Other Important Random Variables - Random Signals and Processes Primer with MATLAB - page 328

We can easily verify that the condition ( 2.83 ) is satisfied.
From ( 5.147 ), we write

1

P X ðkÞ¼ 1
k¼ 0

¼ e lt 1
k¼ 0

k k

ðltÞ ðltÞ

e lt

: (5.154)

k ! k !

k¼ 0

Knowing that

1

x k
k !

¼ e x

; (5.155)

k¼ 0

from ( 5.154 ), we finally get:

1

P X ðkÞ¼ e lt e lt

¼ 1 : (5.156)

k¼ 0

5.7.4 Characteristic Function

We calculate a characteristic function in order to find the variance of the Poisson
random variable.

g¼ 1
k¼ 0

e jok PfX ¼ kg¼ e lt 1
k¼ 0

k

ðltÞ

f X ðoÞ¼Ef e joX e jok

k !

¼ e lt 1
k¼ 0

k

ðlt e jo Þ

: (5.157)

k !

Applying ( 5.155 ), we finally arrive at:

f X ðoÞ¼ e lt e e jo lt ¼ e ltð e jo

1 Þ ¼ e kð e jo

1 Þ :

(5.158)

Using the moment theorem, we have:

o¼ 0 ¼ lt ¼ k:

1
j

d f X ð o Þ
d o

EfXg¼ (5.159)

This result is the same as ( 5.148 ).
The second moment,

o¼ 0 ¼ðltÞ

d 2 f X ð o Þ
d o 2

1
j 2

2

2

EfX 2

g¼ þ lt ¼ k þ k: (5.160)