Random Variable - Random Signals and Processes Primer with MATLAB - page 121

The desired probability corresponds to the shaded area under the density func-
tion as shown in Fig. 2.54 ,

ð ð ð ð

2 3 2 3

1
6 d yþ

1
6 ðyÞ d y ¼

1
3 <

3
4 :

PfjYj 2 g¼ f Y ðyÞ d yþ f Y ðyÞ d y ¼ (2.443)

3 2 3 2

Comparing results ( 2.438 ) and ( 2.443 ), we note that in the absence of more
knowledge of the random variable, the result provided by Chebyshev inequality is
quite crude.
In this example knowing only mean value and the variance, we found that the
desired probability is less than 3/4. Knowing that the variable has a uniform density
we can find the exact value of probability, which is 1/3.

Example 2.8.15 Consider the case that the variance of the random variable X is
s X ¼ 0. We know that in this case there is no random variable any more, because all
values of random variables are equal to its mean value. We will use the Chebyshev
inequality to prove this statement.

From ( 2.430 ), we have:

PfjX m X jeg 0 : (2.444)

However, the probability cannot be less than zero, and the equality stands in
( 2.444 ), i.e.,

PfjX m X jeg¼ 0 : (2.445)

In other words, the probability that the random variable takes the value different
than its mean value m X is zero, i.e., all values of the random variable are equal to its
mean value.

2.9 Numerical Exercises

Exercise E.2.1 A telephone call can come in random during a time interval from
0 P.M. until 8 P.M. Find the PDF and the distribution of the random variable X ,
defined in the following way,

1 call in the interval ½ 2 ; 5 ;

X ¼

(2.446)

0 otherwise

if the probability that the call appears in the interval [2, 5] is:

p ¼ð 5 2 Þ= 8 ¼ 3 = 8 : (2.447)