Random Variable - Random Signals and Processes Primer with MATLAB - page 126

The probability ( 2.463 ) corresponds to the shaded area plus the area of the delta
function in Fig. 2.59b .
The probability ( 2.463 ) can be rewritten as:

ð

1

3
8 ¼

5
8 ¼ 1 F X ð 1 Þ:

PfX> 1 g¼ 1 PfX 1 g¼ 1 f X ðxÞ d x ¼ 1 (2.464)

0

Exercise E.2.6 The PDF of the random variable X is given as:

f X ðxÞ¼a e bjxj ; 1< x <1;

(2.465)

where a and b are constants. Find the relation between a and b .

Answer From the PDF property ( 2.83 ), we have:

2 3

1 1 ð

0

1

2 a
b :

4 5 ¼

a e bjxj d x ¼ a e bx d x þ e bx d x f X ðxÞ d x ¼ 1 ¼ (2.466)

1 1 1

0

From ( 2.465 ), it follows b ¼ 2 a and

f X ðxÞ¼a e 2 ajxj ; 1< x <1: (2.467)

Exercise E.2.7 The random variable X is uniform in the interval [ 1, 1] and with
the probability equal to 0.25, it takes both values at the interval ends (i.e . 1 and 1).
Plot the PDF and distribution and find the probability that X is negative. Show this
result in the PDF and distribution plots.

Answer The random variable X is the mixed variable: continuous in the interval
[ 1, 1] and discrete in the points 1 and 1. It takes the discrete values 1 and 1,
both with a probability of 0.25. As a consequence, the PDF (Fig. 2.60a ) has two

Fig. 2.60 ( a ) PDF and ( b ) distribution of the mixed variable