Probability and Stochastic Processes A Friendly Introduction for Electrical and Computer Engineers Quiz Solutions
Related papers
Theory and Problems of Probability, Random Variables, and Random Processes
Start of Citation[PU]McGraw-Hill Professional[/PU][DP]1997[/DP]End of Citation
Probability and stochastic processes: A friendly introduction for electrical and computer engineers
Computers & Mathematics with Applications, 1999
Environmental upgradation during the lockdown phase of COVID.pdf
Today the whole world is under the appalling shadow of COVID pandemics. The disease has posed severe adverse impacts on almost all sectors of the society encompassing health, education, recreation, food, transportation and economic profile. However, the environmental sector is noted for its improvement due to complete closures of transport systems, industries, tourisms and even pilgrims. The chapter has thrown light on the status of the pandemics in terms of infection and mortality, the common symptoms of the disease and sketched an overview of the adverse impacts of COVID-19 on different sectors, except environment. The annexure at the end of the chapter focuses on an innovative alternative livelihood befitted for the COVID and post-COVID situations.
Χρήστος Μουρατίδης -Αθανάσιος Τζουβάλης ΑΝΑΛΥΣΗ Ι Συναρτήσεις -Όρια -Συνέχεια
2016
Μαθηματικά Γ΄ Λυκείου, Συναρτήσεις-Όρια-Συνέχεια.
Probability and Stochastic Processes
A Friendly Introduction for Electrical and Computer Engineers
Second Edition
Quiz Solutions
Roy D. Yates and David J. Goodman
May 22, 2004
• The M ATLAB section quizzes at the end of each chapter use programs available for
download as the archive matcode.zip. This archive has programs of general purpose programs for solving probability problems as well as specific .m files associated
with examples or quizzes in the text. Also available is a manual probmatlab.pdf
describing the general purpose .m files in matcode.zip.
• We have made a substantial effort to check the solution to every quiz. Nevertheless,
there is a nonzero probability (in fact, a probability close to unity) that errors will be
found. If you find errors or have suggestions or comments, please send email to
ryates@winlab.rutgers.edu.
When errors are found, corrected solutions will be posted at the website.
1
Quiz Solutions – Chapter 1
Quiz 1.1
In the Venn diagrams for parts (a)-(g) below, the shaded area represents the indicated
set.
O
M
T
T
(2) M ∪ O
O
T
(4) R ∪ M
(3) M ∩ O
O
M
T
(4) R ∩ M
O
M
T
(1) R = T c
M
O
M
O
M
T
(6) T c − M
Quiz 1.2
(1) A1 = {vvv, vvd, vdv, vdd}
(2) B1 = {dvv, dvd, ddv, ddd}
(3) A2 = {vvv, vvd, dvv, dvd}
(4) B2 = {vdv, vdd, ddv, ddd}
(5) A3 = {vvv, ddd}
(6) B3 = {vdv, dvd}
(7) A4 = {vvv, vvd, vdv, dvv, vdd, dvd, ddv}
(8) B4 = {ddd, ddv, dvd, vdd}
Recall that Ai and Bi are collectively exhaustive if Ai ∪ Bi = S. Also, Ai and Bi are
mutually exclusive if Ai ∩ Bi = φ. Since we have written down each pair Ai and Bi above,
we can simply check for these properties.
The pair A1 and B1 are mutually exclusive and collectively exhaustive. The pair A2 and
B2 are mutually exclusive and collectively exhaustive. The pair A3 and B3 are mutually
exclusive but not collectively exhaustive. The pair A4 and B4 are not mutually exclusive
since dvd belongs to A4 and B4 . However, A4 and B4 are collectively exhaustive.
2
Quiz 1.3
There are exactly 50 equally likely outcomes: s51 through s100 . Each of these outcomes
has probability 0.02.
(1) P[{s79 }] = 0.02
(2) P[{s100 }] = 0.02
(3) P[A] = P[{s90 , . . . , s100 }] = 11 × 0.02 = 0.22
(4) P[F] = P[{s51 , . . . , s59 }] = 9 × 0.02 = 0.18
(5) P[T ≥ 80] = P[{s80 , . . . , s100 }] = 21 × 0.02 = 0.42
(6) P[T < 90] = P[{s51 , s52 , . . . , s89 }] = 39 × 0.02 = 0.78
(7) P[a C grade or better] = P[{s70 , . . . , s100 }] = 31 × 0.02 = 0.62
(8) P[student passes] = P[{s60 , . . . , s100 }] = 41 × 0.02 = 0.82
Quiz 1.4
We can describe this experiment by the event space consisting of the four possible
events V B, V L, D B, and DL. We represent these events in the table:
V
D
L 0.35 ?
B ?
?
In a roundabout way, the problem statement tells us how to fill in the table. In particular,
P [V ] = 0.7 = P [V L] + P [V B]
P [L] = 0.6 = P [V L] + P [DL]
(1)
(2)
Since P[V L] = 0.35, we can conclude that P[V B] = 0.35 and that P[DL] = 0.6 −
0.35 = 0.25. This allows us to fill in two more table entries:
V
D
L 0.35 0.25
B 0.35 ?
The remaining table entry is filled in by observing that the probabilities must sum to 1.
This implies P[D B] = 0.05 and the complete table is
V
D
L 0.35 0.25
B 0.35 0.05
Finding the various probabilities is now straightforward:
3
(1) P[DL] = 0.25
(2) P[D ∪ L] = P[V L] + P[DL] + P[D B] = 0.35 + 0.25 + 0.05 = 0.65.
(3) P[V B] = 0.35
(4) P[V ∪ L] = P[V ] + P[L] − P[V L] = 0.7 + 0.6 − 0.35 = 0.95
(5) P[V ∪ D] = P[S] = 1
(6) P[L B] = P[L L c ] = 0
Quiz 1.5
(1) The probability of exactly two voice calls is
P [N V = 2] = P [{vvd, vdv, dvv}] = 0.3
(1)
(2) The probability of at least one voice call is
P [N V ≥ 1] = P [{vdd, dvd, ddv, vvd, vdv, dvv, vvv}]
= 6(0.1) + 0.2 = 0.8
(2)
(3)
An easier way to get the same answer is to observe that
P [N V ≥ 1] = 1 − P [N V < 1] = 1 − P [N V = 0] = 1 − P [{ddd}] = 0.8
(4)
(3) The conditional probability of two voice calls followed by a data call given that there
were two voice calls is
1
P [{vvd} , N V = 2]
P [{vvd}]
0.1
=
(5)
P [{vvd} |N V = 2] =
=
=
P [N V = 2]
P [N V = 2]
0.3
3
(4) The conditional probability of two data calls followed by a voice call given there
were two voice calls is
P [{ddv} , N V = 2]
P [{ddv} |N V = 2] =
=0
(6)
P [N V = 2]
The joint event of the outcome ddv and exactly two voice calls has probability zero
since there is only one voice call in the outcome ddv.
(5) The conditional probability of exactly two voice calls given at least one voice call is
P [N V = 2, N V ≥ 1]
P [N V = 2]
0.3
3
=
=
=
(7)
P [N V = 2|Nv ≥ 1] =
P [N V ≥ 1]
P [N V ≥ 1]
0.8
8
(6) The conditional probability of at least one voice call given there were exactly two
voice calls is
P [N V ≥ 1, N V = 2]
P [N V = 2]
P [N V ≥ 1|N V = 2] =
=
=1
(8)
P [N V = 2]
P [N V = 2]
Given that there were two voice calls, there must have been at least one voice call.
4
Quiz 1.6
In this experiment, there are four outcomes with probabilities
P[{vv}] = (0.8)2 = 0.64
P[{vd}] = (0.8)(0.2) = 0.16
P[{dd}] = (0.2)2 = 0.04
P[{dv}] = (0.2)(0.8) = 0.16
When checking the independence of any two events A and B, it’s wise to avoid intuition
and simply check whether P[AB] = P[A]P[B]. Using the probabilities of the outcomes,
we now can test for the independence of events.
(1) First, we calculate the probability of the joint event:
P [N V = 2, N V ≥ 1] = P [N V = 2] = P [{vv}] = 0.64
(1)
Next, we observe that
P [N V ≥ 1] = P [{vd, dv, vv}] = 0.96
(2)
Finally, we make the comparison
P [N V = 2] P [N V ≥ 1] = (0.64)(0.96) = P [N V = 2, N V ≥ 1]
(3)
which shows the two events are dependent.
(2) The probability of the joint event is
P [N V ≥ 1, C1 = v] = P [{vd, vv}] = 0.80
(4)
P [N V ≥ 1] P [C1 = v] = (0.96)(0.8) = 0.768 = P [N V ≥ 1, C1 = v]
(5)
From part (a), P[N V ≥ 1] = 0.96. Further, P[C1 = v] = 0.8 so that
Hence, the events are dependent.
(3) The problem statement that the calls were independent implies that the events the
second call is a voice call, {C2 = v}, and the first call is a data call, {C1 = d} are
independent events. Just to be sure, we can do the calculations to check:
P [C1 = d, C2 = v] = P [{dv}] = 0.16
(6)
Since P[C1 = d]P[C2 = v] = (0.2)(0.8) = 0.16, we confirm that the events are
independent. Note that this shouldn’t be surprising since we used the information that
the calls were independent in the problem statement to determine the probabilities of
the outcomes.
(4) The probability of the joint event is
P [C2 = v, N V is even] = P [{vv}] = 0.64
(7)
Also, each event has probability
P [C2 = v] = P [{dv, vv}] = 0.8,
P [N V is even] = P [{dd, vv}] = 0.68 (8)
Thus, P[C2 = v]P[N V is even] = (0.8)(0.68) = 0.544. Since P[C2 = v, N V is even] =
0.544, the events are dependent.
5
Quiz 1.7
Let Fi denote the event that that the user is found on page i. The tree for the experiment
is
0.8✟✟ F2
0.8✟✟ F3
0.8✟✟ F1
✟
✟
✟
✟
✟✟
✟✟
✟✟
F3c
F1c ✟
F2c ✟
0.2
0.2
0.2
The user is found unless all three paging attempts fail. Thus the probability the user is
found is
P [F] = 1 − P F1c F2c F3c = 1 − (0.2)3 = 0.992
(1)
Quiz 1.8
(1) We can view choosing each bit in the code word as a subexperiment. Each subexperiment has two possible outcomes: 0 and 1. Thus by the fundamental principle of
counting, there are 2 × 2 × 2 × 2 = 24 = 16 possible code words.
(2) An experiment that can yield all possible code words with two zeroes is to choose
which
2 bits (out of 4 bits) will be zero. The other two bits then must be ones. There
are 42 = 6 ways to do this. Hence, there are six code words with exactly two zeroes.
For this problem, it is also possible to simply enumerate the six code words:
1100, 1010, 1001, 0101, 0110, 0011.
(3) When the first bit must be a zero, then the first subexperiment of choosing the first
bit has only one outcome. For each of the next three bits, we have two choices. In
this case, there are 1 × 2 × 2 × 2 = 8 ways of choosing a code word.
(4) For the constant ratio code, we can specify a code word by choosing M of the bits to
be ones. The other
N N − M bits will be zeroes. The number
8 of ways of choosing such
a code word is M . For N = 8 and M = 3, there are 3 = 56 code words.
Quiz 1.9
(1) In this problem, k bits received in error is the same as k failures in 100 trials. The
failure probability is ǫ = 1 − p and the success probability is 1 − ǫ = p. That is, the
probability of k bits in error and 100 − k correctly received bits is
100 k
(1)
ǫ (1 − ǫ)100−k
P Sk,100−k =
k
6
For ǫ = 0.01,
P S0,100 = (1 − ǫ)100 = (0.99)100 = 0.3660
P S1,99 = 100(0.01)(0.99)99 = 0.3700
P S2,98 = 4950(0.01)2 (0.99)9 8 = 0.1849
P S3,97 = 161, 700(0.01)3 (0.99)97 = 0.0610
(2) The probability a packet is decoded correctly is just
P [C] = P S0,100 + P S1,99 + P S2,98 + P S3,97 = 0.9819
(2)
(3)
(4)
(5)
(6)
Quiz 1.10
Since the chip works only if all n transistors work, the transistors in the chip are like
devices in series. The probability that a chip works is P[C] = pn .
The module works if either 8 chips work or 9 chips work. Let Ck denote the event that
exactly k chips work. Since transistor failures are independent of each other, chip failures
are also independent. Thus each P[Ck ] has the binomial probability
9
(1)
P [C8 ] =
(P [C])8 (1 − P [C])9−8 = 9 p 8n (1 − p n ),
8
P [C9 ] = (P [C])9 = p 9n .
(2)
The probability a memory module works is
P [M] = P [C8 ] + P [C9 ] = p 8n (9 − 8 p n )
(3)
Quiz 1.11
R=rand(1,100);
X=(R<= 0.4) ...
+ (2*(R>0.4).*(R<=0.9)) ...
+ (3*(R>0.9));
Y=hist(X,1:3)
For a M ATLAB simulation, we first generate a vector R of 100 random numbers.
Second, we generate vector X as a function of R to represent the 3 possible outcomes of a flip. That is, X(i)=1 if flip i
was heads, X(i)=2 if flip i was tails, and
X(i)=3) is flip i landed on the edge.
To see how this works, we note there are three cases:
• If R(i) <= 0.4, then X(i)=1.
• If 0.4 < R(i) and R(i)<=0.9, then X(i)=2.
• If 0.9 < R(i), then X(i)=3.
These three cases will have probabilities 0.4, 0.5 and 0.1. Lastly, we use the hist function
to count how many occurences of each possible value of X(i).
7
Quiz Solutions – Chapter 2
Quiz 2.1
The sample space, probabilities and corresponding grades for the experiment are
Outcome P[·]
BB
BC
CB
CC
G
0.36 3.0
0.24 2.5
0.24 2.5
0.16 2
Quiz 2.2
(1) To find c, we recall that the PMF must sum to 1. That is,
3
n=1
1 1
PN (n) = c 1 + +
2 3
=1
(1)
This implies c = 6/11. Now that we have found c, the remaining parts are straightforward.
(2) P[N = 1] = PN (1) = c = 6/11
(3) P[N ≥ 2] = PN (2) + PN (3) = c/2 + c/3 = 5/11
(4) P[N > 3] = ∞
n=4 PN (n) = 0
Quiz 2.3
Decoding each transmitted bit is an independent trial where we call a bit error a “success.” Each bit is in error, that is, the trial is a success, with probability p. Now we can
interpret each experiment in the generic context of independent trials.
(1) The random variable X is the number of trials up to and including the first success.
Similar to Example 2.11, X has the geometric PMF
PX (x) =
p(1 − p)x−1 x = 1, 2, . . .
0
otherwise
(1)
(2) If p = 0.1, then the probability exactly 10 bits are sent is
P [X = 10] = PX (10) = (0.1)(0.9)9 = 0.0387
8
(2)
The probability that at least 10 bits are sent is P[X ≥ 10] = ∞
x=10 PX (x). This
sum is not too hard to calculate. However, its even easier to observe that X ≥ 10 if
the first 10 bits are transmitted correctly. That is,
P [X ≥ 10] = P [first 10 bits are correct] = (1 − p)10
(3)
For p = 0.1, P[X ≥ 10] = 0.910 = 0.3487.
(3) The random variable Y is the number of successes in 100 independent trials. Just as
in Example 2.13, Y has the binomial PMF
100 y
PY (y) =
p (1 − p)100−y
(4)
y
If p = 0.01, the probability of exactly 2 errors is
100
P [Y = 2] = PY (2) =
(0.01)2 (0.99)98 = 0.1849
2
(5)
(4) The probability of no more than 2 errors is
P [Y ≤ 2] = PY (0) + PY (1) + PY (2)
(6)
100
100
99
= (0.99) + 100(0.01)(0.99) +
(0.01)2 (0.99)98
2
= 0.9207
(7)
(8)
(5) Random variable Z is the number of trials up to and including the third success. Thus
Z has the Pascal PMF (see Example 2.15)
z−1 3
PZ (z) =
p (1 − p)z−3
(9)
2
Note that PZ (z) > 0 for z = 3, 4, 5, . . ..
(6) If p = 0.25, the probability that the third error occurs on bit 12 is
11
PZ (12) =
(0.25)3 (0.75)9 = 0.0645
2
(10)
Quiz 2.4
Each of these probabilities can be read off the CDF FY (y). However, we must keep in
mind that when FY (y) has a discontinuity at y0 , FY (y) takes the upper value FY (y0+ ).
(1) P[Y < 1] = FY (1− ) = 0
9
(2) P[Y ≤ 1] = FY (1) = 0.6
(3) P[Y > 2] = 1 − P[Y ≤ 2] = 1 − FY (2) = 1 − 0.8 = 0.2
(4) P[Y ≥ 2] = 1 − P[Y < 2] = 1 − FY (2− ) = 1 − 0.6 = 0.4
(5) P[Y = 1] = P[Y ≤ 1] − P[Y < 1] = FY (1+ ) − FY (1− ) = 0.6
(6) P[Y = 3] = P[Y ≤ 3] − P[Y < 3] = FY (3+ ) − FY (3− ) = 0.8 − 0.8 = 0
Quiz 2.5
(1) With probability 0.7, a call is a voice call and C = 25. Otherwise, with probability
0.3, we have a data call and C = 40. This corresponds to the PMF
⎧
⎨ 0.7 c = 25
PC (c) = 0.3 c = 40
(1)
⎩
0
otherwise
(2) The expected value of C is
E [C] = 25(0.7) + 40(0.3) = 29.5 cents
(2)
Quiz 2.6
(1) As a function of N , the cost T is
T = 25N + 40(3 − N ) = 120 − 15N
(1)
(2) To find the PMF of T , we can draw the following tree:
N =0 •T =120
0.1✟✟
✟
✟ ✘✘N =1 •T =105
✟ 0.3
✘
✟✘✘
✘
✘
✟❳
❳
❍
❳❳❳
❍❳
❍❍0.3
❳N =2 •T =90
❍
❍
0.3 ❍
N =3 •T =75
From the tree, we can write down the PMF of T :
⎧
⎨ 0.3 t = 75, 90, 105
PT (t) = 0.1 t = 120
⎩
0
otherwise
(2)
From the PMF PT (t), the expected value of T is
E [T ] = 75PT (75) + 90PT (90) + 105PT (105) + 120PT (120)
= (75 + 90 + 105)(0.3) + 120(0.1) = 62
10
(3)
(4)
Quiz 2.7
(1) Using Definition 2.14, the expected number of applications is
E [A] =
4
a=1
a PA (a) = 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1) = 2
(2) The number of memory chips is M = g(A) where
⎧
⎨ 4 A = 1, 2
g(A) = 6 A = 3
⎩
8 A=4
(1)
(2)
(3) By Theorem 2.10, the expected number of memory chips is
E [M] =
4
a=1
g(A)PA (a) = 4(0.4) + 4(0.3) + 6(0.2) + 8(0.1) = 4.8
(3)
Since E[A] = 2, g(E[A]) = g(2) = 4. However, E[M] = 4.8 = g(E[A]). The two
quantities are different because g(A) is not of the form α A + β.
Quiz 2.8
The PMF PN (n) allows to calculate each of the desired quantities.
(1) The expected value of N is
E [N ] =
2
n=0
n PN (n) = 0(0.1) + 1(0.4) + 2(0.5) = 1.4
(1)
(2) The second moment of N is
E N
2
=
2
n=0
n 2 PN (n) = 02 (0.1) + 12 (0.4) + 22 (0.5) = 2.4
(2)
(3) The variance of N is
Var[N ] = E N
2
(4) The standard deviation is σ N =
− (E [N ])2 = 2.4 − (1.4)2 = 0.44
√
Var[N ] =
11
√
0.44 = 0.663.
(3)
Quiz 2.9
(1) From the problem statement, we learn that the conditional PMF of N given the event
I is
0.02 n = 1, 2, . . . , 50
PN |I (n) =
(1)
0
otherwise
(2) Also from the problem statement, the conditional PMF of N given the event T is
0.2 n = 1, 2, 3, 4, 5
0
otherwise
PN |T (n) =
(2)
(3) The problem statement tells us that P[T ] = 1 − P[I ] = 3/4. From Theorem 1.10
(the law of total probability), we find the PMF of N is
(4) First we find
PN (n) = PN |T (n) P [T ] + PN |I (n) P [I ]
⎧
⎨ 0.2(0.75) + 0.02(0.25) n = 1, 2, 3, 4, 5
n = 6, 7, . . . , 50
= 0(0.75) + 0.02(0.25)
⎩
0
otherwise
⎧
⎨ 0.155 n = 1, 2, 3, 4, 5
= 0.005 n = 6, 7, . . . , 50
⎩
0
otherwise
P [N ≤ 10] =
10
n=1
PN (n) = (0.155)(5) + (0.005)(5) = 0.80
(3)
(4)
(5)
(6)
By Theorem 2.17, the conditional PMF of N given N ≤ 10 is
PN |N ≤10 (n) =
PN (n)
P[N ≤10]
0
n ≤ 10
otherwise
(7)
⎧
⎨ 0.155/0.8 n = 1, 2, 3, 4, 5
= 0.005/0.8 n = 6, 7, 8, 9, 10
⎩
0
otherwise
⎧
⎨ 0.19375 n = 1, 2, 3, 4, 5
= 0.00625 n = 6, 7, 8, 9, 10
⎩
0
otherwise
(8)
(9)
(5) Once we have the conditional PMF, calculating conditional expectations is easy.
E [N |N ≤ 10] =
(10)
n PN |N ≤10 (n)
=
n
5
n=1
n(0.19375) +
= 3.15625
12
10
n(0.00625)
(11)
n=6
(12)
10
10
8
8
6
6
4
4
2
2
0
0
50
0
100
(a) samplemean(100)
0
500
1000
(b) samplemean(1000)
Figure 1: Two examples of the output of samplemean(k)
(6) To find the conditional variance, we first find the conditional second moment
2
E N |N ≤ 10 =
n 2 PN |N ≤10 (n)
(13)
2
(14)
=
n
5
n=1
n (0.19375) +
10
n 2 (0.00625)
n=6
= 55(0.19375) + 330(0.00625) = 12.71875
(15)
The conditional variance is
Var[N |N ≤ 10] = E N 2 |N ≤ 10 − (E [N |N ≤ 10])2
= 12.71875 − (3.15625)2 = 2.75684
(16)
(17)
Quiz 2.10
The function samplemean(k) generates and plots five m n sequences for n = 1, 2, . . . , k.
The ith column M(:,i) of M holds a sequence m 1 , m 2 , . . . , m k .
function M=samplemean(k);
K=(1:k)’;
M=zeros(k,5);
for i=1:5,
X=duniformrv(0,10,k);
M(:,i)=cumsum(X)./K;
end;
plot(K,M);
Examples of the function calls (a) samplemean(100) and (b) samplemean(1000)
are shown in Figure 1. Each time samplemean(k) is called produces a random output.
What is observed in these figures is that for small n, m n is fairly random but as n gets
13
large, m n gets close to E[X ] = 5. Although each sequence m 1 , m 2 , . . . that we generate is
random, the sequences always converges to E[X ]. This random convergence is analyzed
in Chapter 7.
14
Quiz Solutions – Chapter 3
Quiz 3.1
The CDF of Y is
FY(y)
1
⎧
y<0
⎨ 0
y/4 0 ≤ y ≤ 4
FY (y) =
⎩
1
y>4
0.5
0
0
2
y
4
(1)
From the CDF FY (y), we can calculate the probabilities:
(1) P[Y ≤ −1] = FY (−1) = 0
(2) P[Y ≤ 1] = FY (1) = 1/4
(3) P[2 < Y ≤ 3] = FY (3) − FY (2) = 3/4 − 2/4 = 1/4
(4) P[Y > 1.5] = 1 − P[Y ≤ 1.5] = 1 − FY (1.5) = 1 − (1.5)/4 = 5/8
Quiz 3.2
(1) First we willfind the constant c and then we will sketch the PDF. To find c, we use
∞
the fact that −∞ f X (x) d x = 1. We will evaluate this integral using integration by
parts:
∞
∞
f X (x) d x =
cxe−x/2 d x
(1)
−∞
0
∞ ∞
−x/2
2ce−x/2 d x
(2)
= −2cxe
+
0
0
=0
∞
−x/2
= −4ce
(3)
= 4c
0
Thus c = 1/4 and X has the Erlang (n = 2, λ = 1/2) PDF
fX(x)
0.2
0.1
f X (x) =
0
0
5
10
15
x
15
(x/4)e−x/2 x ≥ 0
0
otherwise
(4)
(2) To find the CDF FX (x), we first note X is a nonnegative random variable so that
FX (x) = 0 for all x < 0. For x ≥ 0,
x
x
y −y/2
e
FX (x) =
dy
(5)
f X (y) dy =
0
0 4
x
1
y −y/2 x
(6)
− e−y/2 dy
=− e
−
2
2
0
0
x
= 1 − e−x/2 − e−x/2
(7)
2
The complete expression for the CDF is
FX(x)
1
0.5
1−
0
FX (x) =
0
0
5
10
15
x
2
+ 1 e−x/2 x ≥ 0
otherwise
(8)
x
(3) From the CDF FX (x),
P [0 ≤ X ≤ 4] = FX (4) − FX (0) = 1 − 3e−2 .
(9)
P [−2 ≤ X ≤ 2] = FX (2) − FX (−2) = 1 − 3e−1 .
(10)
(4) Similarly,
Quiz 3.3
The PDF of Y is
fY(y)
3
2
0
−2
3y 2 /2 −1 ≤ y ≤ 1,
0
otherwise.
f Y (y) =
1
0
y
(1)
2
(1) The expected value of Y is
∞
E [Y ] =
y f Y (y) dy =
−∞
1
−1
1
(3/2)y 3 dy = (3/8)y 4 = 0.
−1
(2)
Note that the above calculation wasn’t really necessary because E[Y ] = 0 whenever
the PDF f Y (y) is an even function (i.e., f Y (y) = f Y (−y)).
(2) The second moment of Y is
∞
2
2
E Y =
y f Y (y) dy =
−∞
1
−1
16
1
(3/2)y 4 dy = (3/10)y 5 = 3/5.
−1
(3)
(3) The variance of Y is
Var[Y ] = E Y
(4) The standard deviation of Y is σY =
2
√
− (E [Y ])2 = 3/5.
Var[Y ] =
√
(4)
3/5.
Quiz 3.4
(1) When X is an exponential (λ) random variable, E[X ] = 1/λ and Var[X ] = 1/λ2 .
Since E[X ] = 3 and Var[X ] = 9, we must have λ = 1/3. The PDF of X is
f X (x) =
(1/3)e−x/3 x ≥ 0,
0
otherwise.
(1)
(2) We know X is a uniform (a, b) random variable. To find a and b, we apply Theorem 3.6 to write
E [X ] =
a+b
=3
2
Var[X ] =
This implies
a + b = 6,
The only valid solution with a < b is
√
a = 3 − 3 3,
(b − a)2
= 9.
12
(2)
√
b − a = ±6 3.
(3)
√
b = 3 + 3 3.
(4)
The complete expression for the PDF of X is
√
√
√
1/(6 3) 3 − 3 3 ≤ x < 3 + 3 3,
f X (x) =
0
otherwise.
(5)
Quiz 3.5
Each of the requested probabilities can be calculated using (z) function and Table 3.1
or Q(z) and Table 3.2. We start with the sketches.
fY(y)
(1) The PDFs of X and Y are shown below. The fact that Y has twice the standard
deviation of X is reflected in the greater spread of f Y (y). However, it is important
to remember that as the standard deviation increases, the peak value of the Gaussian
PDF goes down.
0.4
← fX(x)
fX(x)
0.2
← fY(y)
0
−5
0
x
17
5
y
(2) Since X is Gaussian (0, 1),
P [−1 < X ≤ 1] = FX (1) − FX (−1)
= (1) − (−1) = 2(1) − 1 = 0.6826.
(1)
(2)
(3) Since Y is Gaussian (0, 2),
P [−1 < Y ≤ 1] = FY (1) − FY (−1)
−1
1
1
− 1 = 0.383.
−
= 2
=
σY
σY
2
(3)
(4)
(4) Again, since X is Gaussian (0, 1), P[X > 3.5] = Q(3.5) = 2.33 × 10−4 .
(5) Since Y is Gaussian (0, 2), P[Y > 3.5] = Q( 3.5
2 ) = Q(1.75) = 1 − (1.75) =
0.0401.
Quiz 3.6
The CDF of X is
FX(x)
1
⎧
x < −1,
⎨ 0
FX (x) = (x + 1)/4 −1 ≤ x < 1,
⎩
1
x ≥ 1.
0.5
0
−2
0
x
2
(1)
The following probabilities can be read directly from the CDF:
(1) P[X ≤ 1] = FX (1) = 1.
(2) P[X < 1] = FX (1− ) = 1/2.
(3) P[X = 1] = FX (1+ ) − FX (1− ) = 1 − 1/2 = 1/2.
(4) We find the PDF f Y (y) by taking the derivative of FY (y). The resulting PDF is
0.5
fX(x)
0.5
0
−2
0
x
2
⎧
−1 ≤ x < 1,
⎨ 1/4
f X (x) = (1/2)δ(x − 1) x = 1,
⎩
0
otherwise.
Quiz 3.7
18
(2)
(1) Since X is always nonnegative, FX (x) = 0 for x < 0. Also, FX (x) = 1 for x ≥ 2
since its always true that x ≤ 2. Lastly, for 0 ≤ x ≤ 2,
x
x
FX (x) =
f X (y) dy =
(1 − y/2) dy = x − x 2 /4.
(1)
−∞
0
The complete CDF of X is
FX(x)
1
⎧
x < 0,
⎨ 0
2
FX (x) =
x − x /4 0 ≤ x ≤ 2,
⎩
1
x > 2.
0.5
0
−1
0
1
x
2
3
(2)
(2) The probability that Y = 1 is
P [Y = 1] = P [X ≥ 1] = 1 − FX (1) = 1 − 3/4 = 1/4.
(3)
(3) Since X is nonnegative, Y is also nonnegative. Thus FY (y) = 0 for y < 0. Also,
because Y ≤ 1, FY (y) = 1 for all y ≥ 1. Finally, for 0 < y < 1,
FY (y) = P [Y ≤ y] = P [X ≤ y] = FX (y) .
(4)
Using the CDF FX (x), the complete expression for the CDF of Y is
FY(y)
1
⎧
y < 0,
⎨ 0
2
FY (y) =
y − y /4 0 ≤ y < 1,
⎩
1
y ≥ 1.
0.5
0
−1
0
1
y
2
3
(5)
As expected, we see that the jump in FY (y) at y = 1 is exactly equal to P[Y = 1].
(4) By taking the derivative of FY (y), we obtain the PDF f Y (y). Note that when y < 0
or y > 1, the PDF is zero.
fY(y)
1.5
0.25
1
0.5
0
−1
0
1
y
2
3
f Y (y) =
1 − y/2 + (1/4)δ(y − 1) 0 ≤ y ≤ 1
0
otherwise
Quiz 3.8
(1) P[Y ≤ 6] =
6
−∞ f Y (y) dy
=
6
0
(1/10) dy = 0.6 .
19
(6)
(2) From Definition 3.15, the conditional PDF of Y given Y ≤ 6 is
f Y (y)
1/6 0 ≤ y ≤ 6,
P[Y ≤6] y ≤ 6,
=
f Y |Y ≤6 (y) =
0
otherwise.
0
otherwise,
(1)
(3) The probability Y > 8 is
P [Y > 8] =
8
10
1
dy = 0.2 .
10
(4) From Definition 3.15, the conditional PDF of Y given Y > 8 is
f Y (y)
1/2 8 < y ≤ 10,
P[Y >8] y > 8,
f Y |Y >8 (y) =
=
0
otherwise.
0
otherwise,
(2)
(3)
(5) From the conditional PDF f Y |Y ≤6 (y), we can calculate the conditional expectation
6
∞
y
y f Y |Y ≤6 (y) dy =
E [Y |Y ≤ 6] =
dy = 3.
(4)
−∞
0 6
(6) From the conditional PDF f Y |Y >8 (y), we can calculate the conditional expectation
∞
10
y
E [Y |Y > 8] =
y f Y |Y >8 (y) dy =
dy = 9.
(5)
2
−∞
8
Quiz 3.9
A natural way to produce random variables with PDF f T |T >2 (t) is to generate samples
of T with PDF f T (t) and then to discard those samples which fail to satisfy the condition
T > 2. Here is a M ATLAB function that uses this method:
function t=t2rv(m)
i=0;lambda=1/3;
t=zeros(m,1);
while (i<m),
x=exponentialrv(lambda,1);
if (x>2)
t(i+1)=x;
i=i+1;
end
end
A second method exploits the fact that if T is an exponential (λ) random variable, then
T ′ = T + 2 has PDF f T ′ (t) = f T |T >2 (t). In this case the command
t=2.0+exponentialrv(1/3,m)
generates the vector t.
20
Quiz Solutions – Chapter 4
Quiz 4.1
Each value of the joint CDF can be found by considering the corresponding probability.
(1) FX,Y (−∞, 2) = P[X ≤ −∞, Y ≤ 2] ≤ P[X ≤ −∞] = 0 since X cannot take on
the value −∞.
(2) FX,Y (∞, ∞) = P[X ≤ ∞, Y ≤ ∞] = 1. This result is given in Theorem 4.1.
(3) FX,Y (∞, y) = P[X ≤ ∞, Y ≤ y] = P[Y ≤ y] = FY (y).
(4) FX,Y (∞, −∞) = P[X ≤ ∞, Y ≤ −∞] = 0 since Y cannot take on the value −∞.
Quiz 4.2
From the joint PMF of Q and G given in the table, we can calculate the requested
probabilities by summing the PMF over those values of Q and G that correspond to the
event.
(1) The probability that Q = 0 is
P [Q = 0] = PQ,G (0, 0) + PQ,G (0, 1) + PQ,G (0, 2) + PQ,G (0, 3)
= 0.06 + 0.18 + 0.24 + 0.12 = 0.6
(1)
(2)
(2) The probability that Q = G is
P [Q = G] = PQ,G (0, 0) + PQ,G (1, 1) = 0.18
(3)
(3) The probability that G > 1 is
P [G > 1] =
1
3
PQ,G (q, g)
(4)
g=2 q=0
= 0.24 + 0.16 + 0.12 + 0.08 = 0.6
(5)
(4) The probability that G > Q is
P [G > Q] =
3
1
PQ,G (q, g)
(6)
q=0 g=q+1
= 0.18 + 0.24 + 0.12 + 0.16 + 0.08 = 0.78
21
(7)
Quiz 4.3
By Theorem 4.3, the marginal PMF of H is
PH (h) =
PH,B (h, b)
(1)
b=0,2,4
For each value of h, this corresponds to calculating the row sum across the table of the joint
PMF. Similarly, the marginal PMF of B is
PB (b) =
1
PH,B (h, b)
(2)
h=−1
For each value of b, this corresponds to the column sum down the table of the joint PMF.
The easiest way to calculate these marginal PMFs is to simply sum each row and column:
PH,B (h, b) b = 0 b = 2 b = 4 PH (h)
h = −1
0
0.4
0.2
0.6
0.1
0
0.1
0.2
h=0
h=1
0.1
0.1
0
0.2
PB (b)
0.2
0.5
0.3
(3)
Quiz 4.4
∞ ∞
To find the constant c, we apply −∞ −∞ f X,Y (x, y) d x d y = 1. Specifically,
2 1
∞ ∞
f X,Y (x, y) d x d y =
cx y d x dy
0
−∞ −∞
0
2
1
2
=c
y x /2 dy
0
0
2
2
= (c/2)
y dy = (c/4)y 2 = c
0
0
Thus c = 1. To calculate P[A], we write
P [A] =
f X,Y (x, y) d x d y
(1)
(2)
(3)
(4)
A
To integrate over A, we convert to polar coordinates using the substitutions x = r cos θ ,
y = r sin θ and d x d y = r dr dθ , yielding
Y
P [A] =
2
1
=
A
X
1
=
π/2 1
0
0
1
π/2
r 3 dr
0
r 2 sin θ cos θ r dr dθ
sin θ cos θ dθ
0
⎛
⎞
1 sin2 θ π/2
⎠
= 1/8
r 4 /4 ⎝
0
2
0
22
(5)
(6)
(7)
Quiz 4.5
By Theorem 4.8, the marginal PDF of X is
∞
f X (x) =
f X,Y (x, y) dy
(1)
−∞
For x < 0 or x > 1, f X (x) = 0. For 0 ≤ x ≤ 1,
y=1 6
6x + 2
6
6 1
x y + y 3 /3
f X (x) =
= (x + 1/3) =
(x + y 2 ) dy =
y=0
5 0
5
5
5
(2)
The complete expression for the PDf of X is
f X (x) =
(6x + 2)/5 0 ≤ x ≤ 1
0
otherwise
(3)
By the same method we obtain the marginal PDF for Y . For 0 ≤ y ≤ 1,
∞
f Y (y) =
f X,Y (x, y) dy
=
−∞
6 1
5
0
(x + y 2 ) d x =
x=1 6
3 + 6y 2
6 2
x /2 + x y 2
= (1/2 + y 2 ) =
x=0
5
5
5
(4)
(5)
Since f Y (y) = 0 for y < 0 or y > 1, the complete expression for the PDF of Y is
f Y (y) =
(3 + 6y 2 )/5 0 ≤ y ≤ 1
0
otherwise
(6)
Quiz 4.6
(A) The time required for the transfer is T = L/B. For each pair of values of L and B,
we can calculate the time T needed for the transfer. We can write these down on the
table for the joint PMF of L and B as follows:
PL ,B (l, b)
l = 518, 400
l = 2, 592, 000
l = 7, 776, 000
b = 14, 400
0.20 (T =36)
0.05 (T =180)
0.00 (T =540)
b = 21, 600
0.10 (T =24)
0.10 (T =120)
0.10 (T =360)
b = 28, 800
0.05 (T =18)
0.20 (T =90)
0.20 (T =270)
From the table, writing down the PMF of T is straightforward.
⎧
0.05 t = 18
⎪
⎪
⎪
⎪
0.1 t = 24
⎪
⎪
⎪
⎪
0.2 t = 36, 90
⎪
⎪
⎨
0.1 t = 120
PT (t) =
0.05 t = 180
⎪
⎪
⎪
⎪
0.2 t = 270
⎪
⎪
⎪
⎪
0.1 t = 360
⎪
⎪
⎩
0
otherwise
23
(1)
(B) First, we observe that since 0 ≤ X ≤ 1 and 0 ≤ Y ≤ 1, W = X Y satisfies
0 ≤ W ≤ 1. Thus f W (0) = 0 and f W (1) = 1. For 0 < w < 1, we calculate the
CDF FW (w) = P[W ≤ w]. As shown below, integrating over the region W ≤ w
is fairly complex. The calculus is simpler if we integrate over the region X Y > w.
Specifically,
FW (w) = 1 − P [X Y > w]
1 1
=1−
dy dx
Y
XY > w
1
w
(1 − w/x) d x
x=1
= 1 − x − w ln x|x=w
=1−
XY = w
X
w
w w/x
1
1
(2)
(3)
(4)
w
= 1 − (1 − w + w ln w) = w − w ln w
The complete expression for the CDF is
⎧
w<0
⎨ 0
FW (w) = w − w ln w 0 ≤ w ≤ 1
⎩
1
w>1
By taking the derivative of the CDF, we find the PDF is
⎧
0
w<0
d FW (w) ⎨
f W (w) =
= − ln w 0 ≤ w ≤ 1
⎩
dw
0
w>1
(5)
(6)
(7)
(8)
Quiz 4.7
(A) It is helpful to first make a table that includes the marginal PMFs.
PL ,T (l, t)
l=1
l=2
l=3
PT (t)
t = 40
0.15
0.3
0.15
0.6
t = 60
0.1
0.2
0.1
0.4
PL (l)
0.25
0.5
0.25
(1) The expected value of L is
E [L] = 1(0.25) + 2(0.5) + 3(0.25) = 2.
Since the second moment of L is
E L 2 = 12 (0.25) + 22 (0.5) + 32 (0.25) = 4.5,
(1)
(2)
the variance of L is
Var [L] = E L 2 − (E [L])2 = 0.5.
24
(3)
(2) The expected value of T is
E [T ] = 40(0.6) + 60(0.4) = 48.
The second moment of T is
E T 2 = 402 (0.6) + 602 (0.4) = 2400.
Thus
Var[T ] = E T 2 − (E [T ])2 = 2400 − 482 = 96.
(4)
(5)
(6)
(3) The correlation is
E [L T ] =
3
(7)
lt PL T (lt)
t=40,60 l=1
= 1(40)(0.15) + 2(40)(0.3) + 3(40)(0.15)
+ 1(60)(0.1) + 2(60)(0.2) + 3(60)(0.1)
= 96
(8)
(9)
(10)
(4) From Theorem 4.16(a), the covariance of L and T is
Cov [L , T ] = E [L T ] − E [L] E [T ] = 96 − 2(48) = 0
(11)
(5) Since Cov[L , T ] = 0, the correlation coefficient is ρ L ,T = 0.
(B) As in the discrete case, the calculations become easier if we first calculate the marginal
PDFs f X (x) and f Y (y). For 0 ≤ x ≤ 1,
f X (x) =
∞
−∞
2
f X,Y (x, y) dy =
2
f X,Y (x, y) d x =
0
Similarly, for 0 ≤ y ≤ 2,
f Y (y) =
∞
−∞
0
1 2 y=2
= 2x
x y dy = x y
2
y=0
y
1 2 x=1
=
x y d x = x y
2
2
x=0
(12)
(13)
The complete expressions for the marginal PDFs are
f X (x) =
2x 0 ≤ x ≤ 1
0 otherwise
f Y (y) =
y/2 0 ≤ y ≤ 2
0
otherwise
(14)
From the marginal PDFs, it is straightforward to calculate the various expectations.
25
(1) The first and second moments of X are
∞
E [X ] =
x f X (x) d x =
E X
2
=
−∞
∞
−∞
1
0
2
x f X (x) d x =
0
2x 2 d x =
1
2
3
2x 3 d x =
(15)
1
2
(16)
(17)
The variance of X is Var[X ] = E[X 2 ] − (E[X ])2 = 1/18.
(2) The first and second moments of Y are
∞
y f Y (y) dy =
E [Y ] =
2
1 2
4
y dy =
3
−∞
0 2
2
∞
1 3
y 2 f Y (y) dy =
E Y2 =
y dy = 2
−∞
0 2
(18)
(19)
The variance of Y is Var[Y ] = E[Y 2 ] − (E[Y ])2 = 2 − 16/9 = 2/9.
(3) The correlation of X and Y is
∞
E [X Y ] =
=
∞
x y f X,Y (x, y) d x, dy
−∞ −∞
1 2
2 2
0
0
(4) The covariance of X and Y is
1 2
x 3 y 3
8
x y d x, dy =
=
3 0 3 0 9
2
4
8
Cov [X, Y ] = E [X Y ] − E [X ] E [Y ] = −
= 0.
9
3
3
(20)
(21)
(22)
(5) Since Cov[X, Y ] = 0, the correlation coefficient is ρ X,Y = 0.
Quiz 4.8
(A) Since the event V > 80 occurs only for the pairs (L , T ) = (2, 60), (L , T ) = (3, 40)
and (L , T ) = (3, 60),
P [A] = P [V > 80] = PL ,T (2, 60) + PL ,T (3, 40) + PL ,T (3, 60) = 0.45
(1)
By Definition 4.9,
PL ,T |A (l, t) =
26
PL ,T (l,t)
P[A]
0
lt > 80
otherwise
(2)
We can represent this conditional PMF in the following table:
PL ,T |A (l, t) t = 40 t = 60
l=1
0
0
l=2
0
4/9
1/3
2/9
l=3
The conditional expectation of V can be found from the conditional PMF.
E [V |A] =
lt PL ,T |A (l, t)
(3)
t
l
1
2
1
4
= (2 · 60) + (3 · 40) + (3 · 60) = 133
9
3
9
3
(4)
For the conditional variance Var[V |A], we first find the conditional second moment
(5)
E V 2 |A =
(lt)2 PL ,T |A (l, t)
t
l
4
1
2
= (2 · 60)2 + (3 · 40)2 + (3 · 60)2 = 18, 400
9
3
9
(6)
It follows that
2
Var [V |A] = E V 2 |A − (E [V |A])2 = 622
9
(7)
(B) For continuous random variables X and Y , we first calculate the probability of the
conditioning event.
60 3
xy
dx dy
(8)
P [B] =
f X,Y (x, y) d x d y =
B
80/y 4000
40
3
60
y
x 2
=
dy
(9)
2 80/y
40 4000
60
9 3200
y
− 2
dy
(10)
=
y
40 4000 2
9 4 3
= − ln ≈ 0.801
(11)
8 5 2
The conditional PDF of X and Y is
f X,Y |B (x, y) =
=
f X,Y (x, y) /P [B] (x, y) ∈ B
0
otherwise
(12)
K x y 40 ≤ y ≤ 60, 80/y ≤ x ≤ 3
0
otherwise
(13)
27
where K = (4000P[B])−1 . The conditional expectation of W given event B is
∞ ∞
E [W |B] =
(14)
x y f X,Y |B (x, y) d x d y
=
−∞ −∞
60 3
40
K x 2 y2 d x d y
80/y
60
= (K /3)
= (K /3)
x=3
y2 x 3
40
60
x=80/y
2
(15)
dy
(16)
3
27y − 80 /y dy
60
3
3
= (K /3) 9y − 80 ln y ≈ 120.78
40
40
The conditional second moment of K given B is
∞ ∞
2
E W |B =
(x y)2 f X,Y |B (x, y) d x d y
=
−∞ −∞
60 3
40
K x 3 y3 d x d y
80/y
60
= (K /4)
x=3
y3 x 4
40
60
x=80/y
(18)
(19)
(20)
dy
(21)
81y 3 − 804 /y dy
40
60
4
4
= (K /4) (81/4)y − 80 ln y ≈ 16, 116.10
= (K /4)
(17)
40
It follows that the conditional variance of W given B is
Var [W |B] = E W 2 |B − (E [W |B])2 ≈ 1528.30
(22)
(23)
(24)
Quiz 4.9
(A) (1) The joint PMF of A and B can be found from the marginal and conditional
PMFs via PA,B (a, b) = PB|A (b|a)PA (a). Incorporating the information from
the given conditional PMFs can be confusing, however. Consequently, we can
note that A has range S A = {0, 2} and B has range S B = {0, 1}. A table of the
joint PMF will include all four possible combinations of A and B. The general
form of the table is
b=0
b=1
PA,B (a, b)
a=0
PB|A (0|0)PA (0) PB|A (1|0)PA (0)
PB|A (0|2)PA (2) PB|A (1|2)PA (2)
a=2
28
Substituting values from PB|A (b|a) and PA (a), we have
b=0
b=1
PA,B (a, b)
a=0
(0.8)(0.4) (0.2)(0.4)
(0.5)(0.6) (0.5)(0.6)
a=2
or
PA,B (a, b) b = 0 b = 1
a=0
0.32 0.08
0.3
0.3
a=2
(2) Given the conditional PMF PB|A (b|2), it is easy to calculate the conditional
expectation
E [B|A = 2] =
1
b=0
b PB|A (b|2) = (0)(0.5) + (1)(0.5) = 0.5
(1)
(3) From the joint PMF PA,B (a, b), we can calculate the the conditional PMF
⎧
0.32/0.62 a = 0
PA,B (a, 0) ⎨
PA|B (a|0) =
= 0.3/0.62 a = 2
(2)
⎩
PB (0)
0
otherwise
⎧
⎨ 16/31 a = 0
= 15/31 a = 2
(3)
⎩
0
otherwise
(4) We can calculate the conditional variance Var[A|B = 0] using the conditional
PMF PA|B (a|0). First we calculate the conditional expected value
a PA|B (a|0) = 0(16/31) + 2(15/31) = 30/31 (4)
E [A|B = 0] =
a
The conditional second moment is
E A2 |B = 0 =
a 2 PA|B (a|0) = 02 (16/31) + 22 (15/31) = 60/31 (5)
a
The conditional variance is then
960
Var[A|B = 0] = E A |B = 0 − (E [A|B = 0])2 =
961
2
(6)
(B) (1) The joint PDF of X and Y is
f X,Y (x, y) = f Y |X (y|x) f X (x) =
6y 0 ≤ y ≤ x, 0 ≤ x ≤ 1
0 otherwise
(7)
(2) From the given conditional PDF f Y |X (y|x),
f Y |X (y|1/2) =
29
8y 0 ≤ y ≤ 1/2
0 otherwise
(8)
(3) The conditional PDF of Y given X = 1/2 is f X |Y (x|1/2) = f X,Y (x, 1/2)/ f Y (1/2).
To find f Y (1/2), we integrate the joint PDF.
∞
1
f Y (1/2) =
f X,1/2 ( ) d x =
6(1/2) d x = 3/2
(9)
−∞
1/2
Thus, for 1/2 ≤ x ≤ 1,
f X |Y (x|1/2) =
f X,Y (x, 1/2)
6(1/2)
=2
=
f Y (1/2)
3/2
(10)
(4) From the pervious part, we see that given Y = 1/2, the conditional PDF of X
is uniform (1/2, 1). Thus, by the definition of the uniform (a, b) PDF,
Var [X |Y = 1/2] =
(1 − 1/2)2
1
=
12
48
(11)
Quiz 4.10
(A) (1) For random variables X and Y from Example 4.1, we observe that PY (1) =
0.09 and PX (0) = 0.01. However,
PX,Y (0, 1) = 0 = PX (0) PY (1)
(1)
Since we have found a pair x, y such that PX,Y (x, y) = PX (x)PY (y), we can
conclude that X and Y are dependent. Note that whenever PX,Y (x, y) = 0,
independence requires that either PX (x) = 0 or PY (y) = 0.
(2) For random variables Q and G from Quiz 4.2, it is not obvious whether they
are independent. Unlike X and Y in part (a), there are no obvious pairs q, g
that fail the independence requirement. In this case, we calculate the marginal
PMFs from the table of the joint PMF PQ,G (q, g) in Quiz 4.2.
PQ,G (q, g) g = 0 g = 1 g = 2 g = 3 PQ (q)
q=0
0.06 0.18 0.24 0.12
0.60
q=1
0.04 0.12 0.16 0.08
0.40
PG (g)
0.10 0.30 0.40 0.20
Careful study of the table will verify that PQ,G (q, g) = PQ (q)PG (g) for every
pair q, g. Hence Q and G are independent.
(B) (1) Since X 1 and X 2 are independent,
f X 1 ,X 2 (x1 , x2 ) = f X 1 (x1 ) f X 2 (x2 )
=
(1 − x1 /2)(1 − x2 /2) 0 ≤ x1 ≤ 2, 0 ≤ x2 ≤ 2
0
otherwise
30
(2)
(3)
(2) Let FX (x) denote the CDF of both X 1 and X 2 . The CDF of Z = max(X 1 , X 2 )
is found by observing that Z ≤ z iff X 1 ≤ z and X 2 ≤ z. That is,
P [Z ≤ z] = P [X 1 ≤ z, X 2 ≤ z]
= P [X 1 ≤ z] P [X 2 ≤ z] = [FX (z)]2
(4)
(5)
To complete the problem, we need to find the CDF of each X i . From the PDF
f X (x), the CDF is
⎧
x
x <0
⎨ 0
2
FX (x) =
f X (y) dy =
(6)
x − x /4 0 ≤ x ≤ 2
⎩
−∞
1
x >2
Thus for 0 ≤ z ≤ 2,
FZ (z) = (z − z 2 /4)2
The complete expression for the CDF of Z is
⎧
z<0
⎨ 0
2
2
FZ (z) = (z − z /4) 0 ≤ z ≤ 2
⎩
1
z>1
(7)
(8)
Quiz 4.11
This problem just requires identifying the various terms in Definition 4.17 and Theorem 4.29. Specifically, from the problem statement, we know that ρ = 1/2,
µ1 = µ X = 0,
µ2 = µY = 0,
(1)
σ1 = σ X = 1,
σ2 = σY = 1.
(2)
and that
(1) Applying these facts to Definition 4.17, we have
1
2
2
e−2(x −x y+y )/3 .
f X,Y (x, y) = √
3π 2
(3)
(2) By Theorem 4.30, the conditional expected value and standard deviation of X given
Y = y are
E [X |Y = y] = y/2
σ̃ X = σ12 (1 − ρ 2 ) = 3/4.
(4)
When Y = y = 2, we see that E[X |Y = 2] = 1 and Var[X |Y = 2] = 3/4. The
conditional PDF of X given Y = 2 is simply the Gaussian PDF
1
2
f X |Y (x|2) = √
e−2(x−1) /3 .
3π/2
31
(5)
Quiz 4.12
One straightforward method is to follow the approach of Example 4.28. Instead, we use
an alternate approach. First we observe that X has the discrete uniform (1, 4) PMF. Also,
given X = x, Y has a discrete uniform (1, x) PMF. That is,
PX (x) =
1/4 x = 1, 2, 3, 4,
0
otherwise,
PY |X (y|x) =
1/x y = 1, . . . , x
0
otherwise
(1)
Given X = x, and an independent uniform (0, 1) random variable U , we can generate a
sample value of Y with a discrete uniform (1, x) PMF via Y = ⌈xU ⌉. This observation
prompts the following program:
function xy=dtrianglerv(m)
sx=[1;2;3;4];
px=0.25*ones(4,1);
x=finiterv(sx,px,m);
y=ceil(x.*rand(m,1));
xy=[x’;y’];
32
Quiz Solutions – Chapter 5
Quiz 5.1
We find P[C] by integrating the joint PDF over the region of interest. Specifically,
P [C] =
=4
1/2
dy2
1/2
y2
1/2
dy1
0
0
0
dy4
y4
0
1/2
y2 dy2
0
y4 dy4
0
4dy3
= 1/4.
(1)
(2)
Quiz 5.2
By definition of A, Y1 = X 1 , Y2 = X 2 − X 1 and Y3 = X 3 − X 2 . Since 0 < X 1 < X 2 <
X 3 , each Yi must be a strictly positive integer. Thus, for y1 , y2 , y3 ∈ {1, 2, . . .},
PY (y) = P [Y1 = y1 , Y2 = y2 , Y3 = y3 ]
= P [X 1 = y1 , X 2 − X 1 = y2 , X 3 − X 2 = y3 ]
= P [X 1 = y1 , X 2 = y2 + y1 , X 3 = y3 + y2 + y1 ]
= (1 − p)3 p y1 +y2 +y3
(1)
(2)
(3)
(4)
′
By defining the vector a = 1 1 1 , the complete expression for the joint PMF of Y is
PY (y) =
′
(1 − p) p a y y1 , y2 , y3 ∈ {1, 2, . . .}
0
otherwise
(5)
Quiz 5.3
First we note that each marginal PDF is nonzero only if any subset of the xi obeys the
ordering contraints 0 ≤ x 1 ≤ x2 ≤ x3 ≤ 1. Within these constraints, we have
f X 1 ,X 2 (x1 , x2 ) =
f X 2 ,X 3 (x2 , x3 ) =
f X 1 ,X 3 (x1 , x3 ) =
∞
−∞
∞
−∞
∞
−∞
f X (x) d x3 =
f X (x) d x1 =
f X (x) d x2 =
1
6 d x3 = 6(1 − x2 ),
x
2x2
0
(1)
6 d x1 = 6x2 ,
(2)
6 d x2 = 6(x3 − x1 ).
(3)
x3
x1
In particular, we must keep in mind that f X 1 ,X 2 (x1 , x2 ) = 0 unless 0 ≤ x 1 ≤ x2 ≤ 1,
f X 2 ,X 3 (x2 , x3 ) = 0 unless 0 ≤ x2 ≤ x3 ≤ 1, and that f X 1 ,X 3 (x1 , x3 ) = 0 unless 0 ≤ x 1 ≤
33
x3 ≤ 1. The complete expressions are
f X 1 ,X 2 (x1 , x2 ) =
6(1 − x2 ) 0 ≤ x1 ≤ x2 ≤ 1
0
otherwise
(4)
f X 2 ,X 3 (x2 , x3 ) =
6x2 0 ≤ x2 ≤ x3 ≤ 1
0
otherwise
(5)
f X 1 ,X 3 (x1 , x3 ) =
6(x3 − x1 ) 0 ≤ x1 ≤ x3 ≤ 1
0
otherwise
(6)
Now we can find the marginal PDFs. When 0 ≤ xi ≤ 1 for each xi ,
f X 1 (x1 ) =
f X 2 (x2 ) =
f X 3 (x3 ) =
∞
−∞
∞
−∞
∞
−∞
f X 1 ,X 2 (x1 , x2 ) d x2 =
f X 2 ,X 3 (x2 , x3 ) d x3 =
f X 2 ,X 3 (x2 , x3 ) d x2 =
1
x1
1
6(1 − x2 ) d x2 = 3(1 − x1 )2
(7)
6x2 d x3 = 6x2 (1 − x2 )
(8)
x
2x3
0
6x2 d x2 = 3x32
(9)
The complete expressions are
f X 1 (x1 ) =
3(1 − x1 )2 0 ≤ x1 ≤ 1
0
otherwise
(10)
f X 2 (x2 ) =
6x2 (1 − x2 ) 0 ≤ x2 ≤ 1
0
otherwise
(11)
f X 3 (x3 ) =
3x32 0 ≤ x3 ≤ 1
0
otherwise
(12)
Quiz 5.4
In the PDF f Y (y), the components have dependencies as a result of the ordering constraints Y1 ≤ Y2 and Y3 ≤ Y4 . We can separate these constraints by creating the vectors
! "
! "
Y
Y1
(1)
,
W= 3 .
V=
Y4
Y2
The joint PDF of V and W is
f V,W (v, w) =
4 0 ≤ v1 ≤ v2 ≤ 1, 0 ≤ w1 ≤ w2 ≤ 1
0 otherwise
34
(2)
We must verify that V and W are independent. For 0 ≤ v1 ≤ v2 ≤ 1,
f V,W (v, w) dw1 dw2
f V (v) =
1
1
=
=
4 dw2
0
(3)
dw1
(4)
w1
1
4(1 − w1 ) dw1 = 2
0
(5)
Similarly, for 0 ≤ w1 ≤ w2 ≤ 1,
f W (w) =
=
0
1
f V,W (v, w) dv1 dv2
(6)
1
4 dv2
v1
dv1 = 2
(7)
It follows that V and W have PDFs
f V (v) =
2 0 ≤ v1 ≤ v2 ≤ 1
,
0 otherwise
f W (w) =
2 0 ≤ w 1 ≤ w2 ≤ 1
0 otherwise
(8)
It is easy to verify that f V,W (v, w) = f V (v) f W (w), confirming that V and W are independent vectors.
Quiz 5.5
(A) Referring to Theorem 1.19, each test is a subexperiment
with three
′ possible outcomes: L, A and R. In five trials, the vector X = X 1 X 2 X 3 indicating the
number of outcomes of each subexperiment has the multinomial PMF
⎧ 5
⎨ x1 ,x2 ,x3 (0.3)x1 (0.6)x2 (0.1)x3 x1 + x2 + x3 = 5;
PX (x) =
(1)
x1 , x2 , x3 ∈ {0, 1, . . . , 5}
⎩
0
otherwise
We can find the marginal PMF for each X i from the joint PMF PX (x); however it
is simpler to just start from first principles and observe that X 1 is the number of
occurrences of L in five independent tests. If we view each test as a trial with success
probability P[L] = 0.3, we see that X 1 is a binomial (n, p) = (5, 0.3) random
variable. Similarly, X 2 is a binomial (5, 0.6) random variable and X 3 is a binomial
(5, 0.1) random variable. That is, for p1 = 0.3, p2 = 0.6 and p3 = 0.1,
5 x
5−x x = 0, 1, . . . , 5
x pi (1 − pi )
(2)
PX i (x) =
0
otherwise
35
From the marginal PMFs, we see that X 1 , X 2 and X 3 are not independent. Hence, we
must use Theorem 5.6 to find the PMF of W . In particular, since X 1 + X 2 + X 3 = 5
and since each X i is non-negative, PW (0) = PW (1) = 0. Furthermore,
PW (2) = PX (1, 2, 2) + PX (2, 1, 2) + PX (2, 2, 1)
5![0.3(0.6)2 (0.1)2 + 0.32 (0.6)(0.1)2 + 0.32 (0.6)2 (0.1)]
2!2!1!
= 0.1458
=
(3)
(4)
(5)
In addition, for w = 3, w = 4, and w = 5, the event W = w occurs if and only if
one of the mutually exclusive events X 1 = w, X 2 = w, or X 3 = w occurs. Thus,
PW (3) = PX 1 (3) + PX 2 (3) + PX 3 (3) = 0.486
PW (4) = PX 1 (4) + PX 2 (4) + PX 3 (4) = 0.288
PW (5) = PX 1 (5) + PX 2 (5) + PX 3 (5) = 0.0802
(B) Since each Yi = 2X i + 4, we can apply Theorem 5.10 to write
y1 − 4 y2 − 4 y3 − 4
1
,
,
f Y (y) = 3 f X
2
2
2
2
(1/8)e−(y3 −4)/2 4 ≤ y1 ≤ y2 ≤ y3
=
0
otherwise
(6)
(7)
(8)
(9)
(10)
Note that for other matrices A, the constraints on y resulting from the constraints
0 ≤ X 1 ≤ X 2 ≤ X 3 can be much more complicated.
Quiz 5.6
∞
We start by finding the components E[X i ] = −∞ x f X i (x) d x of µ X . To do so, we use
the marginal PDFs f X i (x) found in Quiz 5.3:
E [X 1 ] =
E [X 2 ] =
E [X 3 ] =
1
0
1
0
1
0
3x(1 − x)2 d x = 1/4,
(1)
6x 2 (1 − x) d x = 1/2,
(2)
3x 3 d x = 3/4.
(3)
To find the correlation matrix R X , we need to find E[X i X j ] for all i and j. We start with
36
the second moments:
E
E
E
X 12
X 22
X 32
=
=
=
1
0
1
0
1
0
3x 2 (1 − x)2 d x = 1/10.
(4)
6x 3 (1 − x) d x = 3/10.
(5)
3x 4 d x = 3/5.
(6)
Using marginal PDFs from Quiz 5.3, the cross terms are
∞ ∞
E [X 1 X 2 ] =
x1 x2 f X 1 ,X 2 (x1 , x2 ) , d x1 d x2
−∞ −∞
1
=
(7)
1
6x1 x2 (1 − x2 ) d x2
d x1
(8)
[x1 − 3x13 + 2x14 ] d x1 = 3/20.
0
1 1
E [X 2 X 3 ] =
6x22 x3 d x3 d x2
(9)
=
0
x1
1
0
(10)
x2
1
[3x22 − 3x24 ] d x2 = 2/5
0
1 1
E [X 1 X 3 ] =
6x1 x3 (x3 − x1 ) d x3 d x1 .
(11)
(2x1 x33
(13)
=
=
=
0
0
(12)
x1
0
1
1
x3 =1
− 3x12 x32 )
x3 =x1
d x1
[2x1 − 3x12 + x14 ] d x1 = 1/5.
Summarizing the results, X has correlation matrix
⎡
⎤
1/10 3/20 1/5
R X = ⎣3/20 3/10 2/5⎦ .
1/5 2/5 3/5
(14)
(15)
Vector X has covariance matrix
C X = R X − E [X] E [X]′
⎡
⎤ ⎡ ⎤
1/10 3/20 1/5
1/4
⎣
⎦
⎣
= 3/20 3/10 2/5 − 1/2⎦ 1/4 1/2 3/4
1/5 2/5 3/5
3/4
⎤
⎡
⎤
⎡
⎤ ⎡
3 2 1
1/10 3/20 1/5
1/16 1/8 3/16
1
⎣2 4 2⎦ .
= ⎣3/20 3/10 2/5⎦ − ⎣ 1/8 1/4 3/8 ⎦ =
80
1 2 3
1/5 2/5 3/5
3/16 3/8 9/16
37
(16)
(17)
(18)
This problem shows that even for fairly simple joint PDFs, computing the covariance matrix
by calculus can be a time consuming task.
Quiz 5.7
We observe that X = AZ + b where
"
!
2 1
,
A=
1 −1
! "
2
.
b=
0
(1)
It follows from Theorem 5.18 that µ X = b and that
!
"!
" !
"
2 1
2 1
5 1
′
C X = AA =
=
.
1 −1 1 −1
1 2
(2)
Quiz 5.8
′
First, we observe that Y = AT where A = 1/31 1/31 · · · 1/31 . Since T is a
Gaussian random vector, Theorem 5.16 tells us that Y is a 1 dimensional Gaussian vector,
i.e., just a Gaussian random variable. The expected value of Y is µY = µT = 80. The
covariance matrix of Y is 1 × 1 and is just equal to Var[Y ]. Thus, by Theorem 5.16,
Var[Y ] = ACT A′ .
In julytemps.m, the first two lines generate the 31 × 31 covariance matrix CT, or
CT . Next we calculate Var[Y ]. The final
step is to use the (·) function to calculate
P[Y < T ].
function p=julytemps(T);
[D1 D2]=ndgrid((1:31),(1:31));
CT=36./(1+abs(D1-D2));
A=ones(31,1)/31.0;
CY=(A’)*CT*A;
p=phi((T-80)/sqrt(CY));
Here is the output of julytemps.m:
>> julytemps([70 75 80 85 90 95])
ans =
0.0000
0.0221
0.5000
0.9779
1.0000
1.0000
Note that P[T ≤ 70] is not actually zero and that P[T ≤ 90] is not actually 1.0000. Its
just that the M ATLAB’s short format output, invoked with the command format short,
rounds off those probabilities. Here is the long format output:
>> format long
>> julytemps([70 75 80 85 90 95])
ans =
Columns 1 through 4
0.00002844263128 0.02207383067604
Columns 5 through 6
0.99997155736872 0.99999999922010
38
0.50000000000000
0.97792616932396
The ndgrid function is a useful to way calculate many covariance matrices. However, in
this problem, C X has a special structure; the i, jth element is
CT (i, j) = c|i− j| =
36
.
1 + |i − j|
If we write out the elements of the covariance matrix, we see that
⎡
⎤
c0 c1 · · · c30
. . . .. ⎥
⎢
. ⎥
⎢ c1 c0
CT = ⎢ . .
⎥.
.
. . c1 ⎦
..
⎣ ..
c30 · · · c1 c0
(1)
(2)
This covariance matrix is known as a symmetric Toeplitz matrix. We will see in Chapters 9 and 11 that Toeplitz covariance matrices are quite common. In fact, M ATLAB has a
toeplitz function for generating them. The function julytemps2 use the toeplitz
to generate the correlation matrix CT .
function p=julytemps2(T);
c=36./(1+abs(0:30));
CT=toeplitz(c);
A=ones(31,1)/31.0;
CY=(A’)*CT*A;
p=phi((T-80)/sqrt(CY));
39
Quiz Solutions – Chapter 6
Quiz 6.1
Let K 1 , . . . , K n denote a sequence of iid random variables each with PMF
PK (k) =
1/4 k = 1, . . . , 4
0
otherwise
(1)
We can write Wn in the form of Wn = K 1 + · · · + K n . First, we note that the first two
moments of K i are
E [K i ] = (1 + 2 + 3 + 4)/4 = 2.5
E K i2 = (12 + 22 + 32 + 42 )/4 = 7.5
(2)
(3)
Thus the variance of K i is
Var[K i ] = E K i2 − (E [K i ])2 = 7.5 − (2.5)2 = 1.25
(4)
E [Wn ] = E [K 1 ] + · · · + E [K n ] = n E [K i ] = 2.5n
(5)
Since E[K i ] = 2.5, the expected value of Wn is
Since the rolls are independent, the random variables K 1 , . . . , K n are independent. Hence,
by Theorem 6.3, the variance of the sum equals the sum of the variances. That is,
Var[Wn ] = Var[K 1 ] + · · · + Var[K n ] = 1.25n
(6)
Quiz 6.2
Random variables X and Y have PDFs
f X (x) =
3e−3x x ≥ 0
0
otherwise
2e−2y y ≥ 0
0
otherwise
f Y (y) =
(1)
Since X and Y are nonnegative, W = X + Y is nonnegative. By Theorem 6.5, the PDF of
W = X + Y is
∞
w
f W (w) =
f X (w − y) f Y (y) dy = 6
e−3(w−y) e−2y dy
(2)
−∞
0
Fortunately, this integral is easy to evaluate. For w > 0,
−3w y w
−2w
−3w
e 0 =6 e
f W (w) = e
−e
Since f W (w) = 0 for w < 0, a conmplete expression for the PDF of W is
6e−2w 1 − e−w w ≥ 0,
f W (w) =
0
otherwise.
40
(3)
(4)
Quiz 6.3
The MGF of K is
4
φ K (s) = E es K ==
(0.2)esk = 0.2 1 + es + e2s + e3s + e4s
(1)
k=0
We find the moments by taking derivatives. The first derivative of φ K (s) is
d φ K (s)
= 0.2(es + 2e2s + 3e3s + 4e4s )
ds
Evaluating the derivative at s = 0 yields
d φ K (s)
E [K ] =
= 0.2(1 + 2 + 3 + 4) = 2
ds s=0
To find higher-order moments, we continue to take derivatives:
d 2 φ K (s)
2
s
2s
3s
4s
=
0.2(e
+
4e
+
9e
+
16e
)
=6
E K =
s=0
ds 2 s=0
d 3 φ K (s)
s
2s
3s
4s
3
=
0.2(e
+
8e
+
27e
+
64e
)
E K =
= 20
s=0
ds 3 s=0
4 φ (s)
d
K
4
s
2s
3s
4s
E K =
= 0.2(e + 16e + 81e + 256e )
= 70.8
s=0
ds 4 s=0
(2)
(3)
(4)
(5)
(6)
(7)
Quiz 6.4
(A) Each K i has MGF
es + e2s + · · · + ens
es (1 − ens )
=
φ K (s) = E es K i =
n
n(1 − es )
(1)
Since the sequence of K i is independent, Theorem 6.8 says the MGF of J is
φ J (s) = (φ K (s))m =
ems (1 − ens )m
n m (1 − es )m
(2)
(B) Since the set of α j X j are independent Gaussian random variables, Theorem 6.10
says that W is a Gaussian random variable. Thus to find the PDF of W , we need
only find the expected value and variance. Since the expectation of the sum equals
the sum of the expectations:
E [W ] = α E [X 1 ] + α 2 E [X 2 ] + · · · + α n E [X n ] = 0
41
(3)
Since the α j X j are independent, the variance of the sum equals the sum of the variances:
Var[W ] = α 2 Var[X 1 ] + α 4 Var[X 2 ] + · · · + α 2n Var[X n ]
= α 2 + 2(α 2 )2 + 3(α 2 )3 + · · · + n(α 2 )n
(4)
(5)
Defining q = α 2 , we can use Math Fact B.6 to write
Var[W ] =
α 2 − α 2n+2 [1 + n(1 − α 2 )]
(1 − α 2 )2
(6)
2 = Var[W ], we can write the PDF of W as
With E[W ] = 0 and σW
1
2
2
f W (w) = )
e−w /2σW
2
2π σW
(7)
Quiz 6.5
(1) From Table 6.1, each X i has MGF φ X (s) and random variable N has MGF φ N (s)
where
φ X (s) =
1
,
1−s
φ N (s) =
1 s
5e
.
1 − 45 es
(1)
From Theorem 6.12, R has MGF
φ R (s) = φ N (ln φ X (s)) =
1
5 φ X (s)
1 − 54 φ X (s)
(2)
.
(3)
Substituting the expression for φ X (s) yields
φ R (s) =
1
5
1
5
−s
(2) From Table 6.1, we see that R has the MGF of an exponential (1/5) random variable.
The corresponding PDF is
f R (r ) =
(1/5)e−r/5 r ≥ 0
0
otherwise
(4)
This quiz is an example of the general result that a geometric sum of exponential
random variables is an exponential random variable.
42
Quiz 6.6
(1) The expected access time is
E [X ] =
∞
−∞
x f X (x) d x =
12
x
d x = 6 msec
12
0
(2) The second moment of the access time is
∞
2
2
x f X (x) d x =
E X =
−∞
0
12
(1)
x2
d x = 48
12
(2)
The variance of the access time is Var[X ] = E[X 2 ] − (E[X ])2 = 48 − 36 = 12.
(3) Using X i to denote the access time of block i, we can write
A = X 1 + X 2 + · · · + X 12
(3)
Since the expectation of the sum equals the sum of the expectations,
E [A] = E [X 1 ] + · · · + E [X 12 ] = 12E [X ] = 72 msec
(4)
(4) Since the X i are independent,
Var[A] = Var[X 1 ] + · · · + Var[X 12 ] = 12 Var[X ] = 144
(5)
Hence, the standard deviation of A is σ A = 12
(5) To use the central limit theorem, we write
P [A > 75] = 1 − P [A ≤ 75]
"
!
75 − E [A]
A − E [A]
≤
=1− P
σA
σA
75 − 72
≈1−
12
= 1 − 0.5987 = 0.4013
(6)
(7)
(8)
(9)
Note that we used Table 3.1 to look up (0.25).
(6) Once again, we use the central limit theorem and Table 3.1 to estimate
!
"
48 − E [A]
A − E [A]
P [A < 48] = P
<
σA
σA
48 − 72
≈
12
= 1 − (2) = 1 − 0.9773 = 0.0227
43
(10)
(11)
(12)
Quiz 6.7
Random variable K n has a binomial distribution for n trials and success probability
P[V ] = 3/4.
(1) The expected number of voice calls out of 48 calls is E[K 48 ] = 48P[V ] = 36.
(2) The variance of K 48 is
Var[K 48 ] = 48P [V ] (1 − P [V ]) = 48(3/4)(1/4) = 9
(1)
Thus K 48 has standard deviation σ K 48 = 3.
(3) Using the ordinary central limit theorem and Table 3.1 yields
42 − 36
30 − 36
P [30 ≤ K 48 ≤ 42] ≈
−
= (2) − (−2)
3
3
(2)
Recalling that (−x) = 1 − (x), we have
P [30 ≤ K 48 ≤ 42] ≈ 2(2) − 1 = 0.9545
(3)
(4) Since K 48 is a discrete random variable, we can use the De Moivre-Laplace approximation to estimate
42 + 0.5 − 36
30 − 0.5 − 36
P [30 ≤ K 48 ≤ 42] ≈
−
(4)
3
3
= 2(2.16666) − 1 = 0.9687
(5)
Quiz 6.8
The train interarrival times X 1 , X 2 , X 3 are iid exponential (λ) random variables. The
arrival time of the third train is
W = X 1 + X 2 + X 3.
(1)
In Theorem 6.11, we found that the sum of three iid exponential (λ) random variables is an
Erlang (n = 3, λ) random variable. From Appendix A, we find that W has expected value
and variance
E [W ] = 3/λ = 6
Var[W ] = 3/λ2 = 12
(2)
(1) By the Central Limit Theorem,
!
"
√
W −6
20 − 6
P [W > 20] = P √
> √
≈ Q(7/ 3) = 2.66 × 10−5
12
12
44
(3)
(2) To use the Chernoff bound, we note that the MGF of W is
3
λ
1
=
φW (s) =
λ−s
(1 − 2s)3
(4)
The Chernoff bound states that
P [W > 20] ≤ min e−20s φ X (s) = min
s≥0
s≥0
e−20s
(1 − 2s)3
(5)
To minimize h(s) = e−20s /(1 − 2s)3 , we set the derivative of h(s) to zero:
−20(1 − 2s)3 e−20s + 6e−20s (1 − 2s)2
d h(s)
=
=0
ds
(1 − 2s)6
(6)
This implies 20(1 − 2s) = 6 or s = 7/20. Applying s = 7/20 into the Chernoff
bound yields
e−20s
= (10/3)3 e−7 = 0.0338
(7)
P [W > 20] ≤
(1 − 2s)3 s=7/20
(3) Theorem 3.11 says that for any w > 0, the CDF of the Erlang (λ, 3) random variable
W satisfies
2
(λw)k e−λw
FW (w) = 1 −
(8)
k!
k=0
Equivalently, for λ = 1/2 and w = 20,
P [W > 20] = 1 − FW (20)
10 102
−10
1+
=e
= 61e−10 = 0.0028
+
1!
2!
(9)
(10)
Although the Chernoff bound is relatively weak in that it overestimates the probability by roughly a factor of 12, it is a valid bound. By contrast, the Central Limit
Theorem approximation grossly underestimates the true probability.
Quiz 6.9
One solution to this problem is to follow the approach of Example 6.19:
%unifbinom100.m
sx=0:100;sy=0:100;
px=binomialpmf(100,0.5,sx); py=duniformpmf(0,100,sy);
[SX,SY]=ndgrid(sx,sy); [PX,PY]=ndgrid(px,py);
SW=SX+SY; PW=PX.*PY;
sw=unique(SW); pw=finitepmf(SW,PW,sw);
pmfplot(sw,pw,’\itw’,’\itP_W(w)’);
A graph of the PMF PW (w) appears in Figure 2 With some thought, it should be apparent
that the finitepmf function is implementing the convolution of the two PMFs.
45
0.01
PW(w)
0.008
0.006
0.004
0.002
0
0
20
40
60
80
100
w
120
140
160
180
200
Figure 2: From Quiz 6.9, the PMF PW (w) of the independent sum of a binomial (100, 0.5)
random variable and a discrete uniform (0, 100) random variable.
46
Quiz Solutions – Chapter 7
Quiz 7.1
An exponential random variable with expected value 1 also has variance 1. By Theorem 7.1, Mn (X ) has variance Var[Mn (X )] = 1/n. Hence, we need n = 100 samples.
Quiz 7.2
The arrival time of the third elevator is W = X 1 + X 2 + X 3 . Since each X i is uniform
(0, 30),
(30 − 0)2
= 75.
(1)
Var [X i ] =
E [X i ] = 15,
12
Thus E[W ] = 3E[X i ] = 45, and Var[W ] = 3 Var[X i ] = 225.
(1) By the Markov inequality,
P [W > 75] ≤
E [W ]
45
3
=
=
75
75
5
(2)
(2) By the Chebyshev inequality,
P [W > 75] = P [W − E [W ] > 30]
≤ P [|W − E [W ]| > 30] ≤
(3)
225
Var [W ]
1
=
=
2
900
4
30
(4)
Quiz 7.3
Define the random variable W = (X − µ X )2 . Observe that V100 (X ) = M100 (W ). By
Theorem 7.6, the mean square error is
Var[W ]
E (M100 (W ) − µW )2 =
100
Observe that µ X = 0 so that W = X 2 . Thus,
2
µW = E X =
E W2 = E X4 =
1
−1
1
−1
(1)
x 2 f X (x) d x = 1/3
(2)
x 4 f X (x) d x = 1/5
(3)
Therefore Var[W ] = E[W 2 ] − µ2W = 1/5 − (1/3)2 = 4/45 and the mean square error is
4/4500 = 0.000889.
47
Quiz 7.4
Assuming the number n of samples is large, we can use a Gaussian approximation for
Mn (X ). SinceE[X ] = p and Var[X ] = p(1 − p), we apply Theorem 7.13 which says that
the interval estimate
Mn (X ) − c ≤ p ≤ Mn (X ) + c
(1)
has confidence coefficient 1 − α where
α = 2 − 2
√
c n
.
p(1 − p)
(2)
We must ensure for every value of p that 1 − α ≥ 0.9 or α ≤ 0.1. Equivalently, we must
have
√
c n
≥ 0.95
(3)
p(1 − p)
√
for every value of p. Since (x) is an increasing function of x, we must satisfy c n ≥
1.65 p(1 − p). Since p(1 − p) ≤ 1/4 for all p, we require that
1.65
0.41
c≥ √ = √ .
4 n
n
(4)
The 0.9 confidence interval estimate of p is
0.41
0.41
Mn (X ) − √ ≤ p ≤ Mn (X ) + √ .
n
n
(5)
√
For the 0.99 confidence
interval, we have α ≤ 0.01, implying (c n/( p(1− p))) ≥ 0.995.
√
This implies c n√≥ 2.58 p(1 − p). Since p(1 − p) ≤ 1/4 for all p, we require that
c ≥ (0.25)(2.58)/ n. In this case, the 0.99 confidence interval estimate is
0.645
0.645
Mn (X ) − √ ≤ p ≤ Mn (X ) + √ .
n
n
(6)
Note that if M100 (X ) = 0.4, then the 0.99 confidence interval estimate is
0.3355 ≤ p ≤ 0.4645.
(7)
The interval is wide because the 0.99 confidence is high.
Quiz 7.5
Following the approach of bernoullitraces.m, we generate m = 1000 sample
paths, each sample path having n = 100 Bernoulli traces. at time k, OK(k) counts the
fraction of sample paths that have sample mean within one standard error of p. The program bernoullisample.m generates graphs the number of traces within one standard
error as a function of the time, i.e. the number of trials in each trace.
48
function OK=bernoullisample(n,m,p);
x=reshape(bernoullirv(p,m*n),n,m);
nn=(1:n)’*ones(1,m);
MN=cumsum(x)./nn;
stderr=sqrt(p*(1-p))./sqrt((1:n)’);
stderrmat=stderr*ones(1,m);
OK=sum(abs(MN-p)<stderrmat,2)/m;
plot(1:n,OK,’-s’);
The following graph was generated by bernoullisample(100,5000,0.5):
1
0.9
0.8
0.7
0.6
0.5
0.4
0
10
20
30
40
50
60
70
80
90
100
As we would expect, as m gets large, the fraction of traces within one standard error approaches 2(1) − 1 ≈ 0.68. The unusual sawtooth pattern, though perhaps unexpected, is
examined in Problem 7.5.2.
49
Quiz Solutions – Chapter 8
Quiz 8.1
From the problem statement, each X i has PDF and CDF
f X i (x) =
e−x x ≥ 0
0
otherwise
FX i (x) =
0
x <0
1 − e−x x ≥ 0
(1)
Hence, the CDF of the maximum of X 1 , . . . , X 15 obeys
FX (x) = P [X ≤ x] = P [X 1 ≤ x, X 2 ≤ x, · · · , X 15 ≤ x] = [P [X i ≤ x]]15 .
(2)
This implies that for x ≥ 0,
15
15
FX (x) = FX i (x) = 1 − e−x
(3)
α = P [X ≤ r ] = (1 − e−r )15 = 0.01
(4)
To design a significance test, we must choose a rejection region for X . A reasonable choice
is to reject the hypothesis if X is too small. That is, let R = {X ≤ r }. For a significance
level of α = 0.01, we obtain
It is straightforward to show that
r = − ln 1 − (0.01)1/15 = 1.33
(5)
Hence, if we observe X < 1.33, then we reject the hypothesis.
Quiz 8.2
From the problem statement, the conditional PMFs of K are
4
104k e−10
k = 0, 1, . . .
k!
PK |H0 (k) =
0
otherwise
6
106k e−10
k = 0, 1, . . .
k!
PK |H1 (k) =
0
otherwise
(1)
(2)
Since the two hypotheses are equally likely, the MAP and ML tests are the same. From
Theorem 8.6, the ML hypothesis rule is
k ∈ A0 if PK |H0 (k) ≥ PK |H1 (k) ;
k ∈ A1 otherwise.
(3)
This rule simplifies to
106 − 104
k ∈ A0 if k ≤ k =
= 214, 975.7;
ln 100
∗
k ∈ A1 otherwise.
Thus if we observe at least 214, 976 photons, then we accept hypothesis H1 .
50
(4)
Quiz 8.3
For the QPSK system, a symbol error occurs when si is transmitted but (X 1 , X 2 ) ∈ A j
for some j = i. For a QPSK system, it is easier to calculate the probability of a correct
decision. Given H0 , the conditional probability of a correct decision is
√
√
P [C|H0 ] = P [X 1 > 0, X 2 > 0|H0 ] = P
E/2 + N1 > 0, E/2 + N2 > 0
(1)
Because of the symmetry of the signals, P[C|H0 ] = P[C|Hi ] for all i. This implies the
probability of a correct decision is P[C] = P[C|H0 ]. Since N1 and N2 are iid Gaussian
(0, σ ) random variables, we have
√
√
P [C] = P [C|H0 ] = P
E/2 + N1 > 0 P
E/2 + N2 > 0
(2)
√
2
(3)
= P N1 > − E/2
*
√
+2
− E/2
(4)
= 1−
σ
Since (−x) = 1 − (x), we have P[C] = 2 ( E/2σ 2 ). Equivalently, the probability
of error is
,
E
PERR = 1 − P [C] = 1 − 2
(5)
2σ 2
Quiz 8.4
To generate the ROC, the existing program sqdistor already calculates this miss
probability PMISS = P01 and the false alarm probability PFA = P10 . The modified program, sqdistroc.m is essentially the same as sqdistor except the output is a matrix FM whose columns are the false alarm and miss probabilities. Next, the program
sqdistrocplot.m calls sqdistroc three times to generate a plot that compares the
receiver performance for the three requested values of d. Here is the modified code:
function FM=sqdistroc(v,d,m,T)
%square law distortion recvr
%P(error) for m bits tested
%transmit v volts or -v volts,
%add N volts, N is Gauss(0,1)
%add d(v+N)ˆ2 distortion
%receive 1 if x>T, otherwise 0
%FM = [P(FA) P(MISS)]
x=(v+randn(m,1));
[XX,TT]=ndgrid(x,T(:));
P01=sum((XX+d*(XX.ˆ2)< TT),1)/m;
x= -v+randn(m,1);
[XX,TT]=ndgrid(x,T(:));
P10=sum((XX+d*(XX.ˆ2)>TT),1)/m;
FM=[P10(:) P01(:)];
51
function FM=sqdistrocplot(v,m,T);
FM1=sqdistroc(v,0.1,m,T);
FM2=sqdistroc(v,0.2,m,T);
FM5=sqdistroc(v,0.3,m,T);
FM=[FM1 FM2 FM5];
loglog(FM1(:,1),FM1(:,2),’-k’, ...
FM2(:,1),FM2(:,2),’--k’, ...
FM5(:,1),FM5(:,2),’:k’);
legend(’\it d=0.1’,’\it d=0.2’,...
’\it d=0.3’,3)
ylabel(’P_{MISS}’);
xlabel(’P_{FA}’);
To see the effect of d, the commands
T=-3:0.1:3; sqdistrocplot(3,100000,T);
generated the plot shown in Figure 3.
0
10
−1
10
−2
PMISS
10
−3
10
−4
10
d=0.1
d=0.2
d=0.3
−5
10
−5
10
−4
10
−3
−2
10
10
−1
10
0
10
PFA
T=-3:0.1:3; sqdistrocplot(3,100000,T);
Figure 3: The receiver operating curve for the communications system of Quiz 8.4 with
squared distortion.
52
Quiz Solutions – Chapter 9
Quiz 9.1
(1) First, we calculate the marginal PDF for 0 ≤ y ≤ 1:
y
x=y
f Y (y) =
2(y + x) d x = 2x y + x 2
= 3y 2
x=0
0
This implies the conditional PDF of X given Y is
2
f X,Y (x, y)
+
f X |Y (x|y) =
= 3y
f Y (y)
0
2x
3y 2
0≤x ≤y
otherwise
(2) The minimum mean square error estimate of X given Y = y is
y
2x 2
2x
+
d x = 5y/9
x̂ M (y) = E [X |Y = y] =
3y 3y 2
0
(1)
(2)
(3)
Thus the MMSE estimator of X given Y is X̂ M (Y ) = 5Y /9.
(3) To obtain the conditional PDF f Y |X (y|x), we need the marginal PDF f X (x). For
0 ≤ x ≤ 1,
1
y=1
2
f X (x) =
2(y + x) dy = y + 2x y
= 1 + 2x − 3x 2
(4)
y=x
x
(5)
For 0 ≤ x ≤ 1, the conditional PDF of Y given X is
2(y+x)
x ≤y≤1
f Y |X (y|x) = 1+2x−3x 2
0
otherwise
(6)
(4) The MMSE estimate of Y given X = x is
1
2y 2 + 2x y
dy
2
x 1 + 2x − 3x
y=1
2y 3 /3 + x y 2
=
1 + 2x − 3x 2
ŷ M (x) = E [Y |X = x] =
(7)
(8)
y=x
=
53
2 + 3x − 5x 3
3 + 6x − 9x 2
(9)
Quiz 9.2
(1) Since the expectation of the sum equals the sum of the expectations,
E [R] = E [T ] + E [X ] = 0
(1)
(2) Since T and X are independent, the variance of the sum R = T + X is
Var[R] = Var[T ] + Var[X ] = 9 + 3 = 12
(2)
(3) Since T and R have expected values E[R] = E[T ] = 0,
Cov [T, R] = E [T R] = E [T (T + X )] = E T 2 + E [T X ]
(3)
Since T and X are independent and have zero expected value, E[T X ] = E[T ]E[X ] =
0 and E[T 2 ] = Var[T ]. Thus Cov[T, R] = Var[T ] = 9.
(4) From Definition 4.8, the correlation coefficient of T and R is
ρT,R = √
√
σT
Cov [T, R]
=
= 3/2
σR
Var[R] Var[T ]
(4)
(5) From Theorem 9.4, the optimum linear estimate of T given R is
T̂L (R) = ρT,R
σT
(R − E [R]) + E [T ]
σR
(5)
Since E[R] = E[T ] = 0 and ρT,R = σT /σ R ,
T̂L (R) =
σT2
σT2
σR
σT2 + σ X2
R=
2
R=
3
R
4
(6)
Hence a ∗ = 3/4 and b∗ = 0.
(6) By Theorem 9.4, the mean square error of the linear estimate is
2
) = 9(1 − 3/4) = 9/4
e∗L = Var[T ](1 − ρT,R
(7)
Quiz 9.3
When R = r , the conditional PDF of X = Y −40−40 log10 r is Gaussian with expected
value −40 − 40 log10 r and variance 64. The conditional PDF of X given R is
1
2
f X |R (x|r ) = √
e−(x+40+40 log10 r ) /128
128π
54
(1)
From the conditional PDF f X |R (x|r ), we can use Definition 9.2 to write the ML estimate
of R given X = x as
r̂ML (x) = arg max f X |R (x|r )
(2)
r ≥0
We observe that f X |R (x|r ) is maximized when the exponent (x + 40 + 40 log10 r )2 is
minimized. This minimum occurs when the exponent is zero, yielding
log10 r = −1 − x/40
(3)
r̂ML (x) = (0.1)10−x/40 m
(4)
or
If the result doesn’t look correct, note that a typical figure for the signal strength might be
x = −120 dB. This corresponds to a distance estimate of r̂ML (−120) = 100 m.
For the MAP estimate, we observe that the joint PDF of X and R is
f X,R (x, r ) = f X |R (x|r ) f R (r ) =
1
√
106 32π
2 /128
r e−(x+40+40 log10 r )
(5)
From Theorem 9.6, the MAP estimate of R given X = x is the value of r that maximizes
f X,R (x, r ). That is,
(6)
r̂MAP (x) = arg max f X,R (x, r )
0≤r ≤1000
Note that we have included the constraint r ≤ 1000 in the maximization to highlight the
fact that under our probability model, R ≤ 1000 m. Setting the derivative of f X,R (x, r )
with respect to r to zero yields
"
!
80 log10 e
−(x+40+40 log10 r )2 /128
(7)
1−
(x + 40 + 40 log10 r ) = 0
e
128
Solving for r yields
r = 10
1
25 log10 e −1
10−x/40 = (0.1236)10−x/40
(8)
This is the MAP estimate of R given X = x as long as r ≤ 1000 m. When x ≤ −156.3 dB,
the above estimate will exceed 1000 m, which is not possible in our probability model.
Hence, the complete description of the MAP estimate is
r̂MAP (x) =
1000
x < −156.3
−x/40
x ≥ −156.3
(0.1236)10
(9)
For example, if x = −120dB, then r̂MAP (−120) = 123.6 m. When the measured signal
strength is not too low, the MAP estimate is 23.6% larger than the ML estimate. This reflects the fact that large values of R are a priori more probable than small values. However,
for very low signal strengths, the MAP estimate takes into account that the distance can
never exceed 1000 m.
55
Quiz 9.4
(1) From Theorem 9.4, the LMSE estimate of X 2 given Y2 is X̂ 2 (Y2 ) = a ∗ Y2 + b∗ where
a∗ =
Cov [X 2 , Y2 ]
,
Var[Y2 ]
b ∗ = µ X 2 − a ∗ µ Y2 .
(1)
Because E[X] = E[Y] = 0,
Cov [X 2 , Y2 ] = E [X 2 Y2 ] = E [X 2 (X 2 + W2 )] = E X 22 = 1
Var[Y2 ] = Var[X 2 ] + Var[W2 ] = E X 22 + E W22 = 1.1
(2)
(3)
It follows that a ∗ = 1/1.1. Because µ X 2 = µY2 = 0, it follows that b∗ = 0. Finally,
to compute the expected square error, we calculate the correlation coefficient
ρ X 2 ,Y2 =
1
Cov [X 2 , Y2 ]
=√
σ X 2 σY2
1.1
(4)
The expected square error is
e∗L = Var[X 2 ](1 − ρ X2 2 ,Y2 ) = 1 −
1
1
=
= 0.0909
1.1
11
(5)
(2) Since Y = X + W and E[X] = E[W] = 0, it follows that E[Y] = 0. Thus we can
apply Theorem 9.7. Note that X and W have correlation matrices
!
"
!
"
1
−0.9
0.1 0
RX =
,
RW =
.
(6)
−0.9
1
0 0.1
In terms of Theorem 9.7, n = 2 and we wish to estimate X 2 given the observation
′
vector Y = Y1 Y2 . To apply Theorem 9.7, we need to find RY and RYX 2 .
RY = E YY′ = E (X + W)(X′ + W′ )
(7)
′
′
′
′
(8)
= E XX + XW + WX + WW .
Because X and W are independent, E[XW′ ] = E[X]E[W′ ] = 0. Similarly, E[WX′ ] =
0. This implies
"
!
′
1.1 −0.9
′
.
(9)
RY = E XX + E WW = RX + RW =
−0.9 1.1
In addition, we need to find
RYX 2
!
" !
"
E [Y1 X 2 ]
E [(X 1 + W1 )X 2 ]
= E [YX 2 ] =
=
.
E [Y2 X 2 ]
E [(X 2 + W2 )X 2 ]
56
(10)
Since X and W are independent vectors, E[W1 X 2 ] = E[W1 ]E[X 2 ] = 0 and E[W2 X 2 ] =
0. Thus
!
" !
"
E[X 1 X2 ]
−0.9
RYX 2 =
=
.
(11)
1
E X 22
By Theorem 9.7,
â =
R−1
Y RYX 2
"
!
−0.225
=
0.725
(12)
Therefore, the optimum linear estimator of X 2 given Y1 and Y2 is
X̂ L = â′ Y = −0.225Y1 + 0.725Y2 .
(13)
The mean square error is
Var [X 2 ] − â′ RYX 2 = Var [X ] − a1rY1 ,X 2 − a2rY2 ,X 2 = 0.0725.
(14)
Quiz 9.5
Since X and W have zero expected value, Y also has zero expected value. Thus, by
Theorem 9.7, X̂ L (Y) = â′ Y where â = R−1
Y RYX . Since X and W are independent,
′
′
E[WX ] = 0 and E[X W ] = 0 . This implies
(1)
RYX = E [YX ] = E [(1X + W)X ] = 1E X 2 = 1.
By the same reasoning, the correlation matrix of Y is
RY = E YY′ = E (1X + W)(1′ X + W′ )
= 11′ E X 2 + 1E X W′ + E [WX ] 1′ + E WW′
= 11′ + RW
(2)
(3)
(4)
Note that 11′ is a 20 × 20 matrix with every entry equal to 1. Thus,
−1
′
â = R−1
1
Y RYX = 11 + RW
(5)
−1
Y
X̂ L (Y) = 1′ 11′ + RW
(6)
−1
e∗L = Var[X ] − â′ RYX = 1 − 1′ 11′ + RW
1
(7)
and the optimal linear estimator is
The mean square error is
Now we note that RW has i, jth entry RW (i, j) = c|i− j|−1 . The question we must address
is what value c minimizes e∗L . This problem is atypical in that one does not usually get
57
to choose the correlation structure of the noise. However, we will see that the answer is
somewhat instructive.
We note that the answer is not obviously apparent from Equation (7). In particular, we
observe that Var[Wi ] = RW (i, i) = 1/c. Thus, when c is small, the noises Wi have high
variance and we would expect our estimator to be poor. On the other hand, if c is large
Wi and W j are highly correlated and the separate measurements of X are very dependent.
This would suggest that large values of c will also result in poor MSE. If this argument is
not clear, consider the extreme case in which every Wi and W j have correlation coefficient
ρi j = 1. In this case, our 20 measurements will be all the same and one measurement is as
good as 20 measurements.
To find the optimal value of c, we write a M ATLAB function mquiz9(c) to calculate
the MSE for a given c and second function that finds plots the MSE for a range of values
of c.
function [mse,af]=mquiz9(c);
v1=ones(20,1);
RW=toeplitz(c.ˆ((0:19)-1));
RY=(v1*(v1’)) +RW;
af=(inv(RY))*v1;
mse=1-((v1’)*af);
function cmin=mquiz9minc(c);
msec=zeros(size(c));
for k=1:length(c),
[msec(k),af]=mquiz9(c(k));
end
plot(c,msec);
xlabel(’c’);ylabel(’e_Lˆ*’);
[msemin,optk]=min(msec);
cmin=c(optk);
Note in mquiz9 that v1 corresponds to the vector 1 of all ones. The following commands
finds the minimum c and also produces the following graph:
1
>> c=0.01:0.01:0.99;
>> mquiz9minc(c)
ans =
0.4500
e*L
0.8
0.6
0.4
0.2
0
0.5
c
As we see in the graph, both small values and large values of c result in large MSE.
58
1
Quiz Solutions – Chapter 10
Quiz 10.1
There are many correct answers to this question. A correct answer specifies enough
random variables to specify the sample path exactly. One choice for an alternate set of
random variables that would specify m(t, s) is
• m(0, s), the number of ongoing calls at the start of the experiment
• N , the number of new calls that arrive during the experiment
• X 1 , . . . , X N , the interarrival times of the N new arrivals
• H , the number of calls that hang up during the experiment
• D1 , . . . , D H , the call completion times of the H calls that hang up
Quiz 10.2
(1) We obtain a continuous time, continuous valued process when we record the temperature as a continuous waveform over time.
(2) If at every moment in time, we round the temperature to the nearest degree, then we
obtain a continuous time, discrete valued process.
(3) If we sample the process in part (a) every T seconds, then we obtain a discrete time,
continuous valued process.
(4) Rounding the samples in part (c) to the nearest integer degree yields a discrete time,
discrete valued process.
Quiz 10.3
(1) Each resistor has resistance R in ohms with uniform PDF
f R (r ) =
0.01 950 ≤ r ≤ 1050
0
otherwise
(1)
The probability that a test produces a 1% resistor is
p = P [990 ≤ R ≤ 1010] =
59
1010
990
(0.01) dr = 0.2
(2)
(2) In t seconds, exactly t resistors are tested. Each resistor is a 1% resistor with probability p, independent of any other resistor. Consequently, the number of 1% resistors
found has the binomial PMF
t n
t−n n = 0, 1, . . . , t
n p (1 − p)
(3)
PN (t) (n) =
0
otherwise
(3) First we will find the PMF of T1 . This problem is easy if we view each resistor test
as an independent trial. A success occurs on a trial with probability p if we find a
1% resistor. The first 1% resistor is found at time T1 = t if we observe failures on
trials 1, . . . , t − 1 followed by a success on trial t. Hence, just as in Example 2.11,
T1 has the geometric PMF
PT1 (t) =
(1 − p)t−1 p t = 1, 2, . . .
9
otherwise
(4)
Since p = 0.2, the probability the first 1% resistor is found in exactly five seconds is
PT1 (5) = (0.8)4 (0.2) = 0.08192.
(4) From Theorem 2.5, a geometric random variable with success probability p has expected value 1/ p. In this problem, E[T1 ] = 1/ p = 5.
(5) Note that once we find the first 1% resistor, the number of additional trials needed to
find the second 1% resistor once again has a geometric PMF with expected value 1/ p
since each independent trial is a success with probability p. That is, T2 = T1 + T ′
where T ′ is independent and identically distributed to T1 . Thus
E [T2 |T1 = 10] = E [T1 |T1 = 10] + E T ′ |T1 = 10
(5)
′
(6)
= 10 + E T = 10 + 5 = 15
Quiz 10.4
Since each X i is a N (0, 1) random variable, each X i has PDF
1
2
f X (i) (x) = √ e−x /2
2π
′
By Theorem 10.1, the joint PDF of X = X 1 · · · X n is
f X (x) = f X (1),...,X (n) (x1 , . . . , xn ) =
k
i=1
60
f X (xi ) =
(1)
1
2
2
e−(x1 +···+xn )/2
n/2
(2π )
(2)
Quiz 10.5
The first and second hours are nonoverlapping intervals. Since one hour equals 3600
sec and the Poisson process has a rate of 10 packets/sec, the expected number of packets
in each hour is E[Mi ] = α = 36, 000. This implies M1 and M2 are independent Poisson
random variables each with PMF
PMi (m) =
α m e−α
m!
0
m = 0, 1, 2, . . .
otherwise
Since M1 and M2 are independent, the joint PMF of M1 and M2 is
⎧ α m 1 +m 2 e−2α
m 1 = 0, 1, . . . ;
⎪
⎪
⎨ m 1 !m 2 !
m 2 = 0, 1, . . . ,
PM1 ,M2 (m 1 , m 2 ) = PM1 (m 1 ) PM2 (m 2 ) =
⎪
⎪
⎩
0
otherwise.
(1)
(2)
Quiz 10.6
To answer whether N ′ (t) is a Poisson process, we look at the interarrival times. Let
X 1 , X 2 , . . . denote the interarrival times of the N (t) process. Since we count only evennumbered arrival for N ′ (t), the time until the first arrival of the N ′ (t) is Y1 = X 1 + X 2 .
Since X 1 and X 2 are independent exponential (λ) random variables, Y1 is an Erlang (n =
2, λ) random variable; see Theorem 6.11. Since Yi (t), the ith interarrival time of the N ′ (t)
process, has the same PDF as Y1 (t), we can conclude that the interarrival times of N ′ (t)
are not exponential random variables. Thus N ′ (t) is not a Poisson process.
Quiz 10.7
First, we note that for t > s,
X (t) − X (s) =
W (t) − W (s)
√
α
(1)
Since W (t) − W (s) is a Gaussian random variable, Theorem 3.13 states that W (t) − W (s)
is Gaussian with expected value
E [W (t) − W (s)]
=0
√
α
(2)
E (W (t) − W (s))2
α(t − s)
=
=
α
α
(3)
E [X (t) − X (s)] =
and variance
E (W (t) − W (s))
2
Consider s ′ ≤ s √
< t. Since s ≥ s ′ , W (t) − W√(s) is independent of W (s ′ ). This implies
[W (t) − W (s)]/ α is independent of W (s ′ )/ α for all s ≥ s ′ . That is, X (t) − X (s) is
independent of X (s ′ ) for all s ≥ s ′ . Thus X (t) is a Brownian motion process with variance
Var[X (t)] = t.
61
Quiz 10.8
First we find the expected value
µY (t) = µ X (t) + µ N (t) = µ X (t).
(1)
To find the autocorrelation, we observe that since X (t) and N (t) are independent and since
N (t) has zero expected value, E[X (t)N (t ′ )] = E[X (t)]E[N (t ′ )] = 0. Since RY (t, τ ) =
E[Y (t)Y (t + τ )], we have
RY (t, τ ) = E [(X (t) + N (t)) (X (t + τ ) + N (t + τ ))]
= E [X (t)X (t + τ )] + E [X (t)N (t + τ )]
+ E [X (t + τ )N (t)] + E [N (t)N (t + τ )]
= R X (t, τ ) + R N (t, τ ).
(2)
(3)
(4)
Quiz 10.9
From Definition 10.14, X 1 , X 2 , . . . is a stationary random sequence if for all sets of
time instants n 1 , . . . , n m and time offset k,
f X n1 ,...,X nm (x1 , . . . , xm ) = f X n1 +k ,...,X nm +k (x1 , . . . , xm )
(1)
Since the random sequence is iid,
f X n1 ,...,X nm (x1 , . . . , xm ) = f X (x1 ) f X (x2 ) · · · f X (xm )
(2)
Similarly, for time instants n 1 + k, . . . , n m + k,
f X n1 +k ,...,X nm +k (x1 , . . . , xm ) = f X (x1 ) f X (x2 ) · · · f X (xm )
(3)
We can conclude that the iid random sequence is stationary.
Quiz 10.10
We must check whether each function R(τ ) meets the conditions of Theorem 10.12:
R(τ ) ≥ 0
R(τ ) = R(−τ )
|R(τ )| ≤ R(0)
(1)
(1) R1 (τ ) = e−|τ | meets all three conditions and thus is valid.
2
(2) R2 (τ ) = e−τ also is valid.
(3) R3 (τ ) = e−τ cos τ is not valid because
R3 (−2π ) = e2π cos 2π = e2π > 1 = R3 (0)
(2)
2
(4) R4 (τ ) = e−τ sin τ also cannot be an autocorrelation function because
R4 (π/2) = e−π/2 sin π/2 = e−π/2 > 0 = R4 (0)
62
(3)
Quiz 10.11
(1) The autocorrelation of Y (t) is
RY (t, τ ) = E [Y (t)Y (t + τ )]
= E [X (−t)X (−t − τ )]
= R X (−t − (−t − τ )) = R X (τ )
(1)
(2)
(3)
Since E[Y (t)] = E[X (−t)] = µ X , we can conclude that Y (t) is a wide sense
stationary process. In fact, we see that by viewing a process backwards in time, we
see the same second order statistics.
(2) Since X (t) and Y (t) are both wide sense stationary processes, we can check whether
they are jointly wide sense stationary by seeing if R X Y (t, τ ) is just a function of τ .
In this case,
R X Y (t, τ ) = E [X (t)Y (t + τ )]
= E [X (t)X (−t − τ )]
= R X (t − (−t − τ )) = R X (2t + τ )
(4)
(5)
(6)
Since R X Y (t, τ ) depends on both t and τ , we conclude that X (t) and Y (t) are not
jointly wide sense stationary. To see why this is, suppose R X (τ ) = e−|τ | so that
samples of X (t) far apart in time have almost no correlation. In this case, as t gets
larger, Y (t) = X (−t) and X (t) become less and less correlated.
Quiz 10.12
From the problem statement,
E [X (t)] = E [X (t + 1)] = 0
(1)
E [X (t)X (t + 1)] = 1/2
(2)
Var[X (t)] = Var[X (t + 1)] = 1
(3)
′
The Gaussian random vector X = X (t) X (t + 1) has covariance matrix and corresponding inverse
!
"
!
"
4
1
−1/2
1 1/2
−1
CX =
CX =
(4)
1
1/2 1
3 −1/2
Since
"! "
!
′ 4
4 2
1
−1/2 x0
x0 − x0 x+ x12
=
= x0 x1
x
1
x1
3 −1/2
3
the joint PDF of X (t) and X (t + 1) is the Gaussian vector PDF
1 ′ −1
1
exp − x CX x
f X (t),X (t+1) (x0 , x1 ) =
2
(2π )n/2 [det (CX )]1/2
2 2
1
2
=√
e− 3 x0 −x0 x1 +x1
3π 2
′
C−1
X x
63
(5)
(6)
(7)
120
100
M(t)
80
60
40
20
0
0
10
20
30
40
50
t
60
70
80
90
100
Figure 4: Sample path of 100 minutes of the blocking switch of Quiz 10.13.
Quiz 10.13
The simple structure of the switch simulation of Example 10.28 admits a deceptively
simple solution in terms of the vector of arrivals A and the vector of departures D. With the
introduction of call blocking. we cannot generate these vectors all at once. In particular,
when an arrival occurs at time t, we need to know that M(t), the number of ongoing calls,
satisfies M(t) < c = 120. Otherwise, when M(t) = c, we must block the call. Call
blocking can be implemented by setting the service time of the call to zero so that the call
departs as soon as it arrives.
The blocking switch is an example of a discrete event system. The system evolves via
a sequence of discrete events, namely arrivals and departures, at discrete time instances. A
simulation of the system moves from one time instant to the next by maintaining a chronological schedule of future events (arrivals and departures) to be executed. The program
simply executes the event at the head of the schedule. The logic of such a simulation is
1. Start at time t = 0 with an empty system. Schedule the first arrival to occur at S1 , an
exponential (λ) random variable.
2. Examine the head-of-schedule event.
• When the head-of-schedule event is the kth arrival is at time t, check the state
M(t).
– If M(t) < c, admit the arrival, increase the system state n by 1, and schedule a departure to occur at time t + Sn , where Sk is an exponential (λ)
random variable.
– If M(t) = c, block the arrival, do not schedule a departure event.
• If the head of schedule event is a departure, reduce the system state n by 1.
3. Delete the head-of-schedule event and go to step 2.
After the head-of-schedule event is completed and any new events (departures in this system) are scheduled, we know the system state cannot change until the next scheduled event.
64
Thus we know that M(t) will stay the same until then. In our simulation, we use the vector
t as the set of time instances at which we inspect the system state. Thus for all times t(i)
between the current head-of-schedule event and the next, we set m(i) to the current switch
state.
The complete program is shown in Figure 5. In most programming languages, it is
common to implement the event schedule as a linked list where each item in the list has
a data structure indicating an event timestamp and the type of the event. In M ATLAB, a
simple (but not elegant) way to do this is to have maintain two vectors: time is a list
of timestamps of scheduled events and event is a the list of event types. In this case,
event(i)=1 if the ith scheduled event is an arrival, or event(i)=-1 if the ith scheduled event is a departure.
When the program is passed a vector t, the output [m a b] is such that m(i) is the
number of ongoing calls at time t(i) while a and b are the number of admits and blocks.
The following instructions
t=0:0.1:5000;
[m,a,b]=simblockswitch(10,0.1,120,t);
plot(t,m);
generated a simulation lasting 5,000 minutes. A sample path of the first 100 minutes of
that simulation is shown in Figure 4. The 5,000 minute full simulation produced a=49658
admitted calls and b=239 blocked calls. We can estimate the probability a call is blocked
as
b
= 0.0048.
(1)
P̂b =
a+b
In Chapter 12, we will learn that the exact blocking probability is given by Equation (12.93),
a result known as the “Erlang-B formula.” From the Erlang-B formula, we can calculate
that the exact blocking probability is Pb = 0.0057. One reason our simulation underestimates the blocking probability is that in a 5,000 minute simulation, roughly the first 100
minutes are needed to load up the switch since the switch is idle when the simulation starts
at time t = 0. However, this says that roughly the first two percent of the simulation time
was unusual. Thus this would account for only part of the disparity. The rest of the gap
between 0.0048 and 0.0057 is that a simulation that includes only 239 blocks is not all that
likely to give a very accurate result for the blocking probability.
Note that in Chapter 12, we will learn that the blocking switch is an example of an
M/M/c/c queue, a kind of Markov chain. Chapter 12 develops techniques for analyzing
and simulating systems described by Markov chains that are much simpler than the discrete
event simulation technique shown here. Nevertheless, for very complicated systems, the
discrete event simulation is widely-used and often very efficient simulation method.
65
function [M,admits,blocks]=simblockswitch(lam,mu,c,t);
blocks=0; %total # blocks
admits=0; %total # admits
M=zeros(size(t));
n=0; % # in system
time=[ exponentialrv(lam,1) ];
event=[ 1 ]; %first event is an arrival
timenow=0;
tmax=max(t);
while (timenow<tmax)
M((timenow<=t)&(t<time(1)))=n;
timenow=time(1);
eventnow=event(1);
event(1)=[ ]; time(1)= [ ]; % clear current event
if (eventnow==1) % arrival
arrival=timenow+exponentialrv(lam,1); % next arrival
b4arrival=time<arrival;
event=[event(b4arrival) 1 event(˜b4arrival)];
time=[time(b4arrival) arrival time(˜b4arrival)];
if n<c %call admitted
admits=admits+1;
n=n+1;
depart=timenow+exponentialrv(mu,1);
b4depart=time<depart;
event=[event(b4depart) -1 event(˜b4depart)];
time=[time(b4depart) depart time(˜b4depart)];
else
blocks=blocks+1; %one more block, immed departure
disp(sprintf(’Time %10.3d Admits %10d Blocks %10d’,...
timenow,admits,blocks));
end
elseif (eventnow==-1) %departure
n=n-1;
end
end
Figure 5: Discrete event simulation of the blocking switch of Quiz 10.13.
66
Quiz Solutions – Chapter 11
Quiz 11.1
By Theorem 11.2,
µY = µ X
∞
−∞
h(t)dt = 2
∞
0
e−t dt = 2
Since R X (τ ) = δ(τ ), the autocorrelation function of the output is
∞
∞
∞
RY (τ ) =
h(u)
h(v)δ(τ + u − v) dv du =
h(u)h(τ + u) du
−∞
−∞
(1)
(2)
−∞
For τ > 0, we have
RY (τ ) =
∞
e
−u −τ −u
e
0
du = e
−τ
0
∞
1
e−2u du = e−τ
2
(3)
For τ < 0, we can deduce that RY (τ ) = 21 e−|τ | by symmetry. Just to be safe though, we
can double check. For τ < 0,
∞
∞
1
(4)
h(u)h(τ + u) du =
e−u e−τ −u du = eτ
RY (τ ) =
2
−τ
−τ
Hence,
1
RY (τ ) = e−|τ |
2
(5)
Quiz 11.2
The expected value of the output is
µY = µ X
∞
n=−∞
h n = 0.5(1 + −1) = 0
(1)
The autocorrelation of the output is
RY [n] =
1
1
i=0 j=0
h i h j R X [n + i − j]
= 2R X [n] − R X [n − 1] − R X [n + 1] =
(2)
1 n=0
0 otherwise
Since µY = 0, The variance of Yn is Var[Yn ] = E[Yn2 ] = RY [0] = 1.
67
(3)
x 10
8
6
SX(f)
SX(f)
0.6
0.4
0.2
2
−10
−5
0
f
5
10
0
−1500−1000 −500
15
10
10
5
5
RX(τ)
RX(τ)
0
−15
4
0
−5
−0.2
−0.1
0
τ
0.1
(a) W = 10
500 1000 1500
0
−5
−2
0.2
0
f
−1
0
τ
1
2
−3
x 10
(b) W = 1000
Figure 6: The autocorrelation R X (τ ) and power spectral density S X ( f ) for process X (t) in
Quiz 11.5.
Quiz 11.3
′
By Theorem 11.8, Y = Y33 Y34 Y35 is a Gaussian random vector since X n is
a Gaussian random
process. Moreover, by Theorem 11.5, each Yn has expected value
E[Yn ] = µ X ∞
n=−∞ h n = 0. Thus E[Y] = 0. Fo find the PDF of the Gaussian vector
Y, we need to find the covariance matrix CY , which equals the correlation matrix RY since
Y has zero expected value. One way to find the RY is to observe that RY has the Toeplitz
structure of Theorem 11.6 and to use Theorem 11.5 to find the autocorrelation function
RY [n] =
∞
∞
i=−∞ j=−∞
h i h j R X [n + i − j].
(1)
Despite the fact that R X [k] is an impulse, using Equation (1) is surprisingly tedious because
we still need to sum over all i and j such that n + i − j = 0.
In this problem, it is simpler to observe that Y = HX where
′
X = X 30 X 31 X 32 X 33 X 34 X 35
(2)
and
⎡
⎤
1 1 1 1 0 0
1
H = ⎣0 1 1 1 1 0⎦ .
4 0 0 1 1 1 1
(3)
In this case, following Theorem 11.7, or by directly applying Theorem 5.13 with µX = 0
and A = H, we obtain RY = HRX H′ . Since R X [n] = δn , RX = I, the identity matrix.
68
Thus
⎡
⎤
4 3 2
1 ⎣
3 4 3⎦ .
CY = RY = HH′ =
16 2 3 4
It follows (very quickly if you use M ATLAB for 3 × 3 matrix inversion) that
⎡
⎤
7/12 −1/2 1/12
⎣
1
−1/2⎦ .
C−1
Y = 16 −1/2
1/12 −1/2 7/12
(4)
(5)
Thus, the PDF of Y is
1
1 ′ −1
exp − y CY y .
f Y (y) =
2
(2π )3/2 [det (CY )]1/2
(6)
A disagreeable amount of algebra will show det(CY ) = 3/1024 and that the PDF can be
“simplified” to
"
!
7 2
1
16
7 2
2
y + y34 + y35 − y33 y34 + y33 y35 − y34 y35 . (7)
exp −8
f Y (y) = √
12 33
12
6
6π 3
Equation (7) shows that one of the nicest features of the multivariate Gaussian distribution
is that y′ C−1
Y y is a very concise representation of the cross-terms in the exponent of f Y (y).
Quiz 11.4
This
using Theorem 11.9 for the case of k = 1 and M = 2. In this case,
′
quiz is solved
Xn = X n−1 X n and
" !
"
!
1.1 0.9
R X [0] R X [1]
RXn =
=
(1)
R X [1] R X [0]
0.9 1.1
and
RXn X n+1 = E
!!
" !
"
"
" !
0.81
X n−1
R X [2]
=
.
X n+1 =
Xn
0.9
R X [1]
(2)
The MMSE linear first order filter for predicting X n+1 at time n is the filter h such that
!
"−1 !
"
"
!
1
←
−
1.1 0.9
0.81
81
−1
h = RXn RXn X n+1 =
=
.
(3)
0.9 1.1
0.9
400 261
′
It follows that the filter is h = 261/400 81/400 and the MMSE linear predictor is
X̂ n+1 =
261
81
X n−1 +
Xn.
400
400
(4)
to find the mean square error, one approach is to follow the method of Example 11.13 and
to directly calculate
∗
2
(5)
e L = E (X n+1 − X̂ n+1 ) .
69
This method is workable for this simple problem but becomes increasingly tedious for
higher order filters. Instead, we can derive the mean square error for an arbitary prediction
←
−
filter h. Since X̂ n+1 = h ′ Xn ,
!
"
←
−′ 2
∗
e L = E X n+1 − h Xn
(6)
←
−′
←
−′
′
(7)
= E (X n+1 − h Xn )(X n+1 − h Xn )
←
−
←
−
= E (X n+1 − h ′ Xn )(X n+1 − X′n h )
(8)
After a bit of algebra, we obtain
←
−
←
−
←
−
e∗L = R X [0] − 2 h ′ RXn X n+1 + h ′ RXn h
(9)
(10)
←
−
with the substitution h = R−1
Xn RXn X n+1 , we obtain
e∗L = R X [0] − R′Xn X n+1 R−1
Xn RXn X n+1
←
−
= R X [0] − h ′ RXn X n+1
(11)
(12)
Note that this is essentially the same result as Theorem 9.7 with Y = Xn , X = X n+1 and
←
−
â′ = h ′ . It is noteworthy that the result is derived in a much simpler way in the proof of
Theorem 9.7 by using the orthoginality property of the LMSE estimator.
In any case, the mean square error is
!
"
0.81
1
506
←
−′
∗
81 261
e L = R X [0] − h RXn X n+1 = 1.1 −
= 0.3487. (13)
=
0.9
400
1451
recalling that the blind estimate would yield a mean square error of Var[X ] = 1.1, we see
that observing X n−1 and X n improves the accuracy of our prediction of X n+1 .
Quiz 11.5
(1) By Theorem 11.13(b), the average power of X (t) is
W
∞
5
2
d f = 10 Watts
E X (t) =
SX ( f ) d f =
−∞
−W W
(1)
(2) The autocorrelation function is the inverse Fourier transform of S X ( f ). Consulting
Table 11.1, we note that
1
f
S X ( f ) = 10
rect
(2)
2W
2W
It follows that the inverse transform of S X ( f ) is
sin(2π W τ )
R X (τ ) = 10 sinc(2W τ ) = 10
(3)
2π W τ
(3) For W = 10 Hz and W = 1 kHZ, graphs of S X ( f ) and R X (τ ) appear in Figure 6.
70
Quiz 11.6
In a sampled system, the discrete time impulse δ[n] has a flat discrete Fourier transform.
That is, if R X [n] = 10δ[n], then
S X (φ) =
∞
n=−∞
10δ[n]e− j2π φn = 10
(1)
Thus, R X [n] = 10δ[n]. (This quiz is really lame!)
Quiz 11.7
Since Y (t) = X (t − t0 ),
R X Y (t, τ ) = E [X (t)Y (t + τ )] = E [X (t)X (t + τ − t0 )] = R X (τ − t0 )
(1)
We see that R X Y (t, τ ) = R X Y (τ ) = R X (τ − t0 ). From Table 11.1, we recall the property that g(τ − τ0 ) has Fourier transform G( f )e− j2π f τ0 . Thus the Fourier transform of
R X Y (τ ) = R X (τ − t0 ) = g(τ − t0 ) is
S X Y ( f ) = S X ( f )e− j2π f t0 .
(2)
Quiz 11.8
We solve this quiz using Theorem 11.17. First we need some preliminary facts. Let
a0 = 5,000 so that
1
(1)
R X (τ ) = a0 e−a0 |τ | .
a0
Consulting with the Fourier transforms in Table 11.1, we see that
2a02
2a0
1
= 2
SX ( f ) =
2
2
a0 a0 + (2π f )
a0 + (2π f )2
(2)
The RC filter has impulse response h(t) = a1 e−a1 t u(t), where u(t) is the unit step function
and a1 = 1/RC where RC = 10−4 is the filter time constant. From Table 11.1,
H( f ) =
a1
a1 + j2π f
(3)
(1) Theorem 11.17,
S X Y ( f ) = H ( f )S X ( f ) =
(2) Again by Theorem 11.17,
2a0 a1
.
[a1 + j2π f ] a02 + (2π f )2
SY ( f ) = H ∗ ( f )S X Y ( f ) = |H ( f )|2 S X ( f ).
71
(4)
(5)
Note that
|H ( f )|2 = H ( f )H ∗ ( f ) =
a12
a1
a1
= 2
(a1 + j2π f ) (a1 − j2π f )
a1 + (2π f )2
(6)
Thus,
2a0 a12
SY ( f ) = |H ( f )|2 S X ( f ) = 2
a1 + (2π f )2 a02 + (2π f )2
(7)
(3) To find the
average power at the filter output, we can either use basic calculus and
∞
calculate −∞ SY ( f ) d f directly or we can find RY (τ ) as an inverse transform of
SY ( f ). Using partial fractions and the Fourier transform table, the latter method is
actually less algebra. In particular, some algebra will show that
SY ( f ) =
where
K0 =
a02
K0
K1
+
+ (2π f )2 a1 + (2π f )2
2a0 a12
K1 =
,
2
a12 − a0
−2a0 a12
a12 − a02
.
(8)
(9)
Thus,
2a02
2a12
K0
K1
.
+
SY ( f ) = 2 2
2a0 a0 + (2π f )2 2a12 a1 + (2π f )2
(10)
Consulting with Table 11.1, we see that
RY (τ ) =
K0
K1
a0 e−a0 |τ | + 2 a1 e−a1 |τ |
2
2a0
2a1
(11)
Substituting the values of K 0 and K 1 , we obtain
RY (τ ) =
a12 e−a0 |τ | − a0 a1 e−a1 |τ |
a12 − a02
.
(12)
The average power of the Y (t) process is
RY (0) =
2
a1
= .
a1 + a0
3
(13)
Note that the input signal has average power R X (0) = 1. Since the RC filter has a 3dB
bandwidth of 10,000 rad/sec and the signal X (t) has most of its its signal energy below
5,000 rad/sec, the output signal has almost as much power as the input.
72
Quiz 11.9
This quiz implements an example of Equations (11.146) and (11.147) for a system in
which we filter Y (t) = X (t) + N (t) to produce an optimal linear estimate of X (t). The
solution to this quiz is just to find the filter Ĥ ( f ) using Equation (11.146) and to calculate
the mean square error e L ∗ using Equation (11.147).
Comment: Since the text omitted the derivations of Equations (11.146) and (11.147), we
note that Example 10.24 showed that
RY (τ ) = R X (τ ) + R N (τ ),
RY X (τ ) = R X (τ ).
(1)
SY X ( f ) = S X ( f ).
(2)
Taking Fourier transforms, it follows that
SY ( f ) = S X ( f ) + S N ( f ),
Now we can go on to the quiz, at peace with the derivations.
(1) Since µ N = 0, R N (0) = Var[N ] = 1. This implies
R N (0) =
∞
−∞
SN ( f ) d f =
B
−B
N0 d f = 2N0 B
(3)
Thus N0 = 1/(2B). Because the noise process N (t) has constant power R N (0) = 1,
decreasing the single-sided bandwidth B increases the power spectral density of the
noise over frequencies | f | < B.
(2) Since R X (τ ) = sinc(2W τ ), where W = 5,000 Hz, we see from Table 11.1 that
f
1
.
(4)
S X ( f ) = 4 rect
10
104
The noise power spectral density can be written as
1
f
f
=
,
rect
S N ( f ) = N0 rect
2B
2B
2B
(5)
From Equation (11.146), the optimal filter is
Ĥ ( f ) =
SX ( f )
=
SX ( f ) + SN ( f )
73
rect 10f 4
.
f
1
rect 2B
rect 10f 4 + 2B
1
104
1
104
(6)
(3) We produce the output X̂ (t) by passing the noisy signal Y (t) through the filter Ĥ ( f ).
From Equation (11.147), the mean square error of the estimate is
∞
S X ( f )S N ( f )
∗
eL =
df
(7)
−∞ S X ( f ) + S N ( f )
f
1
∞ 1 rect f
rect 2B
104
104 2B
d f.
(8)
=
f
f
1
−∞ 1 rect
+
rect
2B
2B
104
104
To evaluate the MSE e∗L , we need to whether B ≤ W . Since the problem asks us to
find the largest possible B, let’s suppose B ≤ W . We can go back and consider the
case B > W later. When B ≤ W , the MSE is
B
1
1 1
1
104 2B
104
∗
eL =
(9)
d
f
=
=
1
1
1
1
5,000
+
1
+
−B 104 + 2B
4
2B
B
10
To obtain MSE e∗L ≤ 0.05 requires B ≤ 5,000/19 = 263.16 Hz.
Although this completes the solution to the quiz, what is happening may not be obvious.
The noise power is always Var[N ] = 1 Watt, for all values of B. As B is decreased, the PSD
S N ( f ) becomes increasingly tall, but only over a bandwidth B that is decreasing. Thus as
B descreases, the filter Ĥ ( f ) makes an increasingly deep and narrow notch at frequencies
| f | ≤ B. Two examples of the filter Ĥ ( f ) are shown in Figure 7. As B shrinks, the filter
suppresses less of the signal of X (t). The result is that the MSE goes down.
Finally, we note that we can choose B very large and also achieve MSE e∗L = 0.05. In
particular, when B > W = 5000, S N ( f ) = 1/2B over frequencies | f | < W . In this case,
the Wiener filter Ĥ ( f ) is an ideal (flat) lowpass filter
⎧
1
⎨ 104
| f | < 5,000,
1
+ 1
(10)
Ĥ ( f ) =
104 2B
⎩
0
otherwise.
Thus increasing B spreads the constant 1 watt of power of N (t) over more bandwidth. The
Wiener filter removes the noise that is outside the band of the desired signal. The mean
square error is
e∗L
=
1 1
104 2B
1
1
−5000 104 + 2B
1
2B
5000
df =
1
104
+
1
2B
=
1
B
5000
+1
(11)
In this case, B ≥ 9.5 × 104 guarantees e∗L ≤ 0.05.
Quiz 11.10
It is fairly straightforward to find S X (φ) and SY (φ). The only thing to keep in mind is
to use fftc to transform the autocorrelation R X [ f ] into the power spectral density S X (φ).
The following M ATLAB program generates and plots the functions shown in Figure 8
74
1
H(f)
H(f)
1
0.5
0
−5000
−2000
0
f
2000
0.5
0
5000
B = 500
−5000
−2000
0
f
2000
5000
B = 2500
Figure 7: Wiener filter for Quiz 11.9.
%mquiz11.m
N=32;
rx=[2 4 2]; SX=fftc(rx,N); %autocorrelation and PSD
stem(0:N-1,abs(sx));
xlabel(’n’);ylabel(’S_X(n/N)’);
h2=0.5*[1 1]; H2=fft(h2,N); %impulse/filter response: M=2
SY2=SX.* ((abs(H2)).ˆ2);
figure; stem(0:N-1,abs(SY2)); %PSD of Y for M=2
xlabel(’n’);ylabel(’S_{Y_2}(n/N)’);
h10=0.1*ones(1,10); H10=fft(h10,N); %impulse/filter response: M=10
SY10=sx.*((abs(H10)).ˆ2);
figure; stem(0:N-1,abs(SY10));
xlabel(’n’);ylabel(’S_{Y_{10}}(n/N)’);
Relative to M = 2, when M = 10, the filter H (φ) filters out almost all of the high
frequency components of X (t). In the context of Example 11.26, the low pass moving
average filter for M = 10 removes the high frquency components and results in a filter
output that varies very slowly.
As an aside, note that the vectors SX, SY2 and SY10 in mquiz11 should all be realvalued vectors. However, the finite numerical precision of M ATLAB results in tiny imaginary parts. Although these imaginary parts have no computational significance, they tend
to confuse the stem function. Hence, we generate stem plots of the magnitude of each
power spectral density.
75
SX(n/N)
10
5
0
0
5
10
15
20
25
30
35
20
25
30
35
20
25
30
35
n
5
2
SY (n/N)
10
0
0
5
10
15
n
5
10
SY (n/N)
10
0
0
5
10
15
n
Figure 8: For Quiz 11.10, graphs of S X (φ), SY (n/N ) for M = 2, and Sφ (n/N ) for M = 10
using an N = 32 point DFT.
76
Quiz Solutions – Chapter 12
Quiz 12.1
The system has two states depending on whether the previous packet was received in
error. From the problem statement, we are given the conditional probabilities
P X n+1 = 0|X n = 0 = 0.99
P X n+1 = 1|X n = 1 = 0.9
(1)
Since each X n must be either 0 or 1, we can conclude that
P X n+1 = 1|X n = 0 = 0.01
P X n+1 = 0|X n = 1 = 0.1
(2)
These conditional probabilities correspond to the transition matrix and Markov chain:
0.99
0.9
"
!
0.01
0.99 0.01
(3)
P=
0
1
0.10 0.90
0.1
Quiz 12.2
From the problem statement, the Markov chain and the transition matrix are
⎡
⎤
0.4
0.6
0.4
0.4
0.6
0
0.2
0.6
P = ⎣0.2 0.6 0.2⎦
1
0
1
0 0.6 0.4
0.6
0.2
The eigenvalues of P are
λ1 = 0
λ2 = 0.4
λ3 = 1
We can diagonalize P into
⎤⎡
⎤
⎡
⎤⎡
−0.5 1 −0.5
−0.6 0.5 1
λ1 0 0
0
−1 ⎦
0
1 ⎦ ⎣ 0 λ2 0 ⎦ ⎣ 1
P = S−1 DS = ⎣ 0.4
0 0 λ3
0.2 0.6 0.2
−0.6 −0.5 1
(1)
(2)
(3)
where si , the ith row of S, is the left eigenvector of P satisfying si P = λi si . Algebra will
verify that the n-step transition matrix is
⎡
⎡
⎤
⎤
0.5 0 −0.5
0.2 0.6 0.2
0
0 ⎦
Pn = S−1 Dn S = ⎣0.2 0.6 0.2⎦ + (0.4)n ⎣ 0
(4)
−0.5 0 0.5
0.2 0.6 0.2
Quiz 12.3
The Markov chain describing the factory status and the corresponding state transition
matrix are
77
0.9
0.1
0
⎡
⎤
0.9 0.1 0
0 1⎦
P=⎣ 0
1
0 0
1
1
1
2
(1)
′
With π = π0 π1 π2 , the system of equations π ′ = π ′ P yields π1 = 0.1π0 and
π2 = π1 . This implies
π0 + π1 + π2 = π0 (1 + 0.1 + 0.1) = 1
(2)
It follows that the limiting state probabilities are
π0 = 5/6,
π1 = 1/12,
π2 = 1/12.
(3)
C2 = {2, 3}
C3 = {4, 5, 6}
(1)
Quiz 12.4
The communicating classes are
C1 = {0, 1}
The states in C1 and C3 are aperiodic. The states in C2 have period 2. Once the system
enters a state in C1 , the class C1 is never left. Thus the states in C1 are recurrent. That
is, C1 is a recurrent class. Similarly, the states in C3 are recurrent. On the other hand, the
states in C2 are transient. Once the system exits C2 , the states in C2 are never reentered.
Quiz 12.5
At any time t, the state n can take on the values 0, 1, 2, . . .. The state transition probabilities are
P [K > n]
P [K > n − 1]
P [K = n]
= P [K = n|K > n − 1] =
P [K > n − 1]
Pn−1,n = P [K > n|K > n − 1] =
(1)
Pn−1,0
(2)
(3)
The Markov chain resembles
P[K=5]
P[K=4]
P[K= 1]
P[K=3]
P[K=2]
1
0
1
3
2
1
1
78
4
1
...
1
…
The stationary probabilities satisfy
π0 = π0 P [K = 1] + π1 ,
π1 = π0 P [K = 2] + π2 ,
..
.
πk−1 = π0 P [K = k] + πk ,
(4)
(5)
k = 1, 2, . . .
(6)
From Equation (4), we obtain
π1 = π0 (1 − P [K = 1]) = π0 P [K > 1]
(7)
Similarly, Equation (5) implies
π2 = π1 − π0 P [K = 2] = π0 (P [K > 1] − P [K = 2]) = π0 P [K > 2]
(8)
This suggests that πk = π0 P[K > k]. We verify this pattern by showing that πk =
π0 P[K > k] satisfies Equation (6):
(9)
π0 P [K > k − 1] = π0 P [K = k] + π0 P [K > k] .
∞
∞
When we apply
∞ k=0 πk = 1, we obtain π0 n=0 P[K > k] = 1. From Problem 2.5.11,
we recall that k=0 P[K > k] = E[K ]. This implies
πn =
P [K > n]
E [K ]
(10)
This Markov chain models repeated random countdowns. The system state is the time until
the counter expires. When the counter expires, the system is in state 0, and we randomly
reset the counter to a new value K = k and then we count down k units of time. Since we
spend one unit of time in each state, including state 0, we have k − 1 units of time left after
the state 0 counter reset. If we have a random variable W such that the PMF of W satisfies
PW (n) = πn , then W has a discrete PMF representing the remaining time of the counter at
a time in the distant future.
Quiz 12.6
(1) By inspection, the number of transitions need to return to state 0 is always a multiple
of 2. Thus the period of state 0 is d = 2.
(2) To
3find the stationary probabilities, we solve the system of equations π = πP and
i=0 πi = 1:
π0 = (3/4)π1 + (1/4)π3
π1 = (1/4)π0 + (1/4)π2
π2 = (1/4)π1 + (3/4)π3
1 = π0 + π1 + π2 + π3
79
(1)
(2)
(3)
(4)
Solving the second and third equations for π2 and π3 yields
π2 = 4π1 − π0
π3 = (4/3)π2 − (1/3)π1 = 5π1 − (4/3)π0
(5)
Substituting π3 back into the first equation yields
π0 = (3/4)π1 + (1/4)π3 = (3/4)π1 + (5/4)π1 − (1/3)π0
(6)
This implies π1 = (2/3)π0 . It follows from the first and second equations that
π2 = (5/3)π0 and π3 = 2π0 . Lastly, we choose π0 so the state probabilities sum to
1:
16
2 5
1 = π0 + π1 + π2 + π3 = π0 1 + + + 2 = π0
(7)
3 3
3
It follows that the state probabilities are
π0 =
3
16
π1 =
2
16
π2 =
5
16
π3 =
6
16
(8)
(3) Since the system starts in state 0 at time 0, we can use Theorem 12.14 to find the
limiting probability that the system is in state 0 at time nd:
lim P00 (nd) = dπ0 =
n→∞
3
8
(9)
Quiz 12.7
The Markov chain has the same structure as that in Example 12.22. The only difference
is the modified transition rates:
1
(1/2)
1
0
1- (1/2)
a
(2/3)
a
(3/4)
3
2
a
a
(4/5)
4
…
a
a
1 - (2/3)
a
1 - (3/4)
a
1 - (4/5)
The event T00 > n occurs if the system reaches state n before returning to state 0, which
occurs with probability
α α
α
1
2
n−1 α
1
P [T00 > n] = 1 ×
×
× ··· ×
=
.
(1)
2
3
n
n
Thus the CDF of T00 satisfies FT00 (n) = 1− P[T00 > n] = 1−1/n α . To determine whether
state 0 is recurrent, we observe that for all α > 0
P [V00 ] = lim FT00 (n) = lim 1 −
n→∞
n→∞
80
1
= 1.
nα
(2)
Thus state 0 is recurrent for all α > 0. Since the chain has only one communicating class,
all states are recurrent. ( We also note that if α = 0, then all states are transient.)
To determine whether the chain is null recurrent or positive recurrent, we need to calculate E[T00 ]. In Example 12.24, we did this by deriving the PMF PT00 (n).
∞In this problem,
it will be simpler to use the result of Problem 2.5.11 which says that k=0 P[K > k] =
E[K ] for any non-negative integer-valued random variable K . Applying this result, the
expected time to return to state 0 is
E [T00 ] =
∞
∞
1
.
> n] = 1 +
nα
P [T00
n=0
(3)
n=1
For 0 < α ≤ 1, 1/n α ≥ 1/n and it follows that
∞
1
E [T00 ] ≥ 1 +
n=1
n
= ∞.
(4)
We conclude that the Markov chain is null recurrent for 0 < α ≤ 1. On the other hand, for
α > 1,
∞
1
E [T00 ] = 2 +
.
(5)
nα
n=2
Note that for all n ≥ 2
1
≤
nα
This implies
E [T00 ] ≤ 2 +
=2+
∞
n
n−1
n
(6)
dx
xα
n=2 n−1
∞
dx
1
dx
xα
(7)
(8)
xα
∞
1
x −α+1
<∞
=2+
=2+
−α + 1 1
α−1
(9)
Thus for all α > 1, the Markov chain is positive recurrent.
Quiz 12.8
The number of customers in the ”friendly” store is given by the Markov chain
(1-p)(1-q)
(1-p)(1-q)
(1-p)(1-q)
p
p
1
0
(1-p)q
i+1
i
×××
(1-p)q
(1-p)(1-q)
p
(1-p)q
81
(1-p)q
×××
In the above chain, we note that (1 − p)q is the probability that no new customer arrives,
an existing customer gets one unit of service and then departs the store.
By applying Theorem 12.13 with state space partitioned between S = {0, 1, . . . , i} and
′
S = {i + 1, i + 2, . . .}, we see that for any state i ≥ 0,
πi p = πi+1 (1 − p)q.
(1)
p
πi .
(1 − p)q
(2)
This implies
πi+1 =
Since Equation (2) holds for i = 0, 1, . . ., we have that πi = π0 α i where
p
.
α=
(1 − p)q
(3)
Requiring the state probabilities to sum to 1, we have that for α < 1,
∞
i=0
πi = π0
∞
i=0
π0
= 1.
1−α
(4)
i = 0, 1, 2, . . .
(5)
αi =
Thus for α < 1, the limiting state probabilities are
πi = (1 − α)α i ,
In addition, for α ≥ 1 or, equivalently, p ≥ q/(1 − q), the limiting state probabilities do
not exist.
Quiz 12.9
The continuous time Markov chain describing the processor is
2
2
2
1
0
3.01
2
3
2
3
3
4
3
0.01
0.01
0.01
Note that q10 = 3.1 since the task completes at rate 3 per msec and the processor reboots
at rate 0.1 per msec and the rate to state 0 is the sum of those two rates. From the Markov
chain, we obtain the following useful equations for the stationary distribution.
5.01 p1 = 2 p0 + 3 p2
5.01 p3 = 2 p2 + 3 p4
5.01 p2 = 2 p1 + 3 p3
3.01 p4 = 2 p3
We can solve these equations by working backward and solving for p4 in terms of p3 , p3
in terms of p2 and so on, yielding
p4 =
20
p3
31
p3 =
620
p2
981
p2 =
82
19620
p1
31431
p1 =
628, 620
p0
1, 014, 381
(1)
Applying p0 + p1 + p2 + p3 + p4 = 1 yields p0 = 1, 014, 381/2, 443, 401 and the
stationary probabilities are
p0 = 0.4151 p1 = 0.2573 p2 = 0.1606 p3 = 0.1015 p4 = 0.0655
(2)
Quiz 12.10
The M/M/c/∞ queue has Markov chain
λ
λ
c+1
c
1
0
λ
λ
λ
µ
cµ
2µ
cµ
cµ
From the Markov chain, the stationary probabilities must satisfy
pn =
(ρ/n) pn−1 n = 1, 2, . . . , c
(ρ/c) pn−1 n = c + 1, c + 2, . . .
(1)
It is straightforward to show that this implies
p0 ρ n /n!
n = 1, 2, . . . , c
n−c c
p0 (ρ/c)
ρ /c! n = c + 1, c + 2, . . .
pn =
The requirement that
∞
n=0
(2)
pn = 1 yields
p0 =
c
n=0
ρ c ρ/c
ρ n /n! +
c! 1 − ρ/c
83
−1
(3)
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