Random Variable - Random Signals and Processes Primer with MATLAB - page 117

Given MGF, the moments are derived, by taking derivatives with respect to s ,
and evaluating the result at s ¼ 0,

s¼ 0 ;

d M X ð s Þ
d s

m 1 ¼

s¼ 0 ;

d 2 M X ð s Þ
d s 2

m 2 ¼

s¼ 0 ;

d 3 M X ð s Þ
d s 3

(2.425)

m 3 ¼

...

s¼ 0 :

d n M X ð s Þ
d s n

m n ¼

When the MGF is evaluated at s ¼ jo , the result is the Fourier transform, i.e.,
MGF function becomes the characteristic function. As a difference to MGF, the
characteristic function always converges, and this is the reason why it is often used
instead of MGF, [PEE93, p. 82].

Example 2.8.13 Find the MGF function of the exponential random variable and
find its variance using ( 2.425 ).

Solution

1 1

e sx l e lx d x ¼ l 1
0

M X ðsÞ¼Ef e sX e sx f X ðxÞ d x ¼ e ðlsÞx d x: g¼ (2.426)

1 0

This integral converges for Re{s} < l , where Re{ s } means the real value of s .

From ( 2.426 ), we easily found:

l
l s :

M X ðsÞ¼ (2.427)

The first and second moments are obtained taking the first and second
derivatives of ( 2.427 ), as indicated in ( 2.428 ).

s¼ 0 ¼

s¼ 0 ¼ s¼ 0 ¼

d M X ð s Þ
d s

d
d s

l
l s

l
ðl sÞ

1
l ;

m 1 ¼

2

! s¼ 0 ¼

s¼ 0 ¼

s¼ 0 ¼

d 2 M X ð s Þ
d s 2

d
d s

l
ðl sÞ

2 l
ðl sÞ

2
l 2 :

m 2 ¼ (2.428)

2 3