Page 117 - Fundamentals of Probability and Statistics for Engineers

100 Fundamentals of Probability and Statistics for Engineers

Answer: according to Equation (4.46),
n
n
X jtk k n k
X t†ˆ e p 1 p†
k
kˆ0
n
n
X n k 4:53†
jt k
ˆ pe † 1 p†
k
kˆ0
n
jt
ˆ‰ pe ‡ 1 p†Š :
Using Equation (4.52), we have

1 d n n 1
jt
jt
1 ˆ ‰ pe ‡ 1 p†Š ˆ n‰ pe ‡ 1 p†Š pe †
jt
j dt
tˆ0 tˆ0
ˆ np;
2
d jt n
2 ˆ ‰ pe ‡ 1 p†Š ˆ np‰ n 1†p ‡ 1Š;
dt 2
tˆ0
and
2 2
2
2
ˆ 2 ˆ np‰ n 1†p ‡ 1Š n p ˆ np 1 p†:
X 1
The results for the mean and variance are the same as those obtained in
Examples 4.1 and 4.5.

Example 4.15. Problem: repeat the above when X is exponentially distributed
with density function
ax
ae ; for x 0;
f x†ˆ
X
0; elsewhere:
Answer: the characteristic function X (t) in this case is
Z 1 Z 1 a
jtx
X t†ˆ e ae ax †dx ˆ a e a jt†x dx ˆ : 4:54†
0 0 a jt
The moments are
" #

1 d a 1 ja 1

1 ˆ ˆ ˆ ;
j dt a jt j a jt† 2 a
tˆ0 tˆ0
d 2 a 2
2 ˆ ˆ ;
dt 2 a jt a 2
tˆ0
1
2
2
ˆ 2 ˆ ;
X 1 2
a
which agree with the moment calculations carried out in Examples 4.2 and 4.6.
TLFeBOOK