Mathematical Expectation
Mathematical expectation of a particular random phenomenon basically means
the average value of the random phenomenon E(X) = 𝜇 = 𝑥 . Mathematical
expectation, also known as the expected value, is the summation or integration
of a possible values from a random variable.
(I) Discrete Case: The mathematical expectation of a discrete random variable
X having values 𝑥1 , 𝑥2, … , 𝑥n with respective probability 𝑃(𝑥1 ), P(𝑥2 ), … ,
P(𝑥n ) such that 𝑛𝑖=1 p 𝑥𝑖 = 1 , then the expectation of X denoted by E(X)
is defined as
E(X) = 𝑥1 𝑃(𝑥1 ) + 𝑥2 P(𝑥2 ) + … … + 𝑥n P(𝑥n )
E(X) = 𝑛𝑖=1 𝑥𝑖 p 𝑥𝑖
OR
The mathematical expectation of a discrete random variable X with probability
mass function pmf (probability mass function) p(x) is defined as E(X) =
𝑛
𝑖=1 𝑥𝑖 p 𝑥𝑖
1
Mathematical Expectation
(II) Continuous Case: The mathematical expectation of a continuous random
variable X having probability density function (pdf) f(x) is defined as
E(X) =
∞
𝑥𝑓
−∞
𝑥 𝑑𝑥
2
Expected value of function of a random variable (r.v.)
Consider a random variable X with probability mass function
pmf p(x) / probability density function pdf f(x) and if g(x) is a
function of a random variable X then E g(x) is defined as
E g(x) = g(x)𝑝(𝑥)
; for Discrete Case
E g(x) =
∞
g(x)𝑓
−∞
𝑥 𝑑𝑥
; for Continuous Case
3
Moments of a random variable
(I) Discrete Case: Let X be a discrete random variable then 𝑛𝑡ℎ moment of X
is defined as E(𝑋 𝑛 ) = 𝑥 𝑛 p(x) ; n= 1,2,3 ……
If n=1 , E(X) = 𝑥𝑝(𝑥)
If n=2, E(𝑋 2 ) = 𝑥 2 p(x)
Now, 𝑛𝑡ℎ central moment, E(X−𝜇)𝑛 = (𝑋 − 𝜇)𝑛 p(x)
If n=1, E(X-𝜇) = (𝑥 − 𝜇)𝑝(𝑥) = 𝑥𝑝(𝑥) - 𝜇 𝑝(𝑥) = E(X) - 𝜇 = 𝜇 - 𝜇 = 0
i.e., first central moment is always zero.
4
Moments of a random variable
𝑛𝑡ℎ central moment, E(X−𝜇)𝑛 = (𝑋 − 𝜇)𝑛 p(x)
If n = 2 then E(X−𝜇)2 = (𝑋 − 𝜇)2 p(x)
= (𝑥 2 − 2𝜇𝑥 + 𝜇2 )p(x)
= 𝑥 2 𝑝(𝑥) − 2𝜇 𝑥𝑝(𝑥) + 𝜇2 𝑝(𝑥)
= E(𝑋 2 ) – 2𝜇E(X) + 𝜇2
= E(𝑋 2 ) ) – 2𝜇𝜇 + 𝜇2
= E(𝑋 2 ) - 𝜇2
= E(𝑋 2 ) - 𝐸(𝑋) 2 = Var(X)
i.e., second central moment of a random variable X is the variance
𝑉𝑎𝑟(𝑋)
5
Moments of a random variable
(I) Continuous Case: Let X be a continuous random variable then 𝑛𝑡ℎ moment of
X is defined as
𝑛
E(𝑋 ) =
If n=1 ,
If n=2,
Now, 𝑛
∞
E(X) = −∞ 𝑥𝑓 𝑥
∞ 2
2
E(𝑋 ) = −∞ 𝑥 𝑓
𝑡ℎ
∞ 𝑛
𝑥 𝑓
−∞
𝑥 𝑑𝑥
; n= 1,2,3 ……
𝑑𝑥
; is the first moment of r.v. X i.e., E(X) / 𝑥 / 𝜇
𝑥 𝑑𝑥
; is the second moment of r.v. X
𝑛
central moment of the r.v. X , E(X−𝜇) =
If n=1, E(X-𝜇) =
∞
(𝑥
−∞
− 𝜇)𝑓 𝑥 𝑑𝑥 =
∞
𝑥𝑓
−∞
∞
(𝑥
−∞
𝑥 𝑑𝑥 - 𝜇
− 𝜇)𝑛 𝑓 𝑥 𝑑𝑥
∞
𝑓
−∞
= E(X) - 𝜇.1 = 𝜇 - 𝜇 = 0
i.e., first central moment of a r.v. X is always zero.
𝑥 𝑑𝑥
6
Moments of a random variable
𝑛
𝑡ℎ
𝑛
central moment of the r.v. X , E(X−𝜇) =
∞
(𝑥
−∞
− 𝜇)𝑛 𝑓 𝑥 𝑑𝑥
If n = 2 then E (𝑋 − 𝜇)2
=
=
=
∞
2𝑓 𝑥 𝑑𝑥
𝑥
−
𝜇
−∞
∞
2 − 2 × 𝜇 + 𝜇2 𝑓 𝑥
𝑥
−∞
∞ 2
∞
𝑥 𝑓 𝑥 𝑑𝑥 - 2𝜇 −∞ 𝑋
−∞
𝑑𝑥
𝑓 𝑥 𝑑𝑥 + 𝜇
2
∞
𝑓
−∞
𝑥 𝑑𝑥
= E[X2] - 2 𝜇. 𝜇+ 𝜇2.1
= E[X2] - 𝜇2
= E[X2] – [E[X]]2
= E[X2] - 𝜇2 = 𝜎𝑋2 =Var(X)
∴ 𝜎𝑋 = 𝑉𝑎𝑟(𝑋) = Standard deviation of X (Spread around the mean)
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Example1. If we toss two coins together then r.v., X = no. of heads. Find the variance of
tossing those coins.
Solution.
Sample Space = {TT, TH, HT, HH} ∴ 𝑋 = 0, 1, 2
1
2 1
1
P(X=0) = P(TT) = 4 ; P(X=1) = p(TH or HT) = 4 = 2 ; P(X=2) = P(HH) = 4
X
x1
0
x2
1
x3
2
P(xi)
xip(xi)
X2
𝑥𝑖2p(xi)
1
4
1
2
1
4
0
0
0
1
2
1
2
1
1
2
1
4
xip(xi) = 1
𝑥𝑖2 p(xi) = 1.5
We know, E(X) = 𝑥𝑝(𝑥) , E(𝑋 2 ) = 𝑥 2 p(x)
𝜎𝑋2 = 𝐸 𝑋2 − 𝐸 𝑋 2=1.5-1 = 0.5
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Example2. The random variable X has probability density function given by
f (X) =
3
10
3𝑋 − 𝑋 2
;0 ≤ 𝑥 ≤ 2
0
; 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Find the mean and variance of X.
Solution.
Mean = 𝜇 = 𝐸 𝑋
=
=
=
=
=
∞
𝑥𝑓 𝑥 𝑑𝑥
−∞
2
3
2
𝑥
3𝑥
−
𝑥
𝑑𝑥
0
10
2 9𝑥 2
3𝑥 3
− 10 𝑑𝑥
0 10
2
9 𝑥3
3 𝑥4
(10 . 3 − 10 . 4 )
0
3 3
3 4 2
3 3
(10 𝑥 − 40 𝑥 ) = (10 2
0
−
3 4
2 )
40
−0=
24
(10
−
48
)
40
6
5
= =2.4 – 1.2 = 1.2
9
We know, Variance, 𝜎𝑥2 = Var(X) = E(𝑋 2 ) - 𝐸(𝑋)
E[X2]
=
=
=
∞ 2
𝑥 𝑓 𝑥 𝑑𝑥
−∞
2 2 3
2
𝑥
3𝑥
−
𝑥
0
10
2 9𝑥 3
3𝑥 4
−
𝑑𝑥
0 10
10
2
9
3
= (
40
𝑥4
−
50
𝑥5)
0
𝑑𝑥
=(
9
40
2
24
−
3
50
25)
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒, 𝜎𝑥2 = Var(X) = E(𝑋 ) - 𝐸(𝑋)
= 1.68 – (1.2)2
=
=
2
42
25
6
25
-
–0=
42
25
= 1.68
2
6
5
= 0.24
10