Mathematical Expectation
Mathematical Expectation
Mathematical Expectation
X 1 2 3 4 5 6
P(x 1/ 1/ 1/ 1/ 1/6 1/6
) 6 6 6 6
It satisfies two requirred
conditions for being pmf
(1) All the probabilities are
positive. Probability of
assuming any value is
always positive.
(2) All the probabilites add
up to 1.
E(X)=1X1/6+2X1/6+ ….
6X1/6= 6(6+1)/2X6=3.5
Mathematical expectation
Proof : E(X)= mean of x
Let there be a random
discrete variable X having
vaues x1, x2, x3….. xn.
The following table is
presented
Value frequency prob
x1 f1 f1/N
x2 f2 f2/N
x3 f3
xn fn fn/N
Sum N 1
Mean of x= (f1x1+f2x2+
fnxn)/N=x1(f1/N)+x2(f2/N)+
xn(fn/N)
=x1P(x=x1)+x2P(x=x2)
+xnP(x=xn)=E(X)=Mathema
tical expectation of X.
Formulae
E(a)=a=aX1=a.
Expectation of a constant is
that constant itself
Proof
Using probability
A constant is kind of
variable which assmes a
constant value with
probability 1
(1)E(aX)=aE(X)
Mathematical expectation
of product of a constant
and a variable is equal to
product of the constant and
mathematical expectation
of a variable.
(2)E(aX+b)=aE(X)+b
Mathematical expectation
of a sum of product of a
constant and a variable and
another constant is equal to
sum of product of the
constant and mathematical
expectation of the variable
and another constant.
(3) E(aX+by)=aE(X)+bE(Y)
A=1 and Y=1
E(X+b)= E(X)+b
Mathematical expectation
of a sum of product of a
constant and a variable and
product of another constant
and another variable is
equal to sum of product of
the constant and
mathematical expectation
of the variable and product
of mathematical
expectation of the another
variable and another
constant.
Xn Pn (Xn+c)
E(x+c)=(x1+c)p1+
(xn+c)pn
=(x1p1+x2p2+…..xnpn)
+c(p1+p2+…pn)
=E(X)+c.1
=E(X)+c
It does not a matter
whether one adds a
constant to a variable to
find out the mean of the
sum, or one add constant
to mean of the variable
Application
Mistakenly , a student has
calculated E(ax+b) instead
of E(bx+a). He does not the
have to time to calculate
E(bx+a). What adjustments
he needs to make to
E(ax+b) to derive E(bx+a)?
Explanation
You are supposed to
calculate E(2x+3)
But you calculated E(3x+2).
V(X)=E(Y)
Y=[{X-E(X)}]^2
1 -2 4
2 -1 1
3 0 0
4 1 1
5 2 4
Sum=10
Mean =2
Variance x=E(Y)=2.
Verbal statement of
V(X)=E [{X-E(X)}]^2
Variance of X is equal to
mathematical expectation
of square of difference
between the variable and
mathematical expectation
of the variable.
V(X)=E(X^2)-[E(X)]^2
Since V(X) is always positive.
E(X^2)>[E(X)]^2.
verify
If V(X) is zero, it means
E(X^2)=[E(X)]^2
X will be a constant
In that case x can not be a
variable, it will have to be a
constant.
2 4 2 4
24
2 4
2 4
2 4
4
V(X)= E(X^2)-[E(X)]^2
The verbal statement:
Variance of X is equal to
difference between
mathematical expectation
of square of X and square of
mathematical expectation
of X.
Application variance of a
constant is zero.
Variance of a variable is
defined as positive
Formule
V(cX)=c^2V(X)
Verbal statement
Variance of product of a
variable and a constant is
equal to product of ther
squatre of the variable and
variance of the variable
Proof
V(cX)=E[cX-E(cX)]^2
=c^2E[X-E(X)]^2
=c^2V(X)
V(5x)/V(X)=5 ?
V(ax+b)=V(ax-b)= a^2V(X)
V(ax+b)=V(ax+b)=V(aX)
+V(b)
=a^2V(X)+0=a^2V(X)
V(ax-b)=V(ax-b)=V(aX)-V(b)
=a^2V(X)-0=a^2V(X)
If covariance is zero, it
means that that the two
variables move
independently vis-vis each
other.
V(ax+by)=a^2V(X)+b^2V(Y)
+2abCov(X,Y)
V(ax-by)=a^2V(X)+b^2V(Y)-
2abCov(X,Y)
We will replace Cov(x,y) by
std(x)std(y) cor(X,Y)
V(ax+by)> V(ax-by)
V(ax+b)= V(ax-b)
V(a1x+b1)=V(a2x+b2)
Means a1=a2 but b1 and b2
need not be equal.
V(x)=V(-x)
AM of V(ax+by) and V(ax-
b)=a^2V(x)+b^2V(y)
[V(ax+by)-V(ax-by)]/
4ab=Cov(x,y)
Cov(x,y)=[V(2x+3y)-V(2x-
3y)]/24
Cov(x,y)=[V(a1x+b1y)-
V(a1x-b1y)]/4a1b1
Cov(x,y)=[V(2x+3y)-V(a1x-
b1y)]/24
Implies
a1=2 and b2=3
Cov(x,y)=[V(2x+b1y)-V(a2x-
3y)]/24
Implies
a2=2 and b1=3
V(ax+b)/V(cx+d)=a^2/c^2
V(2x+3)/V(2x-100000)=1
V(2x+3)/V(ax-100000)=1/2
Means 4/a^2=1/2
8/a^2=1
8=a^2
a=2 Xunderoot 2
Comment on Cov(X,Y) and
correlation between x and y
when
1) 4V(x)+9V(Y) > V(2x-3Y)
2) 4V(x)+9V(Y) < V(2x-3Y)
3) 4V(x)+9V(Y) = V(2x-3Y)
Sl no 5962
Unique paper code 227102.
C=10W+5X
Mean of W and X are 400
and 50. V(W)=36 and V(X)=
9. Coefficient of correlation
is(-0.4)
Compute the mean and
variance of C.
Ans.
E(C)= 10E(W)+5 E(X)=4250.
V(10W+5X)
=V(10W)+V(5X)+ 2
cov(10W,5X)
=100.V(W)+25V(X)
+(2.10.5)COV(X,Y)=100.36+
25.9+100(-7.2)=3600+225-
720=3105
Ans
Calculation of Cor(X,Y) is
provided below.
Cor(X,Y)= Cov(X,Y)/std(X)
std(Y)
Cov(X,Y)=
std(X)std(Y)cor(X,Y)=3.6.(--
0.4)=-7.2