Answer: according to Equation (4.46),
U sing Equation (4.52), we have
and
The results for the mean and variance are the same as those obtained in
Examples 4.1 and 4.5.
Ex ample 4. 15. Problem: repeat the above when X is exponentially distributed
with density function
Answer: the characteristic function X (t) in this case is
The moments are
which agree with the moment calculations carried out in Examples 4.2 and 4.6.
100 Fundamentals of Probability and Statistics for Engineers
X
t
Xn
k 0
ejtk
n
k
pk
1 pnk
Xn
k 0
n
k
pejtk
1 pnk
pejt
1 pn:
4 : 53
1
1
j
d
dt
pejt
1 pn
t 0
npejt
1 pn^1
pejt
t 0
np;
2
d^2
dt^2
pejt
1 pn
t 0
np
n 1 p 1 ;
^2 X 2 21 np
n 1 p 1 n^2 p^2 np
1 p:
fX
x
aeax; forx 0 ;
0 ; elsewhere:
X
t
Z 1
0
ejtx
aeaxdxa
Z 1
0
e
ajtxdx
a
ajt
: 4 : 54
1
1
j
d
dt
a
ajt
^
t 0
1
j
ja
ajt^2
"
t 0
1
a
;
2
d^2
dt^2
a
ajt
(^)
t 0
2
a^2
;
^2 X 2 21
1
a^2