Page 222 - Applied Statistics And Probability For Engineers

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5-10

t
n
This last summation is the binomial expansion of [pe (1 p)] , so
t
M 1t2 3pe 11 p24 n
X
Taking the first and second derivatives, we obtain

dM 1t2
t
t
M¿ 1t2 X npe 31 p1e 124 n 1
dt
X
and
2
d M 1t2
t
t
t
M–1t2 X npe 11 p npe 231 p1e 124 n 2
dt
X 2
If we set t 0 in M¿ 1t2 , we obtain
X
M¿ 1t20 t 0 ¿ np
1
X
which is the mean of the binomial random variable X. Now if we set t 0 in M–1t2,
X
M–1t20 t 0 ¿ np11 p np2
2
X
Therefore, the variance of the binomial random variable is
2
2
2
2
¿ np11 p np2 1np2 np np np11 p2
2
EXAMPLE S5-6 Find the moment generating function of the normal random variable and use it to show that
2
the mean and variance of this random variable are and , respectively.
The moment generating function is

1
2
2
M 1t2 e tx 12 e 1x 2
12 2 dx
X

1 3x 21 t 2x 4
12 2
2
2
2
2
12 e dx

If we complete the square in the exponent, we have
2
2 4
2
2
2
2
2
x 21 t 2x 3x 1 t 24 2 t t
and then

1
2 4
2
2 2
2
M 1t2 12 e 53x 1 t 24 2 t t 6
12 2 dx
X

1
2 2
2 2
e t t
2 12 e 11
223x 1 t 24
2 dx