Page 214 - Applied statistics and probability for engineers

192 Chapter 5/Joint Probability Distributions

Assuming that we can differentiate inside the summation and integral signs,
⎧ r tx
(
⎪
r
d M t) = ⎨ ∑ x x e f x), X discretee
(
X
dt r ⎪ ∞ ∫ x e f x dx( ) , X continuous
r tx
⎩ −∞
Now if we set t = 0 in this expression, we i nd that
r
d M t) t= = E X )
(
r
X
(
dt r 0
Example 5-36 Moment-Generating Function for a Binomial Random Variable Suppose that X has a
binomial distribution, that is
⎛ n⎞
−
f x) = ⎜ ⎟ p ( −1 p) n x , x = 0 1 ... n ,
x
, ,
(
⎝
x⎠
Determine the moment-generating function and use it to verify that the mean and variance of the binomial random
variable are μ = np and σ = np( − p).
2
1
From the deinition of a moment-generatingfunction, we have

n ⎛ n⎞ n ⎛ n⎞
M t ( ) = ∑ e tx ⎜ ⎟ p ( −1 p) n x = ∑ ⎜ ⎟ ( pe ) ( −1 p) n−x
−
−
t x
x
X
x=0 ⎝ x⎠ x=0 ⎝ x⎠
This last summation is the binomial expansion of [pe + ( − p )] , so
n
t
1
t
1
M t ( ) [= pe + ( − p)] n
X

Taking the irst and second derivatives, we obtain
′
t
t
X
M t ( ) = dM t ( ) = npe [ +1 p e ( −1 )] n−1
X
dt
and
2
″
X
t
t
M t ( ) = d M t ( ) = npe ( −1 p npe )[+ t 1 + p e ( −1 )] n−2
X
2
dt
′
If we set t = 0 in M t( ), we obtain
X
′
(
M t) | = = μ ′ = μ = np
X
1
t 0
n
which is the mean of the binomial random variable X. Now if we set t = 0 in M t ( ),
X
M t) | = = μ′ 2 = np( − p np)
+
n
1
(
X
t 0
Therefore, the variance of the binomial random variable is
σ = μ ′ 2 − μ 2 = np( 1− + np) − np) 2 = np np 2 = np( 1− p)
−
2
p
(
Example 5-37 Moment-Generating Function for a Normal Random Variable Find the moment-generating
function of the normal random variable and use it to show that the mean and variance of this random
2
variable are μ and σ , respectively.
The moment-generating function is
∞ 1
M t ( ) = ∫ e tx e x ( − −μ) 2 /(2È 2 )
X
−∞ σ 2π
∞
= ∫ 1 e x [ − 2 −2 (μ+ tÈ 2 x ) + μ 2 ]/(2σ ) dx
σ
2
−∞ σ 2π