distribution
will come in handy.
2 Bivariate Distributions
So far we have only considered probability distributions for a single random variable.
In many situations we want to be able to characterize the probabilistic behavior of
two or more random variables simultaneously.
For example, let X denote the monthly return on Microsoft Stock and let Y denote
the monthly return on Apple computer. For simplicity suppose that the sample
spaces for X and Y are SX = {0, 1, 2, 3} and SY = {0, 1} so that the random
variables X and Y are discrete. The joint sample space is the two dimensional
grid SXY = {(0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1)}. The likelihood that
X and Y takes values in the joint sample space is determined by the joint probability
distribution
p(x, y) = Pr(X = x, Y = y).
The function p(x, y) satis&es
/ SXY ;
(ii) p(x, y) = 0 for x, y ∈
P P P
(iii) x,y∈SXY p(x, y) = x∈SX y∈SY p(x, y) = 1.
Table 2
Y
% 0 1 Pr(X)
0 1/8 0 1/8
X 1 2/8 1/8 3/8
2 1/8 2/8 3/8
3 0 1/8 1/8
Pr(Y ) 4/8 4/8 1
18
For example, p(0, 0) = Pr(X = 0, Y = 0) = 1/8. Notice that sum of all the
entries in the table sum to unity. The bivariate distribution is illustrated graphically
in Figure xxx.
Bivariate pdf
0.25
0.2
0.15
p(x,y)
0.1
0.05
0 1
0 y
1 0
x 2
3
What if we want to know only about the likelihood of X occurring? For example,
what is Pr(X = 0) regardless of the value of Y ? Now X can occur if Y = 0 or if
Y = 1 and since these two events are mutually exclusive we have that Pr(X = 0) =
Pr(X = 0, Y = 0) + Pr(X = 0, Y = 1) = 0 + 1/8 = 1/8. Notice that this probability
is equal to the horizontal (row) sum of the probabilities in the table at X = 0. The
probability Pr(X = x) is called the marginal probability of X and is given by
X
Pr(X = x) = Pr(X = x, Y = y).
y∈SY
The marginal probabilities of X = x are given in the last column of Table 2. Notice
that the marginal probabilities sum to unity.
19
We can &nd the marginal probability of Y in a similar fashion. For example, using
the data in Table 2 Pr(Y = 1) = Pr(X = 0, Y = 1) + Pr(X = 1, Y = 1) + Pr(X =
2, Y = 1) + Pr(X = 3, Y = 1) = 0 + 1/8 + 2/8 + 1/8 = 4/8. This probability is the
vertical (column) sum of the probabilities in the table at Y = 1. Hence, the marginal
probability of Y = y is given by
X
Pr(Y = y) = Pr(X = x, Y = y).
x∈SX
The marginal probabilities of Y = y are given in the last row of Table 2. Notice that
these probabilities sum to 1.
For future reference we note that
Pr(X = 0, Y = 0) 1/8
Pr(X = 0|Y = 0) = = = 1/4.
Pr(Y = 0) 4/8
The notation Pr(X = 0|Y = 0) is read as the probability that X = 0 given that
Y = 0 . Notice that the conditional probability that X = 0 given that Y = 0 is
greater than the marginal probability that X = 0. That is, Pr(X = 0|Y = 0) =
1/4 > Pr(X = 0) = 1/8. Hence, knowledge that Y = 0 increases the likelihood that
X = 0. Clearly, X depends on Y.
Now suppose that we know that X = 0. How does this knowledge affect the
probability that Y = 0? To &nd out we compute
Pr(X = 0, Y = 0) 1/8
Pr(Y = 0|X = 0) = = = 1.
Pr(X = 0) 1/8
Notice that Pr(Y = 0|X = 0) = 1 > Pr(Y = 0) = 1/2. That is, knowledge that
X = 0 makes it certain that Y = 0.
In general, the conditional probability that X = x given that Y = y is given by
Pr(X = x, Y = y)
Pr(X = x|Y = y) =
Pr(Y = y)
20
and the conditional probability that Y = y given that X = x is given by
Pr(X = x, Y = y)
Pr(Y = y|X = x) = .
Pr(X = x)
For the example in Table 2, the conditional probabilities along with marginal
probabilities are summarized in Tables 3 and 4. The conditional and marginal distri-
butions of X are graphically displayed in &gure xxx and the conditional and marginal
distribution of Y are displayed in &gure xxx. Notice that the marginal distribution of
X is centered at x = 3/2 whereas the conditional distribution of X|Y = 0 is centered
at x = 1 and the conditional distribution of X|Y = 1 is centered at x = 2.
Table 3
x Pr(X = x) Pr(X|Y = 0) Pr(X|Y = 1)
0 1/8 2/8 0
1 3/8 4/8 2/8
2 3/8 2/8 4/8
3 1/8 0 2/8
Table 4
y Pr(Y = y) Pr(Y |X = 0) Pr(Y |X = 1) Pr(Y |X = 2) Pr(Y |X = 3)
0 1/2 1 2/3 1/3 0
1 1/2 0 1/3 2/3 1
21
Example 27 For the data in Table 2, we have
E[X|Y = 0] = 0 · 1/4 + 1 · 1/2 + 2 · 1/4 + 3 · 0 = 1
E[X|Y = 1] = 0 · 0 + 1 · 1/4 + 2 · 1/2 + 3 · 1/4 = 2
var(X|Y = 0) = (0 − 1)2 · 1/4 + (1 − 1)2 · 1/2 + (2 − 1)2 · 1/2 + (3 − 1)2 · 0 = 1/2
var(X|Y = 1) = (0 − 2)2 · 0 + (1 − 2)2 · 1/4 + (2 − 2)2 · 1/2 + (3 − 2)2 · 1/4 = 1/2.
Using similar calculations gives
E[Y |X = 0] = 0, E[Y |X = 1] = 1/3, E[Y |X = 2] = 2/3, E[Y |X = 3] = 1
var(Y |X = 0) = 0, var(Y |X = 1) = 0, var(Y |X = 2) = 0, var(Y |X = 3) = 0.
22
2.3 Bivariate Distributions for Continuous Random Variables
Let X and Y be continuous random variables de&ned over the real line. We character-
ize the joint probability distribution of X and Y using the joint probability function
(pdf) p(x, y) such that p(x, y) ≥ 0 and
Z ∞Z ∞
p(x, y)dxdy = 1.
−∞ −∞
For example, in Figure xxx we illustrate the pdf of X and Y as a bell-shaped surface
in two dimensions. To compute joint probabilities of x1 ≤ X ≤ x2 and y1 ≤ Y ≤ y2
we need to &nd the volume under the probability surface over the grid where the
intervals [x1 , x2 ] and [y1 , y2 ] overlap. To &nd this volume we must solve the double
integral Z x2 Z y2
Pr(x1 ≤ X ≤ x2 , y1 ≤ Y ≤ y2 ) = p(x, y)dxdy.
x1 y1
Example 29 A standard bivariate normal pdf for X and Y has the form
1 − 1 (x2 +y2 )
p(x, y) = e 2 , −∞ ≤ x, y ≤ ∞
2π
and has the shape of a symmetric bell centered at x = 0 and y = 0 as illustrated in
Figure xxx (insert &gure here). To &nd Pr(−1 < X < 1, −1 < Y < 1) we must solve
Z 1Z 1
1 − 1 (x2 +y2 )
e 2 dxdy
−1 −1 2π
The marginal pdf of X is found by integrating y out of the joint pdf p(x, y) and the
marginal pdf of Y is found by integrating x out of the joint pdf:
Z ∞
p(x) = p(x, y)dy,
−∞
Z ∞
p(y) = p(x, y)dx.
−∞
p(x, y)
p(x|y) =
p(y)
23
and the conditional pdf of Y given that X = x is computed as
p(x, y)
p(y|x) = .
p(x)
The conditional means are computed as
Z
µX|Y =y = E[X|Y = y] = x · p(x|y)dx,
Z
µY |X=x = E[Y |X = x] = y · p(y|x)dy
Z
σ Y |X=x = var(Y |X = x) = (y − µY |X=x )2 p(y|x)dy.
2
2.4 Independence
Let X and Y be two random variables. Intuitively, X is independent of Y if knowledge
about Y does not in! uence the likelihood that X = x for all possible values of x ∈ SX
and y ∈ SY . Similarly, Y is independent of X if knowledge about X does not in! uence
the likelihood that Y = y for all values of y ∈ SY . We represent this intuition formally
for discrete random variables as follows.
De&nition 30 Let X and Y be discrete random variables with sample spaces SX and
SY , respectively. X and Y are independent random variables iff
Pr(X = x|Y = y) = Pr(X = x), for all x ∈ SX , y ∈ SY
Pr(Y = y|X = x) = Pr(Y = y), for all x ∈ SX , y ∈ SY
Example 31 For the data in Table 2, we know that Pr(X = 0|Y = 0) = 1/4 6=
Pr(X = 0) = 1/8 so X and Y are not independent.
Proposition 32 Let X and Y be discrete random variables with sample spaces SX
and SY , respectively. If X and Y are independent then
Pr(X = x, Y = y) = Pr(X = x) · Pr(Y = y), for all x ∈ SX , y ∈ SY
For continuous random variables, we have the following de&nition of independence
De&nition 33 Let X and Y be continuous random variables. X and Y are indepen-
dent iff
p(x|y) = p(x), for − ∞ < x, y < ∞
p(y|x) = p(y), for − ∞ < x, y < ∞
24
Proposition 34 Let X and Y be continuous random variables . X and Y are inde-
pendent iff
p(x, y) = p(x)p(y)
The result in the proposition is extremely useful because it gives us an easy way
to compute the joint pdf for two independent random variables: we simple compute
the product of the marginal distributions.
Example 35 Let X ∼ N(0, 1), Y ∼ N (0, 1) and let X and Y be independent. Then
1 1 2 1 1 2 1 − 1 (x2 +y2 )
p(x, y) = p(x)p(y) = √ e− 2 x √ e− 2 y = e 2 .
2π 2π 2π
This result is a special case of the bivariate normal distribution.
(stuff to add: if X and Y are independent then f (X) and g(Y ) are independent
for any functions f (·) and g(·).)
Let X and Y be two discrete random variables. Figure xxx displays several bivariate
probability scatterplots (where equal probabilities are given on the dots).
In panel (a) we see no linear relationship between X and Y. In panel (b) we see a
perfect positive linear relationship between X and Y and in panel (c) we see a perfect
negative linear relationship. In panel (d) we see a positive, but not perfect, linear
relationship. Finally, in panel (e) we see no systematic linear relationship but we see a
strong nonlinear (parabolic) relationship. The covariance between X and Y measures
the direction of linear relationship between the two random variables. The correlation
between X and Y measures the direction and strength of linear relationship between
the two random variables.
Let X and Y be two random variables with E[X] = µX , var(X) = σ 2X , E[Y ] = µY
and var(Y ) = σ 2Y .
25
De&nition 37 The correlation between two random variables X and Y is given by
cov(X, Y ) σ XY
ρXY = corr(X, Y ) = p =
var(X)var(Y ) σX σY
Notice that the correlation coefficient, ρXY , is just a scaled version of the covari-
ance.
To see how covariance measures the direction of linear association, consider the
probability scatterplot in &gure xxx.
In the plot the random variables X and Y are distributed such that µX = µY = 0.
The plot is separated into quadrants. In the &rst quandrant, the realized values satisfy
x < µX , y > µY so that the product (x − µX )(y − µY ) < 0. In the second quadrant,
the values satisfy x > µX and y > µY so that the product (x − µX )(y − µY ) > 0.
In the third quadrant, the values satisfy x > µX but y < µY so that the product
(x − µX )(y − µY ) < 0. Finally, in the fourth quandrant, x < µX and y < µY so that
the product (x − µX )(y − µY ) > 0. Covariance is then a probability weighted average
all of the product terms in the four quadrants. For the example data, this weighted
average turns out to be positive.
1. cov(X, X) = var(X)
2. cov(X, Y ) = cov(Y, X)
3. cov(aX, bY ) = a · b · cov(X, Y )
4. If X and Y are independent then cov(X, Y ) = 0 (no association =⇒ no linear
association). However, if cov(X, Y ) = 0 then X and Y are not necessarily
independent (no linear association ; no association).
26
The third property above shows that the value of cov(X, Y ) depends on the scaling
of the random variables X and Y. By simply changing the scale of X or Y we can
make cov(X, Y ) equal to any value that we want. Consequently, the numerical value
of cov(X, Y ) is not informative about the strength of the linear association between
X and Y . However, the sign of cov(X, Y ) is informative about the direction of linear
association between X and Y. The fourth property should be intuitive. Independence
between the random variables X and Y means that there is no relationship, linear or
nonlinear, between X and Y. However, the lack of a linear relationship between X and
Y does not preclude a nonlinear relationship. The last result illustrates an important
property of the normal distribution: lack of covariance implies independence.
Some important properties of corr(X, Y ) are
1. −1 ≤ ρXY ≤ 1.
2. If ρXY = 1 then X and Y are perfectly positively linearly related. That is,
Y = aX + b where a > 0.
3. If ρXY = −1 then X and Y are perfectly negatively linearly related. That is,
Y = aX + b where a < 0.
4. If ρXY = 0 then X and Y are not linearly related but may be nonlinearly
related.
The &rst result states that the expected value of a linear combination of two
random variables is equal to a linear combination of the expected values of the random
variables. This result indicates that the expectation operator is a linear operator. In
other words, expectation is additive. The second result states that variance of a
linear combination of random variables is not a linear combination of the variances
of the random variables. In particular, notice that covariance comes up as a term
when computing the variance of the sum of two (not independent) random variables.
27
Hence, the variance operator is not, in general, a linear operator. That is, variance,
in general, is not additive.
It is worthwhile to go through the proofs of these results, at least for the case of
discrete random variables. Let X and Y be discrete random variables. Then,
X X
E[aX + bY ] = (ax + by) Pr(X = x, Y = y)
x∈SX y∈Sy
X X X X
= ax Pr(X = x, Y = y) + bx Pr(X = x, Y = y)
x∈SX y∈Sy x∈SX y∈Sy
X X X X
= a x Pr(X = x, Y = y) + b y Pr(X = x, Y = y)
x∈SX y∈Sy y∈Sy x∈SX
X X
= a x Pr(X = x) + b y Pr(Y = y)
x∈SX y∈Sy
28
This important result states that a linear combination of two normally distributed
random variables is itself a normally distributed random variable. The proof of the
result relies on the change of variables theorem from calculus and is omitted. Not all
random variables have the property that their distributions are closed under addition.
3 Multivariate Distributions
The results for bivariate distributions generalize to the case of more than two random
variables. The details of the generalizations are not important for our purposes.
However, the following results will be used repeatedly.
29
Since each monthly return is normally distributed, the annual return is also normally
distributed. In addition,
" 11 #
X
E[Rt (12)] = E Rt−j
j=0
11
X
= E[Rt−j ] (by linearity of expectation)
j=0
11
X
= µ (by identical distributions)
j=0
= 12 · µ,
so that the expected annual return is equal to 12 times the expected monthly return.
Furthermore,
à 11 !
X
var(Rt (12)) = var Rt−j
j=0
11
X
= var(Rt−j ) (by independence)
j=0
X
11
= σ 2 (by identical distributions)
j=0
= 12 · σ 2 ,
so that the annual variance is also equal to 12 times the monthly variance2 . For the
annual standard deviation, we have
√
SD(Rt (12)) = 12σ.
4 Further Reading
Excellent intermediate level treatments of probability theory using calculus are given
in DeGroot (1986), Hoel, Port and Stone (1971) and Hoag and Craig (19xx). Inter-
mediate treatments with an emphasis towards applications in &nance include Ross
(1999) and Watsom and Parramore (1998). Intermediate textbooks with an emphasis
on econometrics include Amemiya (1994), Goldberger (1991), Ramanathan (1995).
Advanced treatments of probability theory applied to &nance are given in Neftci
(1996). Everything you ever wanted to know about probability distributions is given
Johnson and Kotz (19xx).
2
This result often causes some confusion. It is easy to make the mistake and say that the annual
variance is (12)2 = 144 time the monthly variance. This result would occur if RA = 12Rt , so that
var(RA ) = (12)2 var(Rt ) = 144var(Rt ).
30
5 Problems
Let W, X, Y, and Z be random variables describing next year s annual return on
Weyerhauser, Xerox, Yahoo! and Zymogenetics stock. The table below gives discrete
probability distributions for these random variables based on the state of the economy:
State of Economy W p(w) X p(x) Y p(y) Z p(z)
Depression -0.3 0.05 -0.5 0.05 -0.5 0.15 -0.8 0.05
Recession 0.0 0.2 -0.2 0.1 -0.2 0.5 0.0 0.2
Normal 0.1 0.5 0 0.2 0 0.2 0.1 0.5
Mild Boom 0.2 0.2 0.2 0.5 0.2 0.1 0.2 0.2
Major Boom 0.5 0.05 0.5 0.15 0.5 0.05 1 0.05
• Plot the distributions for each random variable (make a bar chart). Comment
on any differences or similarities between the distributions.
• For each random variable, compute the expected value, variance, standard de-
viation, skewness, kurtosis and brie! y comment.
Suppose X is a normally distributed random variable with mean 10 and variance
24.
• Find Pr(X > 14)
• Find Pr(8 < X < 20)
• Find the probability that X takes a value that is at least 6 away from its mean.
• Suppose y is a constant de&ned such that Pr(X > y) = 0.10. What is the value
of y?
• Determine the 1%, 5%, 10%, 25% and 50% quantiles of the distribution of X.
Let X denote the monthly return on Microsoft stock and let Y denote the monthly
return on Starbucks stock. Suppose X˜N (0.05, (0.10)2 ) and Y ˜N(0.025, (0.05)2 ).
• Plot the normal curves for X and Y
• Comment on the risk-return trade-offs for the two stocks
Let R denote the monthly return on Microsoft stock and let W0 denote ini-
tial wealth to be invested in Microsoft stock over the next month. Assume that
R˜N (0.07, (0.12)2 ) and that W0 = $25, 000.
• What is the distribution of end of month wealth W1 = W0 (1 + R)?
• What is the probability that end of month wealth is less than $20,000?
• What is the Value-at-Risk (VaR) on the investment in Microsoft stock over the
next month with 5% probability?
31
References
[1] Amemiya, T. (1994). Introduction to Statistics and Econometrics. Harvard Uni-
versity Press, Cambridge, MA.
[3] Hoel, P.G., Port, S.C. and Stone, C.J. (1971). Introduction to Probability Theory.
Houghton Mifflin, Boston, MA.
32
Introduction to Financial Econometrics
Chapter 3 The Constant Expected Return Model
Eric Zivot
Department of Economics
University of Washington
January 6, 2000
This version: January 23, 2001
Assumption 1 states that in every time period asset returns are normally dis-
tributed and that the mean and the variance of each asset return is constant over
time. In particular, we have for each asset i
The second assumption states that the contemporaneous covariances between assets
are constant over time. Given assumption 1, assumption 2 implies that the contem-
poraneous correlations between assets are constant over time as well. That is, for all
1
assets
corr(Rit , Rjt ) = ρij for all values of t.
The third assumption stipulates that all of the asset returns are uncorrelated over
time1 . In particular, for a given asset i the returns on the asset are serially uncorre-
lated which implies that
Additionally, the returns on all possible pairs of assets i and j are serially uncorrelated
which implies that
Assumptions 1-3 indicate that all asset returns at a given point in time are jointly
(multivariate) normally distributed and that this joint distribution stays constant
over time. Clearly these are very strong assumptions. However, they allow us to de-
velopment a straightforward probabilistic model for asset returns as well as statistical
tools for estimating the parameters of the model and testing hypotheses about the
parameter values and assumptions.
where µi is a constant and we assume that εit is independent of εjs for all time periods
t 6= s. The notation εit ∼ i.i.d. N (0, σ 2i ) stipulates that the random variable εit is
serially independent and identically distributed as a normal random variable with
mean zero and variance σ 2i . In particular, note that, E[εit ] = 0, var(εit ) = σ 2i and
cov(εit , εjs ) = 0 for i 6= j and t 6= s.
Using the basic properties of expectation, variance and covariance discussed in
chapter 2, we can derive the following properties of returns. For expected returns we
have
2
since µi is constant and E[εit ] = 0. Regarding the variance of returns, we have
which uses the fact that the variance of a constant (µi ) is zero. For covariances of
returns, we have
and
cov(Rit , Rjs ) = cov(µi + εit , µj + εjs ) = cov(εit , εjs ) = 0, t 6= s,
which use the fact that adding constants to two random variables does not affect
the covariance between them. Given that covariances and variances of returns are
constant over time gives the result that correlations between returns over time are
also constant:
cov(Rit , Rjt ) σ ij
corr(Rit , Rjt ) = q = = ρij ,
var(Rit )var(Rjt ) σiσj
cov(Rit , Rjs ) 0
corr(Rit , Rjs ) = q = = 0, i 6= j, t 6= s.
var(Rit )var(Rjs ) σiσj
Finally, since the random variable εit is independent and identically distributed (i.i.d.)
normal the asset return Rit will also be i.i.d. normal:
Hence, the CER model (1) for Rit is equivalent to the model implied by assumptions
1-3.
εit = Rit − µi
= Rit − E[Rit ]
so that εit is de&ned to be the deviation of the random return from its expected value.
If the news is good, then the realized value of εit is positive and the observed return is
3
above its expected value µi . If the news is bad, then εjt is negative and the observed
return is less than expected. The assumption that E[εit ] = 0 means that news, on
average, is neutral; neither good nor bad. The assumption that var(εit ) = σ 2i can be
interpreted as saying that volatility of news arrival is constant over time. The random
news variable affecting asset i, eit , is allowed to be contemporaneously correlated with
the random news variable affecting asset j, εjt , to capture the idea that news about
one asset may spill over and affect another asset. For example, let asset i be Microsoft
and asset j be Apple Computer. Then one interpretation of news in this context is
general news about the computer industry and technology. Good news should lead
to positive values of εit and εjt . Hence these variables will be positively correlated.
The CER model with continuously compounded returns has the following nice
property with respect to the interpretation of εit as news. Consider the default case
where Rit is interpreted as the continuously compounded monthly return. Since mul-
tiperiod continuously compounded returns are additive we can interpret, for example,
Rit as the sum of 30 daily continuously compounded returns2 :
29
X
Rit = Rdit−k
k=0
where Ritd denotes the continuously compounded daily return on asset i. If we assume
that daily returns are described by the CER model then
where
µi = 30 · µdi ,
29
X
εit = εdit−k .
k=0
2
For simplicity of exposition, we will ignore the fact that some assets do not trade over the
weekend.
4
Hence, the monthly expected return, µi , is simply 30 times the daily expected re-
turn. The interpretation of εit in the CER model when returns are continuously
compounded is the accumulation of news between months t − 1 and t. Notice that
à 29
!
X
var(Rit ) = var (µdi + εdit−k )
k=0
29
X
= var(εdit−k )
k=0
X29 ³ ´2
= σ di
k=0
³ ´2
= 30 · σ di
and
à 29 29
!
X X
cov(Rit , Rjt ) = cov εdit−k , εdjt−k
k=0 k=0
29
X
= cov(εdit−k , εdjt−k )
k=0
X29
= σ dij
k=0
= 30 · σ dij ,
so that the monthly variance, σ 2i , is equal to 30 times the daily variance and the
monthly covariance, σ ij , is equal to 30 times the daily covariance.
1.4 The CER Model of Asset Returns and the Random Walk
Model of Asset Prices
The CER model of asset returns (1) gives rise to the so-called random walk (RW)
model of the logarithm of asset prices. To see this, recall that the continuously
compounded return, Rit , is de&ned from asset prices via
à !
Pit
ln = Rit .
Pit−1
Since the log of the ratio of prices is equal to the difference in the logs of prices we
may rewrite the above as
ln(Pit ) − ln(Pit−1 ) = Rit .
Letting pit = ln(Pit ) and using the representation of Rit in the CER model (1), we
may further rewrite the above as
pit − pit−1 = µi + εit . (3)
5
The representation in (3) is know as the RW model for the log of asset prices.
In the RW model, µi represents the expected change in the log of asset prices
(continuously compounded return) between months t − 1 and t and εit represents the
unexpected change in prices. That is,
E[pit − pit−1 ] = E[Rit ] = µi ,
εit = pit − pit−1 − E[pit − pit−1 ].
Further, in the RW model, the unexpected changes in asset prices, εit , are uncorrelated
over time (cov(εit , εis ) = 0 for t 6= s) so that future changes in asset prices cannot be
predicted from past changes in asset prices3 .
The RW model gives the following interpretation for the evolution of asset prices.
Let pi0 denote the initial log price of asset i. The RW model says that the price at
time t = 1 is
pi1 = pi0 + µi + εi1
where εi1 is the value of random news that arrives between times 0 and 1. Notice that
at time t = 0 the expected price at time t = 1 is
E[pi1 ] = pi0 + µi + E[εi1 ] = pi0 + µi
which is the initial price plus the expected return between time 0 and 1. Similarly,
the price at time t = 2 is
pi2 = pi1 + µi + εi2
= pi0 + µi + µi + εi1 + εi2
2
X
= pi0 + 2 · µi + εit
t=1
which is equal to the initial price, pi0 , plus the two period expected return, 2 · µi , plus
P
the accumulated random news over the two periods, 2t=1 εit . By recursive substitu-
tion, the price at time t = T is
T
X
piT = pi0 + T · µi + εit .
t=1
6
Simulated Random Walk
12
E[p(t)]
p(t) - E[p(t)]
p(t)
10
p(t)
8
6
E[p(t)]
pt
p(t) - E[p(t)]
101
1
13
17
21
25
29
33
37
41
45
49
53
57
61
65
69
73
77
81
85
89
93
97
-2
time, t
The term random walk was originally used to describe the unpredictable move-
ments of a drunken sailor staggering down the street. The sailor starts at an initial
position, p0 , outside the bar. The sailor generally moves in the direction described
by µ but randomly deviates from this direction after each step t by an amount equal
P
to εt . After T steps the sailor ends up at position pT = p0 + µ · T + Tt=1 εt .
7
simulated data points must be determined. Here, N = 100. Hence, the model to be
simulated is
Rt = 0.05 + εt , t = 1, . . . , 100
εt ~iid N (0, (0.10)2 )
The key to simulating data from the above model is to simulate N = 100 observations
of the random news variable εt ~iid N(0, (0.10)2 ). Computer algorithms exist which
can easily create such observations. Let {ε1 , . . . , ε100 } denote the 100 simulated values
of εt . The histogram of these values are given in &gure xxx below
16.00%
14.00%
12.00%
10.00%
Frequency
8.00%
6.00%
4.00%
2.00%
0.00%
-0.241
-0.208
-0.175
-0.142
-0.109
-0.076
-0.043
-0.010
0.023
0.056
0.089
0.122
0.155
0.188
0.221
e(t)
1 100 P
The sample averageqof the simulated errors is 100 t=1 εt = −0.004 and the sample
1 P 100 2
standard deviation is 99 t=1 (εt − (−0.004)) = 0.109. These values are very close
to the population values E[εt ] = 0 and SD(εt ) = 0.10, respectively.
Once the simulated values of εt have been created, the simulated values of Rt are
constructed as Rt = 0.05 + εt , t = 1, . . . , 100. A time plot of the simulated values of
Rt is given in &gure xxx below
8
Monte Carlo Simulation of CER Model
R(t) = 0.05 + e(t), e(t) ~ iid N(0, (0.10)^2)
0.400
0.300
0.200
0.100
Return
0.000
-0.100
-0.200
-0.300
100
1
4
7
10
13
16
19
22
25
28
31
34
37
40
43
46
49
52
55
58
61
64
67
70
73
76
79
82
85
88
91
94
97
time
The simulated return data ! uctuates randomly about the expected return value
E[Rt ] = µ = 0.05. The typical size of the ! uctuation is approximately equal to
SE(εt ) = 0.10. Notice that the simulated return data looks remarkably like the
actual return data of Microsoft.
Monte Carlo simulation of a model can be used as a &rst pass reality check of the
model. If simulated data from the model does not look like the data that the model is
supposed to describe then serious doubt is cast on the model. However, if simulated
data looks reasonably close to the data that the model is suppose to describe then
con&dence is instilled on the model.
The CER model of asset returns gives us a rigorous way of interpreting the time
series behavior of asset returns. At the beginning of every month t, Rit is a random
9
variable representing the return to be realized at the end of the month. The CER
model states that Rit ∼ i.i.d. N (µi , σ 2i ). Our best guess for the return at the end of the
month qis E[Rit ] = µi , our measure of uncertainty about our best guess is captured by
σ i = var(Rit ) and our measure of the direction of linear association between Rit and
Rjt is σ ij = cov(Rit , Rjt ). The CER model assumes that the economic environment
is constant over time so that the normal distribution characterizing monthly returns
is the same every month.
Our life would be very easy if we knew the exact values of µi , σ 2i and σ ij , the
parameters of the CER model. In actuality, however, we do not know these values
with certainty. A key task in &nancial econometrics is estimating the values of µi , σ 2i
and σ ij from a history of observed data.
Suppose we observe monthly returns on N different assets over the horizon t =
1, . . . , T. Let ri1 , . . . , riT denote the observed history of T monthly returns on asset
i for i = 1, . . . , N. It is assumed that the observed returns are realizations of the
random variables Ri1 , . . . , RiT , where Rit is described by the CER model (1). We
call Ri1 , . . . , RiT a random sample from the CER model (1) and we call ri1 , . . . , riT
the realized values from the random sample. In this case, we can use the observed
returns to estimate the unknown parameters of the CER model
10
Introduction to Financial Econometrics
Chapter 4 Introduction to Portfolio Theory
Eric Zivot
Department of Economics
University of Washington
January 26, 2000
This version: February 20, 2001
µA = E[RA ], σ 2A = V ar(RA ),
µB = E[RB ], σ 2B = V ar(RB ),
σ AB = Cov(RA , RB ).
We assume that these values are taken as given. We might wonder where such values
come from. One possibility is that they are estimated from historical return data for
the two stocks. Another possibility is that they are subjective guesses.
The expected returns, µA and µB , are our best guesses for the monthly returns on
each of the stocks. However, since the investments are random we must recognize that
the realized returns may be different from our expectations. The variances, σ 2A and
σ 2B , provide measures of the uncertainty associated with these monthly returns. We
can also think of the variances as measuring the risk associated with the investments.
Assets that have returns with high variability (or volatility) are often thought to
be risky and assets with low return volatility are often thought to be safe. The
covariance σ AB gives us information about the direction of any linear dependence
between returns. If σ AB > 0 then the returns on assets A and B tend to move in the
1
same direction; if σ AB < 0 the returns tend to move in opposite directions; if σ AB = 0
then the returns tend to move independently. The strength of the dependence between
the returns is measured by the correlation coefficient ρAB = σσAAB σB
. If ρAB is close to
one in absolute value then returns mimic each other extremely closely whereas if ρAB
is close to zero then the returns may show very little relationship.
The portfolio problem is set-up as follows. We have a given amount of wealth and
it is assumed that we will exhaust all of our wealth between investments in the two
stocks. The investor s problem is to decide how much wealth to put in asset A and
how much to put in asset B. Let xA denote the share of wealth invested in stock A
and xB denote the share of wealth invested in stock B. Since all wealth is put into
the two investments it follows that xA + xB = 1. (Aside: What does it mean for xA
or xB to be negative numbers?) The investor must choose the values of xA and xB .
Our investment in the two stocks forms a portfolio and the shares xA and xB are
referred to as portfolio shares or weights. The return on the portfolio over the next
month is a random variable and is given by
Rp = xA RA + xB RB , (1)
which is just a simple linear combination or weighted average of the random return
variables RA and RB . Since RA and RB are assumed to be normally distributed, Rp
is also normally distributed.
µp = E[Rp ] = xA µA + xB µB (2)
σ 2p = var(Rp ) = x2A σ 2A + x2B σ 2B + 2xA xB σ AB (3)
These results are so important to portfolio theory that it is worthwhile to go
through the derivations. For the &rst result (2), we have
E[Rp ] = E[xA RA + xB RB ] = xA E[RA ] + xB E[RB ] = xA µA + xB µB
by the linearity of the expectation operator. For the second result (3), we have
var(Rp ) = var(xA RA + xB RB ) = E[(xA RA + xB RB ) − E[xA RA + xB RB ])2 ]
= E[(xA (RA − µA ) + xB (RB − µB ))2 ]
= E[x2A (RA − µA )2 + x2B (RB − µB )2 + 2xA xB (RA − µA )(RB − µB )]
= x2A E[(RA − µA )2 ] + x2B E[(RB − µB )2 ] + 2xA xB E[(RA − µA )(RB − µB )],
2
and the result follows by the de&nitions of var(RA ), var(RB ) and cov(RA , RB )..
Notice that the variance of the portfolio is a weighted average of the variances
of the individual assets plus two times the product of the portfolio weights times
the covariance between the assets. If the portfolio weights are both positive then a
positive covariance will tend to increase the portfolio variance, because both returns
tend to move in the same direction, and a negative covariance will tend to reduce the
portfolio variance. Thus &nding negatively correlated returns can be very bene&cial
when forming portfolios. What is surprising is that a positive covariance can also be
bene&cial to diversi&cation.
• Returns are jointly normally distributed. This implies that means, variances
and covariances of returns completely characterize the joint distribution of re-
turns.
• Investors only care about portfolio expected return and portfolio variance. In-
vestors like portfolios with high expected return but dislike portfolios with high
return variance.
Given the above assumptions we set out to characterize the set of portfolios that
have the highest expected return for a given level of risk as measured by portfolio
variance. These portfolios are called efficient portfolios and are the portfolios that
investors are most interested in holding.
For illustrative purposes we will show calculations using the data in the table
below.
Table 1: Example Data
µA µB σ 2A
σ 2B σA σB σ AB ρAB
0.175 0.055 0.067 0.013 0.258 0.115 -0.004875 -0.164
The collection of all feasible portfolios (the investment possibilities set) in the
case of two assets is simply all possible portfolios that can be formed by varying
the portfolio weights xA and xB such that the weights sum to one (xA + xB = 1).
We summarize the expected return-risk (mean-variance) properties of the feasible
portfolios in a plot with portfolio expected return, µp , on the vertical axis and portfolio
standard-deviation, σ p , on the horizontal axis. The portfolio standard deviation is
used instead of variance because standard deviation is measured in the same units as
the expected value (recall, variance is the average squared deviation from the mean).
3
Portfolio Frontier with 2 Risky Assets
0.250
0.200
0.100
0.050
0.000
0.000 0.100 0.200 0.300 0.400
Portfolio std. deviation
Figure 1
The investment possibilities set or portfolio frontier for the data in Table 1 is
illustrated in Figure 1. Here the portfolio weight on asset A, xA , is varied from
-0.4 to 1.4 in increments of 0.1 and, since xB = 1 − xA , the weight on asset is
then varies from 1.4 to -0.4. This gives us 18 portfolios with weights (xA , xB ) =
(−0.4, 1.4), (−0.3, 1.3), ..., (1.3, −0.3), (1.4, −0.4). For
q each of these portfolios we use
the formulas (2) and (3) to compute µp and σ p = σ 2p . We then plot these values1 .
Notice that the plot in (µp , σ p ) space looks like a parabola turned on its side (in
fact it is one side of a hyperbola). Since investors desire portfolios with the highest
expected return for a given level of risk, combinations that are in the upper left corner
are the best portfolios and those in the lower right corner are the worst. Notice that
the portfolio at the bottom of the parabola has the property that it has the smallest
variance among all feasible portfolios. Accordingly, this portfolio is called the global
minimum variance portfolio.
It is a simple exercise in calculus to &nd the global minimum variance portfolio.
We solve the constrained optimization problem
4
Substituting xB = 1 − xA into the formula for σ 2p reduces the problem to
The &rst order conditions for a minimum, via the chain rule, are
dσ 2p
0= = 2xmin 2 min 2 min
A σ A − 2(1 − xA )σ B + 2σ AB (1 − 2xA )
dxA
and straightforward calculations yield
σ 2B − σ AB
xmin
A = , xmin = 1 − xmin
A . (4)
σ 2A + σ 2B − 2σ AB B
For our example, using the data in table 1, we get xmin min
A = 0.2 and xB = 0.8.
Efficient portfolios are those with the highest expected return for a given level
of risk. Inefficient portfolios are then portfolios such that there is another feasible
portfolio that has the same risk (σ p ) but a higher expected return (µp ). From the
plot it is clear that the inefficient portfolios are the feasible portfolios that lie below
the global minimum variance portfolio and the efficient portfolios are those that lie
above the global minimum variance portfolio.
The shape of the investment possibilities set is very sensitive to the correlation
between assets A and B. If ρAB is close to 1 then the investment set approaches a
straight line connecting the portfolio with all wealth invested in asset B, (xA , xB ) =
(0, 1), to the portfolio with all wealth invested in asset A, (xA , xB ) = (1, 0). This
case is illustrated in Figure 2. As ρAB approaches zero the set starts to bow toward
the µp axis and the power of diversi&cation starts to kick in. If ρAB = −1 then
the set actually touches the µp axis. What this means is that if assets A and B
are perfectly negatively correlated then there exists a portfolio of A and B that has
positive expected return and zero variance! To &nd the portfolio with σ 2p = 0 when
ρAB = −1 we use (4) and the fact that σ AB = ρAB σ A σ B to give
σB
xmin
A = , xmin = 1 − xA
σA + σB B
The case with ρAB = −1 is also illustrated in Figure 2.
5
Portfolio Frontier with 2 Risky Assets
0.250
0.200
0.100
0.050
0.000
0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450
P ortfolio std. de via tion
correlation=1 correlation=-1
Figure 2
Given the efficient set of portfolios, which portfolio will an investor choose? Of
the efficient portfolios, investors will choose the one that accords with their risk
preferences. Very risk averse investors will choose a portfolio very close to the global
minimum variance portfolio and very risk tolerant investors will choose portfolios
with large amounts of asset A which may involve short-selling asset B.
1.3.1 Efficient portfolios with one risky asset and one risk free asset
Continuing with our example, consider an investment in asset B and the risk free
asset (henceforth referred to as a T-bill) and suppose that rf = 0.03. Since the risk
free rate is &xed over the investment horizon it has some special properties, namely
µf = E[rf ] = rf
6
var(rf ) = 0
cov(RB , rf ) = 0
Let xB denote the share of wealth in asset B and xf = 1 − xB denote the share of
wealth in T-bills. The portfolio expected return is
Rp = xB RB + (1 − xB )rf
= xB (RB − rf ) + rf
The quantity RB − rf is called the excess return (over the return on T-bills) on asset
B. The portfolio expected return is then
µp = xB (µB − rf ) + rt
where the quantity (µB − rf ) is called the expected excess return or risk premium
on asset B. We may express the risk premium on the portfolio in terms of the risk
premium on asset B:
µp − rf = xB (µB − rf )
The more we invest in asset B the higher the risk premium on the portfolio.
The portfolio variance only depends on the variability of asset B and is given by
σ 2p = x2B σ 2B .
7
Portfolio Frontier with 1 Risky Asset and T-Bill
0.200
0.180
Figure 3
Notice that expected return-risk trade off of these portfolios is linear. Also, notice
that the portfolios which are combinations of asset A and T-bills have expected
returns uniformly higher than the portfolios consisting of asset B and T-bills. This
occurs because the Sharpe s slope for asset A is higher than the slope for asset B:
µA − rf 0.175 − 0.03 µ − rf 0.055 − 0.03
= = 0.562, B = = 0.217.
σA 0.258 σB 0.115
Hence, portfolios of asset A and T-bills are efficient relative to portfolios of asset B
and T-bills.
1.3.2 Efficient portfolios with two risky assets and a risk-free asset
Now we expand on the previous results by allowing our investor to form portfolios of
assets A, B and T-bills. The efficient set in this case will still be a straight line in
(µp , σ p )− space with intercept rf . The slope of the efficient set, the maximum Sharpe
ratio, is such that it is tangent to the efficient set constructed just using the two risky
assets A and B. Figure 5 illustrates why this is so.
8
Portfolio Frontier with 2 Risky Assets and T-Bills
0.350
Figure 4
µ −r
If we invest in only in asset B and T-bills then the Sharpe ratio is BσB f = 0.217
and the CAL intersects the parabola at point B. This is clearly not the efficient set
of portfolios. For example, we could do uniformly better if we instead invest only
µ −r
in asset A and T-bills. This gives us a Sharpe ratio of AσA f = 0.562 and the new
CAL intersects the parabola at point A. However, we could do better still if we invest
in T-bills and some combination of assets A and B. Geometrically, it is easy to see
that the best we can do is obtained for the combination of assets A and B such that
the CAL is just tangent to the parabola. This point is marked T on the graph and
represents the tangency portfolio of assets A and B.
We can determine the proportions of each asset in the tangency portfolio by &nding
the values of xA and xB that maximize the Sharpe ratio of a portfolio that is on the
envelope of the parabola. Formally, we solve
µp − rf
max s.t.
A Bx ,x σp
µp = xA µA + xB µB
σ 2p = x2A σ 2A + x2B σ 2B + 2xA xB σ AB
1 = xA + xB
9
This is a straightforward, albeit very tedious, calculus problem and the solution can
be shown to be
T (µA − rf )σ 2B − (µB − rf )σ AB
xA = 2 2
, xTB = 1 − xTA .
(µA − rf )σ B + (µB − rf )σ A − (µA − rf + µB − rf )σ AB
For the example data using rf = 0.03, we get xTA = 0.542 and xTB = 0.458. The
expected return on the tangency portfolio is
µT = xTA µA + xTB µB
= (0.542)(0.175) + (0.458)(0.055) = 0.110,
the variance of the tangency portfolio is
³ ´2 ³ ´2
σ 2T = xTA σ 2A + xTB σ 2B + 2xTA xTB σ AB
= (0.542)2 (0.067) + (0.458)2 (0.013) + 2(0.542)(0.458) = 0.015,
and the standard deviation of the tangency portfolio is
q √
σ T = σ 2T = 0.015 = 0.124.
The efficient portfolios now are combinations of the tangency portfolio and the
T-bill. This important result is known as the mutual fund separation theorem. The
tangency portfolio can be considered as a mutual fund of the two risky assets, where
the shares of the two assets in the mutual fund are determined by the tangency
portfolio weights, and the T-bill can be considered as a mutual fund of risk free
assets. The expected return-risk trade-off of these portfolios is given by the line
connecting the risk-free rate to the tangency point on the efficient frontier of risky
asset only portfolios. Which combination of the tangency portfolio and the T-bill
an investor will choose depends on the investor s risk preferences. If the investor is
very risk averse, then she will choose a combination with very little weight in the
tangency portfolio and a lot of weight in the T-bill. This will produce a portfolio
with an expected return close to the risk free rate and a variance that is close to zero.
For example, a highly risk averse investor may choose to put 10% of her wealth in
the tangency portfolio and 90% in the T-bill. Then she will hold (10%) × (54.2%) =
5.42% of her wealth in asset A, (10%) × (45.8%) = 4.58% of her wealth in asset B
and 90% of her wealth in the T-bill. The expected return on this portfolio is
µp = rf + 0.10(µT − rf )
= 0.03 + 0.10(0.110 − 0.03)
= 0.038.
and the standard deviation is
σ p = 0.10σ T
= 0.10(0.124)
= 0.012.
10
A very risk tolerant investor may actually borrow at the risk free rate and use these
funds to leverage her investment in the tangency portfolio. For example, suppose the
risk tolerant investor borrows 10% of her wealth at the risk free rate and uses the
proceed to purchase 110% of her wealth in the tangency portfolio. Then she would
hold (110%)×(54.2%) = 59.62% of her wealth in asset A, (110%)×(45.8%) = 50.38%
in asset B and she would owe 10% of her wealth to her lender. The expected return
and standard deviation on this portfolio is
11
Efficient Portfolios
0.250
Efficient portfolios of T-
bills and assets A and B
0.200
Asset A
0.150
Tangency
Portfolio ER
Portfolio
0.055 Asset B
0.050
Combinations of tangency
rf portfolio and T-bills that has
same ER as asset B
Figure 5
To illustrate, consider &gure 5 which shows the portfolio frontier for two risky
assets and the efficient frontier for two risky assets plus a risk-free asset. Suppose
an investor initially holds all of his wealth in asset A. The expected return on this
portfolio is µB = 0.055 and the standard deviation (risk) is σ B = 0.115. An efficient
portfolio (combinations of the tangency portfolio and T-bills) that has the same
standard deviation (risk) as asset B is given by the portfolio on the efficient frontier
that is directly above σ B = 0.115. To &nd the shares in the tangency portfolio and
T-bills in this portfolio recall from (xx) that the standard deviation of a portfolio with
xT invested in the tangency portfolio and 1 − xT invested in T-bills is σ p = xT σ T .
Since we want to &nd the efficient portfolio with σ p = σ B = 0.115, we solve
σB 0.115
xT = = = 0.917, xf = 1 − xT = 0.083.
σT 0.124
That is, if we invest 91.7% of our wealth in the tangency portfolio and 8.3% in T-bills
we will have a portfolio with the same standard deviation as asset B. Since this is an
efficient portfolio, the expected return should be higher than the expected return on
12
asset B. Indeed it is since
µp = rf + xT (µT − rf )
= 0.03 + 0.917(0.110 − 0.03)
= 0.103
Notice that by diversifying our holding into assets A, B and T-bills we can obtain a
portfolio with the same risk as asset B but with almost twice the expected return!
Next, consider &nding an efficient portfolio that has the same expected return
as asset B. Visually, this involves &nding the combination of the tangency portfo-
lio and T-bills that corresponds with the intersection of a horizontal line with in-
tercept µB = 0.055 and the line representing efficient combinations of T-bills and
the tangency portfolio. To &nd the shares in the tangency portfolio and T-bills in
this portfolio recall from (xx) that the expected return of a portfolio with xT in-
vested in the tangency portfolio and 1 − xT invested in T-bills has expected return
equal to µp = rf + xT (µT − rf ). Since we want to &nd the efficient portfolio with
µp = µB = 0.055 we use the relation
µp − rf = xT (µT − rF )
That is, if we invest 31.3% of wealth in the tangency portfolio and 68.7% of our
wealth in T-bills we have a portfolio with the same expected return as asset B. Since
this is an efficient portfolio, the standard deviation (risk) of this portfolio should be
lower than the standard deviation on asset B. Indeed it is since
σ p = xT σ T
= 0.313(0.124)
= 0.039.
Notice how large the risk reduction is by forming an efficient portfolio. The standard
deviation on the efficient portfolio is almost three times smaller than the standard
deviation of asset B!
The above example illustrates two ways to interpret the bene&ts from forming
efficient portfolios. Starting from some benchmark portfolio, we can &x standard de-
viation (risk) at the value for the benchmark and then determine the gain in expected
return from forming a diversi&ed portfolio2 . The gain in expected return has concrete
2
The gain in expected return by investing in an efficient portfolio abstracts from the costs asso-
ciated with selling the benchmark portfolio and buying the efficient portfolio.
13
meaning. Alternatively, we can &x expected return at the value for the benchmark
and then determine the reduction in standard deviation (risk) from forming a diver-
si&ed portfolio. The meaning to an investor of the reduction in standard deviation
is not as clear as the meaning to an investor of the increase in expected return. It
would be helpful if the risk reduction bene&t can be translated into a number that is
more interpretable than the standard deviation. The concept of Value-at-Risk (VaR)
provides such a translation.
Recall, the VaR of an investment is the expected loss in investment value over a
given horizon with a stated probability. For example, consider an investor who invests
W0 = $100, 000 in asset B over the next year. Assume that RB represents the annual
(continuously compounded) return on asset B and that RB ~N(0.055, (0.114)2 ). The
5% annual VaR of this investment is the loss that would occur if return on asset B is
equal to the 5% left tail quantile of the normal distribution of RB . The 5% quantile,
q0.05 is determined by solving
Using the inverse cdf for a normal random variable with mean 0.055 and standard
deviation 0.114 it can be shown that q0.05 = −0.133.That is, with 5% probability the
return on asset B will be −13.3% or less. If RB = −0.133 then the loss in portfolio
value3 , which is the 5% VaR, is
loss in portfolio value = V aR = |W0 · (eq0.05 − 1)| = |$100, 000(e−0.133 − 1)| = $12, 413.
To reiterate, if the investor hold $100,000 in asset B over the next year then the 5%
VaR on the portfolio is $12, 413. This is the loss that would occur with 5% probability.
Now suppose the investor chooses to hold an efficient portfolio with the same
expected return as asset B. This portfolio consists of 31.3% in the tangency portfolio
and 68.7% in T-bills and has a standard deviation equal to 0.039. Let Rp denote the
annual return on this portfolio and assume that Rp ~N (0.055, 0.039). Using the inverse
cdf for this normal distribution, the 5% quantile can be shown to be q0.05 = −0.009.
That is, with 5% probability the return on the efficient portfolio will be −0.9% or
less. This is considerably smaller than the 5% quantile of the distribution of asset B.
If Rp = −0.009 the loss in portfolio value (5% VaR) is
loss in portfolio value = V aR = |W0 · (eq0.05 − 1)| = |$100, 000(e−0.009 − 1)| = $892.
Notice that the 5% VaR for the efficient portfolio is almost &fteen times smaller than
the 5% VaR of the investment in asset B. Since VaR translates risk into a dollar &gure
it is more interpretable than standard deviation.
3
To compute the VaR we need to convert the continuous compounded return (quantile) to a
simple return (quantile). Recall, if Rct is a continuously compounded return and Rt is a somple
c
return then Rct = ln(1 + Rt ) and Rt = eRt − 1.
14
3 Further Reading
The classic text on portfolio optimization is Markowitz (1954). Good intermediate
level treatments are given in Benninga (2000), Bodie, Kane and Marcus (1999) and
Elton and Gruber (1995). An interesting recent treatment with an emphasis on
statistical properties is Michaud (1998). Many practical results can be found in the
Financial Analysts Journal and the Journal of Portfolio Management. An excellent
overview of value at risk is given in Jorian (1997).
y = f (x) = x2
which is illustrated in Figure xxx. Clearly the minimum of this function occurs at
the point x = 0. Using calculus, we &nd the minimum by solving
min
x
y = x2 .
d d 2
0= f (x) = x = 2x
dx dx
and solving for x gives x = 0. The second order condition for a minimum is
d2
0< f (x)
dx
and this condition is clearly satis&ed for f (x) = x2 .
Next, consider the function of two variables
y = f(x, z) = x2 + z 2 (6)
15
y = x^2 + z^2
y 4
1.75
1 1
0.25
0 z
-0.5
-2
-1.75
-1.5
-1.25
-1.25
-1
-0.75
-0.5
-0.25
0
0.25
0.5
-2
0.75
1
1.25
1.5
1.75
2
x
Figure 6
This function looks like a salad bowl whose bottom is at x = 0 and z = 0. To &nd
the minimum of (6), we solve
min y = x2 + z 2
x,z
and the &rst order necessary conditions are
∂y
0= = 2x
∂x
and
∂y
0= = 2z.
∂z
Solving these two equations gives x = 0 and z = 0.
Now suppose we want to minimize (6) subject to the linear constraint
x + z = 1. (7)
The minimization problem is now a constrained minimization
min
x,z
y = x2 + z 2 subject to (s.t.)
x+z = 1
16
and is illustrated in Figure xxx. Given the constraint x + z = 1, the function (6) is
no longer minimized at the point (x, z) = (0, 0) because this point does not satisfy
x + z = 1. The One simple way to solve this problem is to substitute the restriction
(7) into the function (6) and reduce the problem to a minimization over one variable.
To illustrate, use the restriction (7) to solve for z as
z = 1 − x. (8)
The function (9) satis&es the restriction (7) by construction. The constrained mini-
mization problem now becomes
min y = x2 + (1 − x)2 .
x
18
5 Problems
Exercise 1 Consider the problem of investing in two risky assets A and B and a
risk-free asset (T-bill). The optimization problem to &nd the tangency portfolio may
be reduced to
xA (µA − rf ) + (1 − xA )(µB − rf )
max 1/2
xA
(x2A σ 2A + (1 − xA )2 σ 2B + 2xA (1 − xA )σ AB )
References
[1] Benninga, S. (2000), Financial Modeling, Second Edition. Cambridge, MA: MIT
Press.
[3] Elton, E. and G. Gruber (1995). Modern Portfolio Theory and Investment Anal-
ysis, Fifth Edition. New York: Wiley.
19
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H
Introduction to Financial Econometrics
Chapter 6 The Single Index Model and Bivariate
Regression
Eric Zivot University of Washington
Department of Economics
March 1, 2001
1
market risk, cannot be eliminated in a well diversi&ed portfolio. The random error
term εit has a similar interpretation as the error term in the CER model. In the single
index model, εit represents random news that arrives between time t − 1 and t that
captures micro or &rm-speci&c risk factors that affect an individual asset s return
that are not related to macro events. For example, εit may capture the news effects
of new product discoveries or the death of a CEO. This type of risk is often called
&rm speci&c risk, idiosyncratic risk, residual risk or non-market risk. This type of
risk can be eliminated in a well diversi&ed portfolio.
The single index model can be expanded to capture multiple factors. The single
index model then takes the form a k−variable linear regression model
where Fjt denotes the j th systematic factorm, β i,j denotes asset i0 s loading on the j th
factor and εit denotes the random component independent of all of the systematic
factors. The single index model results when F1t = RMt and β i,2 = · · · = β i,k = 0. In
the literature on multiple factor models the factors are usually variables that capture
speci&c characteristics of the economy that are thought to affect returns - e.g. the
market index, GDP growth, unexpected in! ation etc., and &rm speci&c or industry
speci&c characteristics - &rm size, liquidity, industry concentration etc. Multiple
factor models will be discussed in chapter xxx.
The single index model is heavily used in empirical &nance. It is used to estimate
expected returns, variances and covariances that are needed to implement portfolio
theory. It is used as a model to explain the normal or usual rate of return on an
asset for use in so-called event studies2 . Finally, the single index model is often used
the evaluate the performance of mutual fund and pension fund managers.
2
5. cov(εit , εjs ) = 0 for all t, s and i 6= j
6. εit is normally distributed
The normality assumption is justi&ed on the observation that returns are fairly
well characterized by the normal distribution. The error term having mean zero
implies that &rm speci&c news is, on average, neutral and the constant variance
assumptions implies that the magnitude of typical news events is constant over time.
Assumption 4 states that &rm speci&c news is independent (since the random variables
are normally distributed) of macro news and assumption 5 states that news affecting
asset i in time t is independent of news affecting asset j in time s.
That εit is unrelated to RMs and εjs implies that any correlation between asset i
and asset j is solely due to their common exposure to RMt throught the values of β i
and β j .
The proofs of these results are straightforward and utilize the properties of linear
combinations of random variables. Results 1 and 4 are trivial. For 2, note that
var(Rit ) = var(αi + β i,M RMt + εit )
= β 2i,M var(RMt ) + var(εit ) + 2cov(RMt , εit )
= β 2i,M σ 2M + σ 2ε,i
since, by assumption 4, cov(εit , RMt ) = 0. For 3, by the additivity property of
covariance and assumptions 4 and 5 we have
cov(Rit , Rjt ) = cov(αi + β i,M RMt + εit , αj + β j,M RMt + εjt )
= cov(β i,M RMt + εit , β j,M RMt + εjt )
= cov(β i,M RMt , β j,M RMt ) + cov(β i,M RMt , εjt ) + cov(εit , β j,M RMt ) + cov(εit , εjt )
= β i,M β j,M cov(RMt , RMt ) = β i,M β j,M σ 2M
3
Last, for 5 note that
Remarks:
• σ ij = 0 if β i = 0 or β j = 0 or both
• σ ij > 0 if β i and β j are of the same sign
• σ ij < 0 if β i and β j are of opposite signs.
4. The expression for the expected return can be used to provide an unconditional
interpretation of αi . Subtracting β i,M µM from both sides of the expression for
µi gives
αi = µi − β i,M µM .
4
1.1.2 Decomposing Total Risk
The independence assumption between RMt and εit allows the unconditional vari-
ability of Rit , var(Rit ) = σ 2i , to be decomposed into the variability due to the market
index, β 2i,M σ 2M , plus the variability of the &rm speci&c component, σ 2ε,i . This decom-
position is often called analysis of variance (ANOVA). Given the ANOVA, it is useful
to de&ne the proportion of the variability of asset i that is due to the market index
and the proportion that is unrelated to the index. To determine these proportions,
divide both sides of σ 2i = β 2i,M σ 2M + σ 2ε,i to give
5
Property 1 states that the expected return on asset i conditional on RMt = rMt
is allowed to vary with the level of the market index. Property 2 says conditional
on the value of the market index, the variance of the return on asset is equal to the
variance of the random news component. Property 3 shows that once movements in
the market are controlled for, assets are uncorrelated.
Σ = σ 2M ββ 0 + ∆
Consider forming a portfolio of these two assets. Let x1 denote the share of wealth
in asset 1, x2 the share of wealth in asset 2 and suppose that x1 + x2 = 1. The return
6
on this portfolio using (2) and (3) is then
Example 2 To be completed
The additivity result of the single index model above holds for portfolios of any
size. To illustrate, suppose the single index model holds for a collection of N assets:
Consider forming a portfolio of these N assets. Let xi denote the share of wealth
P
invested in asset i and assume that Ni=1 = 1. Then the return on the portfolio is
N
X
Rpt = xi (αi + β i,M RMt + εit )
i=1
N
ÃN ! N
X X X
= xi αi + xi β i,M RMt + xi εit
i=1 i=1 i=1
= αp + β p RMt + εpt
PN ³P ´ PN
N
where αp = i=1 xi α i , β p = i=1 xi β i,M and εpt = i=1 xi εit .
7
the portfolio cannot be completely diversi&ed away. The so-called beta of an asset
captures this covariance contribution and so is a measure of the contribution of the
asset to overall portfolio variability.
To illustrate, consider an equally weighted portfolio of 99 stocks and let R99 denote
the return on this portfolio and σ 299 denote the variance. Now consider adding one
stock, say IBM, to the portfolio. Let RIBM and σ 2IBM denote the return and variance
of IBM and let σ 99,IBM = cov(R99 , RIBM ). What is the contribution of IBM to the
risk, as measured by portfolio variance, of the portfolio? Will the addition of IBM
make the portfolio riskier (increase portfolio variance)? Less risky (decrease portfolio
variance)? Or have no effect (not change portfolio variance)? To answer this question,
consider a new equally weighted portfolio of 100 stocks constructed as
R100 = (0.99) · R99 + (0.01) · RIBM .
The variance of this portfolio is
σ 2100 = var(R100 ) = (0.99)2 σ 299 + (0.01)2 σ 2IBM + 2(0.99)(0.01)σ 99,IBM
= (0.98)σ 299 + (0.0001)σ 2IBM + (0.02)σ 99,IBM
≈ (0.98)σ 299 + (0.02)σ 99,IBM .
Now if
• σ 2100 = σ 299 then adding IBM does not change the variability of the portfolio;
• σ 2100 > σ 299 then adding IBM increases the variability of the portfolio;
• σ 2100 < σ 299 then adding IBM decreases the variability of the portfolio.
Consider the &rst case where σ 2100 = σ 299 . This implies (approximately) that
(0.98)σ 299 + (0.02)σ 99,IBM = σ 299
which upon rearranging gives the condition
σ 99,IBM cov(R99 , RIBM )
= =1
σ 299 var(R99 )
De&ning
cov(R99 , RIBM )
β 99,IBM =
var(R99 )
then adding IBM does not change the variability of the portfolio as long as β 99,IBM =
1. Similarly, it is easy to see that σ 2100 > σ 299 implies that β 99,IBM > 1 and σ 2100 < σ 299
implies that β 99,IBM < 1.
In general, let Rp denote the return on a large diversi&ed portfolio and let Ri
denote the return on some asset i. Then
cov(Rp , Ri )
β p,i =
var(Rp )
measures the contribution of asset i to the overall risk of the portfolio.
8
2.1 The single index model and Portfolio Theory
To be completed
rbit = α b
bi + β i,M rMt , t = 1, . . . , T
b
εit = rit − rbit = rit − α b
bi − β i,M rMt , t = 1, . . . , T
Now some lines will &t better for some observations and some lines will &t better for
others. The least squares regression line is the one that minimizes the error sum of
squares (ESS)
T
X T
X
b
b i, β
SSR(α εb2it = b
bi − β
(rit − α 2
i,M ) = i,M rMt )
t=1 t=1
9
These are two linear equations in two unknowns and by straightforward substitution
the solution is
b
b i = r̄i − β
α i,M r̄M
PT
t=1 (rit − r̄i )(rMt − r̄M )
βb i,M = PT 2
t=1 (rMt − r̄M )
where
T T
1X 1X
r̄i = rit , r̄M = rMt .
T t=1 T t=1
The equation for βb i,M can be rewritten slightly to show that βb i,M is a simple
function of variances and covariances. Divide the numerator and denominator of the
expression for βb i,M by T −1
1
to give
1 PT
t=1 (rit − r̄i )(rMt − r̄M ) d it , RMt )
cov(R
βb i,M = T −1
1 PT =
t=1 (rMt − r̄M )
2 d Mt )
var(R
T −1
which shows that βb i,M is the ratio of the estimated covariance between Rit and RMt
to the estimated variance of RMt .
The least squares estimate of σ 2ε,i = var(εit ) is given by
T T
1 X 1 X b
σb 2ε,i = eb2it = bi − β
(rt − α i,M rMt )
2
T − 2 t=1 T − 2 t=1
The divisor T − 2 is used to make σb 2ε,i an unbiased estimator of σ 2ε,ι .
The least squares estimate of R2 is given by
2
b 2 βb i,M σb 2M σb 2ε,i
Ri = = 1− ,
d it )
var(R d it )
var(R
where
T
1 X
d it ) =
var(R (rit − r̄i )2 ,
T − 1 t=1
and gives a measure of the goodness of &t of the regression equation. Notice that
b 2 = 1 whenever σ
R b 2ε,i = 0 which occurs when bεit = 0 for all values of t. In other
i
b 2
words, Ri = 1 whenever the regression line has a perfect &t. Conversely, R b2 = 0
i
2
when σb ε,i = var(R
d it ); that is, when the market does not explain any of the variability
of Rit . In this case, the regression has the worst possible &t.
10
3.2 Testing the Restriction α = 0.
Using the single index model regression,
Rt = α + βRMt + εt , t = 1, ..., T
εt ∼ iid N(0, σ 2ε ), εt is independent of RMt (4)
consider testing the null or maintained hypothesis α = 0 against the alternative that
α 6= 0
H0 : α = 0 vs. H1 : α 6= 0.
If H0 is true then the single index model regression becomes
Rt = βRMt + εt
and E[Rt |RMt = rMt ] = βrMt . We will reject the null hypothesis, H0 : α = 0, if
the estimated value of α is either much larger than zero or much smaller than zero.
Assuming H0 : α = 0 is true, α̂ ∼ N (0, SE(α̂)2 ) and so is fairly unlikely that α̂ will
be more than 2 values of SE(α̂) from zero. To determine how big the estimated value
of α needs to be in order to reject the null hypothesis we use the t-statistic
αb −0
tα=0 = d ,
SE(αb)
where α d α)
b is the least squares estimate of α and SE( b is its estimated standard error.
The value of the t-statistic, tα=0 , gives the number of estimated standard errors that
b is from zero. If the absolute value of tα=0 is much larger than 2 then the data cast
α
considerable doubt on the null hypothesis α = 0 whereas if it is less than 2 the data
are in support of the null hypothesis3 . To determine how big | tα=0 | needs to be to
reject the null, we use the fact that under the statistical assumptions of the single
index model and assuming the null hypothesis is true
If we set the signi&cance level (the probability that we reject the null given that the
null is true) of our test at, say, 5% then our decision rule is
11
Consider the estimated MM regression equation for IBM using monthly data from
January 1978 through December 1982:
b 2
b ε = 0.0524
R IBM,t =−0.0002 + 0.3390 ·RMt , R = 0.20, σ
(0.0068) (0.0888)
where the estimated standard errors are in parentheses. Here α b = −0.0002, which is
d
very close to zero, and the estimated standard error, SE(α̂) = 0.0068, is much larger
than αb . The t-statistic for testing H0 : α = 0 vs. H1 : α 6= 0 is
−0.0002 − 0
tα=0 = = −0.0363
0.0068
so that α b is only 0.0363 estimated standard errors from zero. Using a 5% signi&cance
level, |t58 (0.025)| ≈ 2 and
|tα=0 | = 0.0363 < 2
so we do not reject H0 : α = 0 at the 5% level.
H0 : β = 1 vs. H1 : β 6= 1.
The data cast doubt on this hypothesis if the estimated value of β is much different
from one. This hypothesis can be tested using the t-statistic
βb − 1
tβ=1 = d b
SE(β)
which measures how many estimated standard errors the least squares estimate of β
is from one. The null hypothesis is reject at the 5% level, say, if |tβ=1 | > |tT −2 (0.025)|.
Notice that this is a two-sided test.
Alternatively, one might want to test the hypothesis that the risk of an asset is
strictly less than the risk of the market index against the alternative that the risk is
greater than or equal to that of the market:
H0 : β = 1 vs. H1 : β ≥ 1.
Notice that this is a one-sided test. We will reject the null hypothesis only if the
estimated value of β much greater than one. The t-statistic for testing this null
12
hypothesis is the same as before but the decision rule is different. Now we reject the
null at the 5% level if
tβ=1 < −tT −2 (0.05)
where tT −2 (0.05) is the one-sided 5% critical value of the Student-t distribution with
T − 2 degrees of freedom.
13
Monthly Returns on IBM Monthly Returns on Market Index
0.2 0.2
0.1 0.1
0.0 0.0
-0.1 -0.1
-0.2 -0.2
-0.3 -0.3
78 79 80 81 82 83 84 85 86 87 78 79 80 81 82 83 84 85 86 87
IBM MARKET
Figure 1
Notice that the IBM and the market index have similar behavior over the sample
with the market index looking a little more volatile than IBM. Both returns dropped
sharply during the October 1987 crash but there were a few times that the market
dropped sharply whereas IBM did not. Sample descriptive statistics for the returns
are displayed in &gure 2.
The mean monthly returns on IBM and the market index are 0.9617% and 1.3992%
per month and the sample standard deviations are 5.9024% and 6.8353% per month,
respectively.. Hence the market index on average had a higher monthly return and
more volatility than IBM.
14
Monthly Returns on IBM Monthly Returns on Market Index
12 30
10 25
8 20
6 15
4 10
2 5
0 0
-0.15 -0.10 -0.05 0.00 0.05 0.10 0.15 -0.2 -0.1 0.0 0.1
Figure 2
Notice that the histogram of returns on the market are heavily skewed left whereas
the histogram for IBM is much more sysingle index modeletric about the mean. Also,
the kurtosis for the market is much larger than 3 (the value for normally distributed
returns) and the kurtosis for IBM is just slightly larger than 3. The negative skewness
and large kurtosis of the market returns is caused by several large negative returns.
The Jarque-Bera statistic for the market returns is 67.97, with a p-value 0.0000, and
so we can easily reject the hypothesis that the market data are normally distributed.
However, the Jarque-Bera statistic for IBM is only 0.1068, with a p-value of 0.9479,
and we therefore cannot reject the hypothesis of normality.
The single index model regression is
Rt = α + βRMt + εt , t = 1, . . . , T
15
Figure 3
where the estimated standard errors are reported underneath the estimated coeffi-
cients. The estimated intercept is close to zero at 0.0053, with a standard error of
d α)),
0.0069 (= SE( b and the estimated value of β is 0.3278, with an standard error of
d b
0.0890 (= SE(β)). Notice that the estimated standard error of βb is much smaller
than the estimated coefficient and indicates that β is estimated reasonably precisely.
The estimated regression equation is displayed graphically in &gure 4 below.
16
Market Model Regression
0.2
0.1
IBM
0.0
-0.1
-0.2
-0.3 -0.2 -0.1 0.0 0.1 0.2
MARKET
Figure 4
To evaluate the overall &t of the single index model regression we look at the R2 of
the regression, which measures the percentage of variability of Rt that is attributable
to the variability in RMt , and the estimated standard deviation of the residuals, σb ε .
From the table, R2 = 0.190 so the market index explains only 19% of the variability
of IBM and 81% of the variability is not explained by the market. In the single index
model regression, we can also interpret R2 as the proportion of market risk in IBM
and 1 − R2 as the proportion of &rm speci&c risk. The standard error (S.E.) of the
regression is the square root of the least squares estimate of σ 2ε = var(εt ). From the
above table, σb ε = 0.052. Recall, εt captures the &rm speci&c risk of IBM and so σb ε is
an estimate of the typical magnitude of the &rm speci&c risk. In order to interpret the
magnitude of σb ε it is useful to compare it to the estimate of the standard deviation
of Rt , which measures the total risk of IBM. This is reported in the table by the
standard deviation (S.D.) of the dependent variable which equals 0.057. Notice that
σb ε = 0.052 is only slightly smaller than 0.057 so that the &rm speci&c risk is a large
proportion of total risk (which is also reported by 1 − R2 ).
Con&dence intervals for the regression parameters are easily computed using the
reported regression output. Since εt is assumed to be normally distributed 95%
con&dence intervals for α and β take the form
α d α
b ± 2 · SE( b)
b d b
β ± 2 · SE(β)
17
The 95% con&dence intervals are then
Our best guess of α is 0.0053 but we wouldn t be too surprised if it was as low as
-0.0085 or as high as 0.0191. Notice that there are both positive and negative values
in the con&dence interval. Similarly, our best guess of β is 0.3278 but it could be as
low as 0.1498 or as high as 0.5058. This is a fairly wide range given the interpretation
of β as a risk measure. The interpretation of these intervals are as follows. In
repeated samples, 95% of the time the estimated con&dence intervals will cover the
true parameter values.
The t-statistic given in the computer output is calculated as
estimated coefficient − 0
t-statistic =
std. error
and it measures how many estimated standard errors the estimated coefficient is away
from zero. This t-statistic is often referred to as a basic signi&cance test because it
tests the null hypothesis that the value of the true coefficient is zero. If an estimate is
several standard errors from zero, so that it s t-statistic is greater than 2, then it is a
good bet that the true coefficient is not equal to zero. From the data in the table, the
b = 0.0053 is 0.767 standard errors from zero. Hence
t-statistic for α is 0.767 so that α
it is quite likely that the true value of α equals zero. The t-statistic for β is 3.684,
βb is more than 3 standard errors from zero, and so it is very unlikely that β = 0.
The Prob Value (p-value of the t-statistic) in the table gives the likelihood (computed
from the Student-t curve) that, given the true value of the coefficient is zero, the data
would generate the observed value of the t-statistic. The p-value for the t-statistic
testing α = 0 is 0.4465 so that it is quite likely that α = 0. Alternatively, the p-value
for the t-statistic testing β = 0 is 0.001 so it is very unlikely that β = 0.
18
Market Model Regression for IBM
0.2
0.1
0.0
0.15
0.10 -0.1
0.05
-0.2
0.00
-0.05
-0.10
-0.15
1978 1979 1980 1981 1982
Figure 5
Notice that the &tted values do not track the actual values very closely and that
the residuals are fairly large. This is due to low R2 of the regression. The residuals
appear to be fairly random by sight. We will develop explicit tests for randomness
later on. The histogram of the residuals, displayed below, can be used to investigate
the normality assumption. As a result of the least squares algorithm the residuals
have mean zero as long as a constant is included in the regression. The standard
deviation of the residuals is essentially equal to the standard error of the regression
- the difference being due to the fact that the formula for the standard error of the
regression uses T − 2 as a divisor for the error sum of squares and the standard
deviation of the residuals uses the divisor T − 1.
19
Residuals from Market Model Regression for IBM
8
Series: Residuals
Sample 1978:01 1982:12
Observations 60
6
Mean -2.31E-19
Median -0.000553
Maximum 0.139584
4
Minimum -0.104026
Std. Dev. 0.051567
Skewness 0.493494
2 Kurtosis 2.821260
Jarque-Bera 2.515234
Probability 0.284331
0
-0.10 -0.05 0.00 0.05 0.10
Figure 6
The skewness of the residuals is slightly positive and the kurtosis is a little less
than 3. The hypothesis that the residuals are normally distributed can be tested
using the Jarque-Bera statistic. This statistic is a function of the estimated skewness
and kurtosis and is give by
à !
T c − 3)2
(K
JB = Sb2 +
6 4
where Sb denotes the estimated skewness and K c denotes the estimated kurtosis. If
the residuals are normally distribued then Sb ≈ 0 and Kc ≈ 3 and JB ≈ 0. Therefore,
b c
if S is moderately different from zero or K is much different from 3 then JB will get
large and suggest that the data are not normally distributed. To determine how large
JB needs to be to be able to reject the normality assumption we use the result that
under the maintained hypothesis that the residuals are normally distributed JB has
a chi-square distribution with 2 degrees of freedom:
JB ∼ χ22 .
For a test with signi&cance level 5%, the 5% right tail critical value of the chi-square
distribution with 2 degrees of freedom, χ22 (0.05), is 5.99 so we would reject the null
that the residuals are normally distributed if JB > 5.99. The Probability (p-value)
reported by Eviews is the probability that a chi-square random variable with 2 degrees
of freedom is greater than the observed value of JB :
P (χ22 ≥ JB) = 0.2843.
For the IBM residuals this p-value is reasonably large and so there is not much data
evidence against the normality assumption. If the p-value was very small, e.g., 0.05 or
smaller, then the data would suggest that the residuals are not normally distributed.
20
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49
Introduction to Financial Econometrics
Hypothesis Testing in the Market Model
Eric Zivot
Department of Economics
University of Washington
February 29, 2000
Rt = α + βRMt + εt , t = 1, ..., T
εt ∼ iid N (0, σ 2ε ), εt is independent of RMt (1)
consider testing the null or maintained hypothesis α = 0 against the alternative that
α 6= 0
H0 : α = 0 vs. H1 : α 6= 0.
If H0 is true then the market model regression becomes
Rt = βRMt + εt
and E[Rt |RMt = rMt ] = βrMt . We will reject the null hypothesis, H0 : α = 0, if
the estimated value of α is either much larger than zero or much smaller than zero.
Assuming H0 : α = 0 is true, α̂ ∼ N (0, SE(α̂)2 ) and so is fairly unlikely that α̂ will
1
be more than 2 values of SE(α̂) from zero. To determine how big the estimated value
of α needs to be in order to reject the null hypothesis we use the t-statistic
αb −0
tα=0 = d ,
SE(αb)
where α d α
b is the least squares estimate of α and SE( b ) is its estimated standard error.
The value of the t-statistic, tα=0 , gives the number of estimated standard errors that
b is from zero. If the absolute value of tα=0 is much larger than 2 then the data cast
α
considerable doubt on the null hypothesis α = 0 whereas if it is less than 2 the data
are in support of the null hypothesis1 . To determine how big | tα=0 | needs to be to
reject the null, we use the fact that under the statistical assumptions of the market
model and assuming the null hypothesis is true
If we set the significance level (the probability that we reject the null given that the
null is true) of our test at, say, 5% then our decision rule is
Consider the estimated MM regression equation for IBM using monthly data from
January 1978 through December 1982:
b 2
b ε = 0.0524
R IBM,t =−0.0002 + 0.3390 ·RMt , R = 0.20, σ
(0.0068) (0.0888)
where the estimated standard errors are in parentheses. Here α b = −0.0002, which is
d
very close to zero, and the estimated standard error, SE(α̂) = 0.0068, is much larger
than αb . The t-statistic for testing H0 : α = 0 vs. H1 : α 6= 0 is
−0.0002 − 0
tα=0 = = −0.0363
0.0068
so that αb is only 0.0363 estimated standard errors from zero. Using a 5% significance
level, t58 (0.025) ≈ 2 and
|tα=0 | = 0.0363 < 2
so we do not reject H0 : α = 0 at the 5% level.
1
This interpretation of the t-statistic relies on the fact that, assuming the null hypothesis is true
b is normally distributed with mean 0 and estimated variance SE(b
so that α = 0, α d α)2 .
2
1.3 Testing Hypotheses about β
In the market model regression β measures the contribution of an asset to the vari-
ability of the market index portfolio. One hypothesis of interest is to test if the asset
has the same level of risk as the market index against the alternative that the risk is
different from the market:
H0 : β = 1 vs. H1 : β 6= 1.
The data cast doubt on this hypothesis if the estimated value of β is much different
from one. This hypothesis can be tested using the t-statistic
βb − 1
tβ=1 = d b
SE(β)
which measures how many estimated standard errors the least squares estimate of β
is from one. The null hypothesis is reject at the 5% level, say, if |tβ=1 | > tT −2 (0.025).
Notice that this is a two-sided test.
Alternatively, one might want to test the hypothesis that the risk of an asset is
strictly less than the risk of the market index against the alternative that the risk is
greater than or equal to that of the market:
H0 : β = 1 vs. H1 : β ≥ 1.
Notice that this is a one-sided test. We will reject the null hypothesis only if the
estimated value of β much greater than one. The t-statistic for testing this null
hypothesis is the same as before but the decision rule is different. Now we reject the
null at the 5% level if
tβ=1 < −tT −2 (0.05)
where tT −2 (0.05) is the one-sided 5% critical value of the Student-t distribution with
T − 2 degrees of freedom.
3
1.4 Testing Joint Hypotheses about α and β
Often it is of interest to formulate hypothesis tests that involve both α and β. For
example, consider the joint hypothesis that α = 0 and β = 1 :
H0 : α = 0 and β = 1.
Recall, this is the quantity that is minimized during the least squares algorithm. Now,
the market model imposing the restrictions under H0 is
Rt = 0 + 1 · (RMt − rf ) + εt
= RMt + εt .
Notice that there are no parameters to be estimated in this model which can be seen
by subtracting RMt from both sides of the restricted model to give
Rt − RMt = eεt
The fit of the restricted (R) model is then measured by the restricted sum of squared
residuals
T
X T
X
SSRR = SSR(α = 0, β = 1) = εe2t = (Rt − RMt )2 .
t=1 t=1
Now since the least squares algorithm works to minimize SSR, the restricted error
sum of squares, SSRR , must be at least as big as the unrestricted error sum of squares,
SSRU R . If the restrictions imposed under the null are true then SSRR ≈ SSRU R
(with SSRR always slightly bigger than SSRU R ) but if the restrictions are not true
then SSRR will be quite a bit bigger than SSRU R . The F-statistic measures the
(adjusted) percentage difference in fit between the restricted and unrestricted models
and is given by
(SSRR − SSRU R )/q (SSRR − SSRU R )
F = = ,
SSRU R /(T − k) q · σb 2ε,U R
4
where q equals the number of restrictions imposed under the null hypothesis, k denotes
the number of regression coefficients estimated under the unrestricted model and
σb 2ε,UR denotes the estimated variance of εt under the unrestricted model. Under the
assumption that the null hypothesis is true, the F-statistic is distributed as an F
random variable with q and T − 2 degrees of freedom:
F ∼ Fq,T −2 .
Notice that an F random variable is always positive since SSRR > SSRUR . The null
hypothesis is rejected, say at the 5% significance level, if
Consider testing the hypothesis H0 : α = 0 and β = 1 for the IBM data. The
unrestricted error sum of squares, SSRU R , is obtained from the unrestricted regression
output in figure 2 and is called Sum Square Resid:
SSRU R = 0.159180.
To form the restricted sum of squared residuals, we create the new variable eεt =
P
Rt − RMt and form the sum of squares SSRR = Tt=1 eε2t = 0.31476. Notice that
SSRR > SSRU R . The F-statistic is then
(0.31476 − 0.159180)/2
Fα=0,β=1 = = 28.34.
0.159180/58
The 95% quantile of the F-distribution with 2 and 58 degrees of freedom is about
3.15. Since Fα=0,β=1 = 28.34 > 3.15 = F2,58 (0.05) we reject H0 : α = 0 and β = 1 at
the 5% level.
5
assets may change over time it is of interest to know if α and β change over time. To
illustrate, suppose we have a ten year sample of monthly data (T = 120) on returns
that we split into two five year subsamples. Denote the first five years as t = 1, ..., TB
and the second five years as t = TB+1 , ..., T. The date t = TB is the “break date” of
the sample and it is chosen arbitrarily in this context. Since the samples are of equal
size (although they do not have to be) T − TB = TB or T = 2 · TB . The market model
regression which assumes that both α and β are constant over the entire sample is
Rt = α + βRMt + εt , t = 1, . . . , T
εt ∼ iid N(0, σ 2 ) independent of RMt .
There are three main cases of interest: (1) β may differ over the two subsamples; (2) α
may differ over the two subsamples; (3) α and β may differ over the two subsamples.
Rt = α + β 1 RMt + εt , t = 1, . . . , TB
Rt = α + β 2 RMt + εt , t = TB+1 , . . . , T
that share the same intercept α but have different slopes β 1 6= β 2 . We can capture
such a model very easily using a step dummy variable defined as
Dt = 0, t ≤ TB
= 1, t > TB
Rt = α + βRMt + Dt RMt + εt .
Rt = α + βRMt + εt , t = 1, . . . , TB
Notice that the “beta” in the first sample is β 1 = β and the beta in the second
subsample is β 2 = β + δ. If δ < 0 the second sample beta is smaller than the first
sample beta and if δ > 0 the beta is larger.
6
We can test the constancy of beta over time by testing δ = 0:
H0 : (beta is constant over two sub-samples) δ = 0 vs. H1 : (beta is not constant over two sub-samples
δb − 0 b
δ
tδ=0 = d b = d b
SE(δ) SE(δ)
and we reject the hypothesis δ = 0 at the 5% level, say, if |tδ=0 | > tT −3 (0.025).
Consider the estimated MM regression equation for IBM using ten years of monthly
data from January 1978 through December 1987. We want to know if the beta on
IBM is using the first five years of data (January 1978 - December 1982) is different
from the beta on IBM using the second five years of data (January 1983 - December
1987). We define the step dummy variable
R2 = 0.311, σb ε = 0.0496.
The estimated value of β is 0.3388, with a standard error of 0.0837, and the estimated
value of δ is 0.3158, with a standard error of 0.1366. The t-statistic for testing δ = 0
is given by
0.3158
tδ=0 = = 2.312
0.1366
which is greater than t117 (0.025) = 1.98 so we reject the null hypothesis (at the
5% significance level) that beta is the same over the two subsamples. The implied
estimate of beta over the period January 1983 - December 1987 is
7
1.5.2 Testing Structural Change in α and β
Now consider the case where both α and β are allowed to be different over the two
subsamples:
Rt = α1 + β 1 RMt + εt , t = 1, . . . , TB
Rt = α2 + β 2 RMt + εt , t = TB+1 , . . . , T
Rt = α + βRMt + δ 1 · Dt + δ 2 · Dt RMt + εt , t = 1, . . . , T.
Rt = α + βRMt + εt , t = 1, . . . , TB ,
Rt = (α + δ 1 ) + (β + δ 2 )RMt + εt , t = TB+1 , . . . , T,
since there are two restrictions and four regression parameters estimated under the
unrestricted model. The unrestricted (UR) model is the dummy variable regression
that allows the intercepts and slopes to differ in the two subsamples and the restricted
model (R) is the regression where these parameters are constrained to be the same
in the two subsamples.
The unrestricted error sum of squares, SSRU R , can be computed in two ways.
The first way is based on the dummy variable regression. The second is based on
estimating separate regression equations for the two subsamples and adding together
the resulting error sum of squares. Let SSR1 and SSR2 denote the error sum of
squares from separate regressions. Then
8
The unrestricted regression is
d
RIBM,t = −0.0001 + 0.3388 ·RMt
(0.0065) (0.0845)
Then Dtup divides the sample into “up market” movements and “down market” move-
ments. The regression that allows beta to differ depending on the market cycle is
then
Rt = α + βRMt + δDtup · RMt + εt .
In the down cycle, when Dtup = 0, the model is
Rt = α + βRMt + εt
9
and β captures the down market beta, and in the up market, when Dtup = 1, the
model is
Rt = α + (β + δ)RMt + εt
so that β + δ capture the up market beta. The hypothesis that β does not vary over
the market cycle is
H0 : δ = 0 vs. H1 : δ 6= 0 (2)
bδ−0
and can be tested with the simple t-statistic tδ=0 = c .
SE(bδ)
If the estimated value of δ is found to be statistically greater than zero we might
then want to go on to test the hypothesis that the up market beta is greater than
one. Since the up market beta is equal to β + δ this corresponds to testing
H0 : β + δ = 1 vs. H1 : β + δ ≥ 1
βb + δb − 1
tβ+δ=1 = d b b .
SE(β + δ)
Since this is a one-sided test we will reject the null hypothesis at the 5% level if
tβ+δ=1 < −t0.05,T −3 .
For IBM the CAPM regression allowing β to vary over the market cycle (1978.01
- 1982.12) is
d
RIBM,t = −0.0019 + 0.3163 ·RMt + 0.0552 ·Dtup · RMt
(0.0111) (0.1476) (0.2860)
2
R = 0.201, σb = 0.053
Notice that bδ = 0.0552, with a standard error of 0.2860, is close to zero and not
estimated very precisely. Consequently, tδ=0 = 0.0552
0.2860
= 0.1929 is not significant at
any reasonable significance level and we therefore reject the hypothesis that beta
varies over the market cycle. However, the results are very different for DEC (Digital
Electronics):
d
RDEC,t = −0.0248 + 0.3689 ·RMt + 0.8227 ·Dtup · RMt
(0.0134) (0.1779) (0.3446)
2
R = 0.460, σb = 0.064.
Here δb = 0.8227, with a standard error of 0.3446, is statistically different from zero
at the 5% level since tδ=0 = 2.388. The estimate of the down market beta is 0.3689,
which is less than one, and the up market beta is 0.3689 + 0.8227 = 1.1916, which
10
is greater than one. The estimated standard error for βb + δb requires the estimated
variances of βb and bδ and the estimated covariance between βb and bδ and is given by
d β
var( b+b d β)
δ) = var( b + var(
d b d β,
δ) + 2 · cov( b δ)
b
11
Chapter 1
The Constant Expected Return
Model
The first model of asset returns we consider is the very simple constant ex-
pected return (CER) model. This model assumes that an asset’s return over
time is normally distributed with a constant (time invariant) mean and vari-
ance The model also assumes that the correlations between asset returns
are constant over time. Although this model is very simple, it allows us to
discuss and develop several important econometric topics such as estimation,
hypothesis testing, forecasting and model evaluation.
1
2CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
t
E[Rit ] = µi
var(Rit ) = σ 2i
The second assumption states that the contemporaneous covariances between
assets are constant over time. Given assumption 1, assumption 2 implies that
the contemporaneous correlations between assets are constant over time as
well. That is, for all assets and time periods
corr(Rit , Rjt ) = ρij
The third assumption stipulates that all of the asset returns are uncorrelated
over time1 . In particular, for a given asset i the returns on the asset are
serially uncorrelated which implies that
corr(Rit , Ris ) = cov(Rit , Ris ) = 0 for all t 6= s.
Additionally, the returns on all possible pairs of assets i and j are serially
uncorrelated which implies that
corr(Rit , Rjs ) = cov(Rit , Rjs ) = 0 for all i 6= j and t 6= s.
Assumptions 1-3 indicate that all asset returns at a given point in time
are jointly (multivariate) normally distributed and that this joint distribution
stays constant over time. Clearly these are very strong assumptions. How-
ever, they allow us to development a straightforward probabilistic model for
asset returns as well as statistical tools for estimating the parameters of the
model and testing hypotheses about the parameter values and assumptions.
which use the fact that adding constants to two random variables does not
affect the covariance between them. Given that covariances and variances
of returns are constant over time gives the result that correlations between
returns over time are also constant:
cov(Rit , Rjt ) σ ij
corr(Rit , Rjt ) = p = = ρij ,
var(Rit )var(Rjt ) σiσj
cov(Rit , Rjs ) 0
corr(Rit , Rjs ) = p = = 0, i 6= j, t 6= s.
var(Rit )var(Rjs ) σiσj
Finally, since the random variable εit is independent and identically distrib-
uted (i.i.d.) normal the asset return Rit will also be i.i.d. normal:
Hence, the CER model (1.1) for Rit is equivalent to the model implied by
assumptions 1-3.
compounded returns are additive, Rit (12) is the sum of 12 monthly continu-
ously compounded returns2 :
11
X
RitA = Rit (12) = Rit−k = Rit + Rit−1 + · · · + Rit−11
t=0
Using the CER model representation (1.1) for the monthly return Rit we may
express the annual return Rit (12) as
11
X
Rit (12) = (µi + εit )
t=0
11
X
= 12 · µi + εit
t=0
= µA
i + εA
it
where µA
P i = 12 · µi is the annual expected return on asset i and εit =
A
11
k=0 εit−k is the annual random news component. Hence, the annual ex-
pected return, µAi , is simply 12 times the monthly expected return, µi . The
annual random news component, εA it , is the accumulation of news over the
year. Using the results from chapter 2 about the variance of a sum of ran-
dom variables, the variance of the annual news component is just 12 time
the variance of the monthly new component:
à 11 !
X
var(εA
it ) = var εit−k )
k=0
11
X
= var(εit−k ) since εit is uncorrelated over time
k=0
X11
= σ 2i since var(εit ) is constant over time
k=0
= 12 · σ 2i
= var(RitA )
2
For simplicity of exposition, we will ignore the fact that some assets do not trade over
the weekend.
6CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
12 · σ ij
= q
12σ 2i · 12σ 2j
σ ij
= = ρij
σi σj
= corr(εit , εjt )
Letting pit = ln(Pit ) and using the representation of Rit in the CER model
(1.1), we may further rewrite the above as
The representation in (1.3) is know as the RW model for the log of asset
prices.
In the RW model, µi represents the expected change in the log of asset
prices (continuously compounded return) between months t − 1 and t and εit
represents the unexpected change in prices. That is,
Further, in the RW model, the unexpected changes in asset prices, εit , are
uncorrelated over time (cov(εit , εis ) = 0 for t 6= s) so that future changes in
asset prices cannot be predicted from past changes in asset prices3 .
The RW model gives the following interpretation for the evolution of asset
prices. Let pi0 denote the initial log price of asset i. The RW model says
that the price at time t = 1 is
where εi1 is the value of random news that arrives between times 0 and 1.
Notice that at time t = 0 the expected price at time t = 1 is
which is the initial price plus the expected return between time 0 and 1.
Similarly, the price at time t = 2 is
which is equal to the initial price, pi0 , plus the two period expected
P2 return,
2 · µi , plus the accumulated random news over the two periods, t=1 εit . By
recursive substitution, the price at time t = T is
T
X
piT = pi0 + T · µi + εit .
t=1
E[piT ] = pi0 + T · µi
The actual price, piT , deviates from the expected price by the accumulated
random news
X
T
piT − E[piT ] = εit .
t=1
Figure 1.1 illustrates the random walk model of asset prices based on the
CER model with µ = 0.05, σ = 0.10 and p0 = 1. The plot shows the log
price, P
pt , the expected price E[pt ] = p0 + 0.05t and the accumulated random
news tt=1 εt .
The term random walk was originally used to describe the unpredictable
movements of a drunken sailor staggering down the street. The sailor starts
at an initial position, p0 , outside the bar. The sailor generally moves in the
direction described by µ but randomly deviates from this direction after each
step t by an amount PTequal to εt . After T steps the sailor ends up at position
pT = p0 + µ · T + t=1 εt .
p(t)
E[p(t)]
p(t)-E[p(t)]
4
2
0
0 20 40 60 80 100
To mimic the monthly return data on Microsoft, the values µ = 0.05 and
σ = 0.10 are used as the model’s parameters and T = 100 is the number of
simulated values (sample size). The key to simulating data from the above
model is to simulate T = 100 observations of the random news variable εt
~iid N(0, (0.10)2 ). Computer algorithms exist which can easily create such
observations..Let {ε∗1 , . . . , ε∗100 } denote the 100 simulated values of εt .The
simulated returns are then computed as
A time plot and histogram of the simulated Rt∗ values are given in figure
.The simulated return data fluctuates randomly about the expected return
10CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
30
0.3
25
0.2
20
0.1
frequency
return
15
0.0
10
-0.1
5
-0.2
0
0 20 40 60 80 100 -0.2 -0.1 0.0 0.1 0.2 0.3
months return
Figure 1.2: Simulated returns from the CER model Rt = 0.05 + εt , εt ~iid
N(0, (0.10)2 )
from the random sample. Under these assumptions, we can use the observed
returns to estimate the unknown parameters of the CER model
1X
T
µ̂(R1 , . . . , RT ) = Rt
T t=1
1.2 ESTIMATING THE PARAMETERS OF THE CER MODEL13
Bias
Bias concerns the location or center of p(θ̂). If p(θ̂) is centered away from θ
then we say θ̂ is biased. If p(θ̂) is centered at θ then we say that θ̂ is unbiased.
Formally we have the following definitions:
error = θ̂ − θ.
bias(θ̂, θ) = E[θ̂] − θ.
0.8
0.7
0.6
0.5
pdf 1
pdf
0.4
pdf 2
0.3
0.2
0.1
0
-10 -5 0 5 10
estimator value
Precision
An estimate is, hopefully, our best guess of the true (but unknown) value of
θ. Our guess most certainly will be wrong but we hope it will not be too far
1.2 ESTIMATING THE PARAMETERS OF THE CER MODEL15
off. A precise estimate, loosely speaking, is one that has a small estimation
error. The magnitude of the estimation error is usually captured by the mean
squared error:
0.2
0.1
0.0
returns
-0.1
-0.2
-0.3
-0.4
Q3 Q1 Q3 Q1 Q3 Q1 Q3 Q1 Q3 Q1 Q3 Q1 Q3 Q1 Q3 Q1 Q3
1992 1993 1994 1995 1996 1997 1998 1999 2000
This is the estimated news component for month t based on the estimate µ̂i .
Now the CER model imposes the condition that the expected value of the
true error is zero
E[εit ] = 0
The method of moments estimator of µi is the value of µ̂i that makes the
average of the sample errors equal to the expected value of the population
errors. That is, the method of moments estimator solves
1X
T
1X
T
ε̂it = (rit − µ̂i ) = E[εit ] = 0 (1.4)
T t=1 T t=1
5
In this book, quantities with a “ˆ” denote an estimate.
1.2 ESTIMATING THE PARAMETERS OF THE CER MODEL17
Returns on Microsoft
0.0
returns
-0.4
Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4
1992 1993 1994 1995 1996 1997 1998 1999 2000
Returns on Starbucks
0.0
returns
-0.4
Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4
1992 1993 1994 1995 1996 1997 1998 1999 2000
-0.15
Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4
1992 1993 1994 1995 1996 1997 1998 1999 2000
1X
T
µ̂i = rit = r̄. (1.5)
T t=1
1 X
100
µ̂msf t = rmsf t,t = 0.0276
100 t=1
1 X
100
µ̂sbux = rsbux,t = 0.0278
100 t=1
1 X
100
µ̂sp500 = rsp500,t = 0.0125
100 t=1
The mean returns for MSFT and SBUX are very similar at about 2.8%
per month whereas the mean return for SP500 is smaller at only 1.25% per
month.
1 X
T
σ̂ 2i= (rit − r̄i )2 , (1.6)
T − 1 t=1
q
σ̂ i = σ̂ 2i , (1.7)
1 X
T
σ̂ ij = (rit − r̄i )(rjt − r̄j ), (1.8)
T − 1 t=1
σ̂ ij
ρ̂ij = (1.9)
σ̂ i σ̂ j
P
where r̄i = T1 Tt=1 rit = µ̂i is the sample average of the returns on asset.i.
Notice that (1.6) is simply the sample variance of the observed returns for
asset i, (1.7) is the sample standard deviation, (1.8) is the sample covariance
of the observed returns on assets i and j and (1.9) is the sample correlation
of returns on assets i and j.
Example 11 Consider again the monthly continuously compounded returns
on Microsoft, Starbucks and the S&P 500 index over the period July 1992
1.2 ESTIMATING THE PARAMETERS OF THE CER MODEL19
0.3
0.1
sbux -0.1
-0.3
-0.5
0.3
0.1
msft
-0.1
-0.3
-0.5
0.10
0.05
0.00
sp500
-0.05
-0.10
-0.15
-0.20
-0.5 -0.3 -0.1 0.1 0.3 -0.20 -0.15 -0.10 -0.05 0.00 0.05 0.10
SBUX has the most variable monthly returns and SP500 has the smallest.
The scatterplots of the returns are illustrated in figure 1.6. All returns appear
to be positively related. The pairs (MSFT,SP500) and (SBUX,SP500) appear
to be the most correlated.The estimates of σ ij and ρij using (1.8) and (1.9)
are
These estimates confirm the visual results from the scatterplot matrix.
20CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
1X
T
µ̂i = µ̂i (Ri1 , . . . , RiT ) = Rit (1.10)
T t=1
Bias
In the CER model, the random variables Rit (t = 1, . . . , T ) are iid normal
with mean µi and variance σ 2i . Since the method of moments estimator
(1.10) is an average of these normal random variables it is also normally
distributed. That is, p(µ̂i ) is a normal density. ToPdetermine the mean of
−1 T
this distribution we must compute E[µ̂i ] = E[T t=1 Rit ]. Using results
from chapter 2 about the expectation of a linear combination of random
variables it is straightforward to show (details are given in the appendix)
that
E[µ̂i ] = µi
Hence, the mean of the distribution of µ̂i is equal to µi . In other words, µ̂i
an unbiased estimator for µi .
Precision
P
To determine the variance of µ̂i we must compute var(µ̂i ) = var(T −1 Tt=1 Rit ).
Using the results from chapter 2 about the variance of a linear combination
of uncorrelated random variables it is easy to show (details in the appendix)
that
σ2
var(µ̂i ) = . (1.11)
T
Notice that the variance of µ̂i is equal to the variance of Rit divided by the
sample size and is therefore much smaller than the variance of Rit .
The standard deviation of µ̂i is just the square root of var(µ̂it )
p σi
SD(µ̂i ) = var(µ̂i ) = √ . (1.12)
T
1.3 STATISTICAL PROPERTIES OF ESTIMATES 21
0.8
pdf 1
pdf 2
0.6
0.4
pdf
0.2
0.0
-3 -2 -1 0 1 2 3
estimate value
Figure 1.7: Pdfs for µ̂i with small and large values of SE(µ̂i ). True value of
µi = 0.
The standard deviation of µ̂i is most often referred to as the standard error
of the estimate µ̂i :
σi
SE(µ̂i ) = SD(µ̂i ) = √ . (1.13)
T
SE(µ̂i ) is in the same units as µ̂i and measures the precision of µ̂i as an
estimate. If SE(µ̂i ) is small relative to µ̂i then µ̂i is a relatively precise of
µi because p(µ̂i ) will be tightly concentrated around µi ; if SE(µ̂i ) is large
relative to µi then µ̂i is a relatively imprecise estimate of µi because p(µ̂i )
will be spread out about µi . Figure 1.7 illustrates these relationships
Unfortunately, SE(µ̂i ) is not a practically useful measure of the precision
of µ̂i because it depends on the unknown value of σ i . To get a practically
useful measure of precision for µ̂i we compute the estimated standard error
p bi
σ
c i ) = vd
SE(µ̂ ar(µ̂i ) = √ (1.14)
T
which is just (1.13) withq
the unknown value of σ i replaced by the method of
moments estimate σ bi = σ b2i .
22CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
Example 12 For the Microsoft, Starbucks and S&P 500 return data, the
c i ) are
values of SE(µ̂
c msf t ) = 0.1068
SE(µ̂ √ = 0.01068
100
c sbux ) = 0.1359
SE(µ̂ √ = 0.01359
100
c sp500 ) = 0.0379
SE(µ̂ √ = 0.003785
100
Clearly, the mean return µi is estimated more precisely for the S&P 500 index
than it is for Microsoft and Starbucks.
and simulate N = 1000 samples of size T = 50 values from the above model
using the technique of Monte Carlo simulation. This gives j = 1, . . . , 1000
sample realizations {r1j∗ , . . . , r50
j∗
}. The first 10 of these sample realizations
are illustrated in figure 1.8.Notice that there is considerable variation in the
simulated samples but that all of the simulated samples fluctuates about
the true mean value of µ = 0.05. For each of the 1000 simulated samples the
estimate µ̂ is formed giving 1000 mean estimates {µ̂1 , . . . , µ̂1000 }. A histogram
of these 1000 mean values is illustrated in figure 1.9.The histogram of the
estimated means, µ̂j , can be thought of as an estimate of the underlying pdf,
p(µ̂), of the estimator µ̂ which we know is a Normal pdf centered at µ = 0.05
0.10
with SE(µ̂i ) = √ 50
= 0.01414. Notice that the center of the histogram is
very close to the true mean value µ = 0.05. That is, on average over the
1.3 STATISTICAL PROPERTIES OF ESTIMATES 23
0.3
0.2
0.1
returns
0.0
-0.1
-0.2
0 10 20 30 40 50
Figure 1.8: Ten simulated samples of size T = 50 from the CER model
Rt = 0.05 + εt , εt ~iid N(0.(0.10)2 )
1000 Monte Carlo samples the value of µ̂ is about 0.05. In some samples,
the estimate is too big and in some samples the estimate is too small but on
average the estimate is correct. In fact, the average value of {µ̂1 , . . . , µ̂1000 }
from the 1000 simulated samples is
1 X j
1000
µ̂ = 0.05045
1000 j=1
which is very close to the true value. If the number of simulated samples is
allowed to go to infinity then the sample average of µ̂j will be exactly equal
to µ :
1 X j
N
lim µ̂ = µ
N→∞ N
j=1
The typical size of the spread about the center of the histogram represents
SE(µ̂i ) and gives an indication of the precision of µ̂i .The value of SE(µ̂i ) may
be approximated by computing the sample standard deviation of the 1000
24CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
250
200
150
100
50
0
µ̂j values v
u
u 1 X 1000
t (µ̂j − 0.05045)2 = 0.01383
999 j=1
0.10
Notice that this value is very close to SE(µ̂i ) = √ 50
= 0.01414. If the number
of simulated sample goes to infinity then
v
u
u 1 X N
1 X j 2
N
lim t j
(µ̂ − µ̂ ) = SE(µ̂i )
N →∞ N − 1 j=1 N j=1
2.5
pdf T=1
pdf T=10
pdf T=50
2.0
1.5
pdf
1.0
0.5
0.0
-3 -2 -1 0 1 2 3
estimate value
The distribution for µ̂i is centered at the true value µi and the spread about
the average depends on the magnitude of σ 2i , the variability of Rit , and the
sample size. For a fixed sample size, T , the uncertainty in µ̂i is larger for
larger values of σ 2i . Notice that the variance of µ̂i is inversely related to
the sample size T. Given σ 2i , var(µ̂i ) is smaller for larger sample sizes than
for smaller sample sizes. This makes sense since we expect to have a more
precise estimator when we have more data. If the sample size is very large (as
T → ∞) then var(µ̂i ) will be approximately zero and the normal distribution
of µ̂i given by (1.16) will be essentially a spike at µi . In other words, if the
sample size is very large then we essentially know the true value of µi . In the
statistics language we say that µ̂i is a consistent estimator of µi .
c i ).
µ̂i ± 2 · SE(µ̂ (1.17)
6 d i ) is equal
This resut follows from the fact that µ̂i is normally distributed and SE(µ̂
to the square root of a chi-square random variable divided by its degrees of freedom.
1.3 STATISTICAL PROPERTIES OF ESTIMATES 27
The above formula for a 95% confidence interval is often used as a rule of
thumb for computing an approximate 95% confidence interval for moderate
sample sizes. It is easy to remember and does not require the computation
of quantile tT −1 (α/2) from the Student-t distribution.
With probability .95, the above intervals will contain the true mean values
assuming the CER model is valid. The approximate 95% confidence inter-
vals for MSFT and SBUX are fairly wide. The widths are almost 5% with
lower limits near 0 and upper limits near 5%. In contrast, the 95% con-
fidence interval for SP500 is about half the width of the MSFT or SBUX
confidence interval. The lower limit is near .5% and the upper limit is near
2%. This clearly shows that the mean return for SP500 is estimated much
more precisely than the mean return for MSFT or SBUX.
1 X
T
σ̂ 2i = σ̂ 2i (Ri1 , . . . RiT )
= (Rit − µ̂i )2 ,
T − 1 t=1
q
σ̂ i = σ̂ i (Ri1 , . . . RiT ) = σ̂ 2i (Ri1 , . . . RiT ).
1 X
T
σ̂ ij = σ̂ ij (Ri1 , . . . , RiT ; Rj1 , . . . , RjT ) = (Rit − µ̂i )(Rjt − µ̂j ),
T − 1 t=1
σ̂ ij (Ri1 , . . . , RiT ; Rj1 , . . . , RjT )
ρ̂ij = σ̂ ij (Ri1 , . . . , RiT ; Rj1 , . . . , RjT ) = .
σ̂ i (Ri1 , . . . RiT ) · σ̂ j (Rj1 , . . . RjT )
Bias
Assuming that returns are generated by the CER model (1.1), the sample
variances and covariances are unbiased estimators,
E[σ̂ 2i ] = σ 2i ,
E[σ̂ ij ] = σ ij ,
but the sample standard deviations and correlations are biased estimators,
6 σi ,
E[σ̂ i ] =
E[ρ̂ij ] =6 ρij .
The proofs of these results are beyond the scope of this book. However, they
may be easily be evaluated using Monte Carlo methods.
Precision
The derivations of the variances of σ̂ 2i , σ̂ i , σ̂ ij and ρ̂ij are complicated and the
exact results are extremely messy and hard to work with. However, there are
simple approximate formulas for the variances of σ̂ 2i , σ̂ i and ρ̂ij that are valid
if the sample size, T, is reasonably large 7 . These large sample approximate
formulas are given by
σ2
SE(σ̂ 2i ) ≈ p i , (1.18)
T /2
σi
SE(σ̂ i ) ≈ √ , (1.19)
2T
(1 − ρ2ij )
SE(ρij ) ≈ √ , (1.20)
T
7
The large sample approximate formula for the variance of σ̂ ij is too messy to work
with so we omit it here.
1.3 STATISTICAL PROPERTIES OF ESTIMATES 29
where “≈” denotes approximately equal. The approximations are such that
the approximation error goes to zero as the sample size T gets very large.
As with the formula for the standard error of the sample mean, the formulas
for the standard errors above are inversely related to the square root of the
sample size. Interestingly, SE(σ̂ i ) goes to zero the fastest and SE(σ̂ 2i ) goes to
zero the slowest. Hence, for a fixed sample size, it appears that σ i is generally
estimated more precisely than σ 2i and ρij , and ρij is estimated generally more
precisely than σ 2i .
The above formulas are not practically useful, however, because they
depend on the unknown quantities σ 2i , σ i and ρij . Practically useful formulas
replace σ 2i , σ i and ρij by the estimates σ̂ 2i , σ̂ i and ρ̂ij and give rise to the
estimated standard errors
2
c 2i ) ≈ pσ̂ i ,
SE(σ̂ (1.21)
T /2
c i ) ≈ √σ̂ i ,
SE(σ̂ (1.22)
2T
(1 − ρ̂2ij )
c ij ) ≈
SE(ρ √ . (1.23)
T
Example 14 To be completed
Sampling distribution
To be completed
200
200
150
150
100
100
50
50
0
0
0.004 0.006 0.008 0.010 0.012 0.014 0.016 0.08 0.10 0.12
1 X 2
1000
σ̂ = 0.009952
1000 j=1
1 X
1000
σ̂ = 0.09928
1000 j=1
2 (0.10)2
SE(σ̂ ) = p = 0.002
50/2
0.10
SE(σ̂) = √ = 0.010
2 · 50
To be completed
1.5 Appendix
Result: E[µ̂i ] = µi
32CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
PT
Proof. Using the fact that µ̂i = T −1 t=1 Rit and Rit = µi + εit we have
" #
1X
T
E[µ̂i ] = E Rit
T t=1
" #
1X
T
= E (µ + εit )
T t=1 i
1X
T
1X
T
= µ + E[εit ] (by the linearity of E[·])
T t=1 i T t=1
1X
T
= µ (since E[εit ] = 0, t = 1, . . . , T )
T t=1 i
1
= T · µi
T
= µi .
σ2
Result: var(µi ) = Ti .
P
Proof. Using the fact that µ̂i = T −1 Tt=1 Rit and Rit = µi + εit we have
à !
1X
T
var(µ̂i ) = var Rit
T t=1
à !
1 XT
= var (µ + εit ) (in the CER model Rit = µi + εit )
T t=1 i
à !
1X
T
= var εit (since µi is a constant)
T t=1
1 X
T
= 2 var(εit ) (since εit is independent over time)
T t=1
1 X 2
T
= 2 σ (since var(εit ) = σ 2i , t = 1, . . . , T )
T t=1 i
1
= 2
T σ 2i
T
σ2
= i.
T
1.5 APPENDIX 33
that Z and X are independent. Define a new random variable t such that
Z
t= p .
X/T
1.6 Problems
To be completed
Bibliography
35
Introduction to Financial Econometrics
Appendix Matrix Algebra Review
Eric Zivot
Department of Economics
University of Washington
January 3, 2000
This version: February 6, 2001
1
1 £ ¤
x = 2 , x0 = 1 2 3 .
(3×1) (1×3)
3
Clearly, · ¸
0 1 2
A =A= .
2 1
Then
· ¸ · ¸ · ¸ · ¸
4 9 2 0 4+2 9+0 6 9
A+B = + = = ,
2 1 0 7 2+0 1+7 2 8
· ¸ · ¸ · ¸ · ¸
4 9 2 0 4−2 9−0 2 9
A−B = − = = .
2 1 0 7 2−0 1−7 2 −6
Then · ¸ · ¸
2 · 3 2 · (−1) 6 −2
c·A= = .
2 · (0) 2·5 0 10
2
1.1.3 Matrix Multiplication
Matrix multiplication only applies to conformable matrices. A and B are conformable
matrices of the number of columns in A is equal to the number of rows in B. For
example, if A is m× n and B is m × p then A and B are conformable. The mechanics
of matrix multiplication is best explained by example. Let
· ¸ · ¸
1 2 1 2 1
A = and B = .
(2×2) 3 4 (2×3) 3 4 2
Then
· ¸ · ¸
1 2 1 2 1
A · B = ·
(2×2) (2×3) 3 4 3 4 2
· ¸
1·1+2·3 1·2+2·4 1·1+2·2
=
3·1+4·3 3·2+4·4 3·1+4·2
· ¸
7 10 5
= = C
15 22 11 (2×3)
Then
· ¸ · ¸
1 2 5
A · B = ·
(2×2) (2×1) 3 4 6
· ¸
1·5+2·6
=
3·5+4·6
· ¸
17
= .
39
Then
£ ¤ 4
x0 y = 1 2 3 · 5 = 1 · 4 + 2 · 5 + 3 · 6 = 32
6
3
1.2 The Identity Matrix
The identity matrix plays a similar role as the number 1. Multiplying any number by
1 gives back that number. In matrix algebra, pre or post multiplying a matrix A by
a conformable identity matrix gives back the matrix A. To illustrate, let
· ¸
1 0
I=
0 1
and
· ¸ · ¸
a11 a12 1 0
A·I = ·
a21 a22 0 1
· ¸
a11 a12
= = A.
a21 a22
4
and
x1 Xn
0
£ ¤ .
1 x = 1 . . . 1 · .. = x1 + · · · + xn = xk .
xn k=1
Note that x0 y = y0 x.
x+y = 1 (1)
2x − y = 1 (2)
which is illustrated in Figure xxx. Equations (1) and (2) represent two straight lines
which intersect at the point x = 23 and y = 13 . This point of intersection is determined
by solving for the values of x and y such that x + y = 2x − y 1 .
1
Soving for x gives x = 2y. Substituting this value into the equation x + y = 1 gives 2y + y = 1
and solving for y gives y = 1/3. Solving for x then gives x = 2/3.
5
The two linear equations can be written in matrix form as
· ¸· ¸ · ¸
1 1 x 1
=
2 −1 y 1
or
A·z=b
where · ¸ · ¸ · ¸
1 1 x 1
A= , z= and b = .
2 −1 y 1
If there was a (2 × 2) matrix B, with elements bij , such that B · A = I, where I
is the (2 × 2) identity matrix, then we could solve for the elements in z as follows. In
the equation A · z = b, pre-multiply both sides by B to give
B·A·z = B·b
=⇒ I · z = B · b
=⇒ z = B · b
or · ¸ · ¸· ¸ · ¸
x b11 b12 1 b11 · 1 + b12 · 1
= =
y b21 b22 1 b21 · 1 + b22 · 1
If such a matrix B exists it is called the inverse of A and is denoted A−1 . In-
tuitively, the inverse matrix A−1 plays a similar role as the inverse of a number.
Suppose a is a number; e.g., a = 2. Then we know that a1 · a = a−1 a = 1. Similarly,
in matrix algebra A−1 A = I where I is the identity matrix. Next, consider solving
the equation ax = 1. By simple division we have that x = a1 x = a−1 x. Similarly, in
matrix algebra if we want to solve the system of equation Ax = b we multiply by
A−1 and get x = A−1 b.
Using B = A−1 , we may express the solution for z as
z = A−1 b.
As long as we can determine the elements in A−1 then we can solve for the values of
x and y in the vector z. Since the system of linear equations has a solution as long as
the two lines intersect, we can determine the elements in A−1 provided the two lines
are not parallel.
There are general numerical algorithms for &nding the elements of A−1 and typical
spreadsheet programs like Excel have these algorithms available. However, if A is a
(2 × 2) matrix then there is a simple formula for A−1 . Let A be a (2 × 2) matrix such
that · ¸
a11 a12
A= .
a21 a22
6
Then · ¸
−1 1 a22 −a12
A = .
a11 a22 − a21 a12 −a21 a11
By brute force matrix multiplication we can verify this formula
· ¸· ¸
−1 1 a22 −a12 a11 a12
A A =
a11 a22 − a21 a12 −a21 a11 a21 a22
· ¸
1 a22 a11 − a12 a21 a22 a12 − a12 a22
=
a11 a22 − a21 a12 −a21 a11 + a11 a21 −a21 a12 + a11 a22
· ¸
1 a22 a11 − a12 a21 0
=
a11 a22 − a21 a12 0 −a21 a12 + a11 a22
· a22 a11 −a12 a21 ¸
a11 a22 −a21 a12
0
= −a21 a12 +a11 a22
0 a11 a22 −a21 a12
· ¸
1 0
= .
0 1
Let s apply the above rule to &nd the inverse of A in our example:
· ¸ · 1 1 ¸
−1 1 −1 −1
A = = 32 −1 3 .
−1 − 2 −2 1 3 3
Notice that · ¸· ¸ · ¸
1 1
−1 3 3
1 1 1 0
A A= 2 −1 = .
3 3
2 −1 0 1
Our solution for z is then
A−1 b ¸ · ¸
z = ·
1 1
3 3
1
= 2 −1
1
· 32 ¸ 3 · ¸
3
x
= 1 =
3
y
so that x = 23 and y = 13 .
In general, if we have n linear equations in n unknown variables we may write the
system of equations as
7
which we may then express in matrix form as
a11 a12 · · · a1n x1 b1
a21 a22 · · · a2n x2 b2
.. .. .. = ..
. . . .
an1 an2 · · · ann xn bn
or
A · x = b.
(n×n) (n×1) (n×1)
x = A−1 b
2 Further Reading
Excellent treatments of portfolio theory using matrix algebra are given in Ingersol
(1987), Huang and Litzenberger (1988) and Campbell, Lo and MacKinlay (1996).
3 Problems
To be completed
References
[1] Campbell, J.Y., Lo, A.W., and MacKinlay, A.C. (1997). The Econometrics of
Financial Markets. Priceton, New Jersey: Princeton University Press.
[2] Huang, C.-F., and Litzenbeger, R.H. (1988). Foundations for Financial Eco-
nomics. New York: North-Holland.
[3] Ingersoll, J.E. (1987). Theory of Financial Decision Making. Totowa, New Jersey:
Rowman & Little&eld.
8
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