Other Important Random Variables - Random Signals and Processes Primer with MATLAB - page 335

Applying the moment's theorem we can obtain the first and second moments:

o¼ 0 ¼

1
j

d f x ð o Þ
d o

1
p ;

EfXg¼ (5.190)

o¼ 0 ¼

d 2 f x ð o Þ
d o 2

1
j 2

1 þ q
p 2

EfX 2 g¼ : (5.191)

From ( 5.190 ) and ( 5.191 ), we get a variance:

1 þ q
p 2

1
p 2 ¼

q
p 2 :

2

s 2 ¼ EfX 2

gEfXg ¼ (5.192)

Example 5.8.1 The message sent over the binary channel has symbols denoted as
“0” and “1,” where the probabilities of the occurrence of “0” or “1” are independent
and equal. For example, a sequence of symbols is given as:

0001101111110001111111000001110000000011000

... : ...

It is necessary to find the distribution and density functions for the random
variable which presents the number of identical symbols in a randomly chosen
group of identical symbols in the message. Additionally, find the probability that
will have at least k ¼ 5 identical symbols in a group of identical symbols.

Solution The number of successive identical symbols in a group is a discrete random
variable with discrete values of 1, 2, , m , , and corresponding probabilities of

... ...

PfX ¼ 1 g¼ 0 : 5

5 2
... ... ...
PfX ¼ mg¼ 0

PfX ¼ 2 g¼ 0 :

: (5.193)

5 m

:

... ... ...

It follows that the random variable X is a geometrical random variable with the
distribution and density functions,

F X ðxÞ¼ 1
m¼ 1

5 m uðx mÞ; 0 : (5.194)

f X ðxÞ¼ 1
m¼ 1

5 m dðx mÞ:

0 : (5.195)

The distribution and density functions are shown in Fig. 5.19 . From ( 5.190 ) the
average number of identical symbols in a randomly chosen group is: