Random Variable - Random Signals and Processes Primer with MATLAB - page 36

Solution Let us observe all values of x keeping in mind the following intervals:

(a) 1 < x <1:

(b) x ¼ 1.
(c) 0 < x < 1 :

(d) x ¼ 0.
(e) 1< x <

:

0

(a) For the values of X in the interval:

1 < x <1; (2.12)

the event

f X x g (2.13)

is a certain event because the values of the random variable are 0 and 1, and
thus always less than x- defined in ( 2.12 )-yielding:

PfX xg¼ 1 for 1 < x <1: (2.14)

(b) For x ¼ 1 (from ( 2.11 )),

PðX ¼ 1 Þ¼p ¼ 3 = 4 : (2.15)

(c) For the values of x in the interval

0 < x < 1 ; (2.16)

the probability of event X x f g corresponds to the probability that the r.v.

takes the value 0,

PfX xg¼PfX ¼ 0 g¼ 1 = 4 for 0 < x < 1 : (2.17)

(d) For x ¼ 0 (from ( 2.11 )),

PðX ¼ 0 Þ¼ 1 = 4 : (2.18)

(e) For the interval

1< x < 0 ; (2.19)

the event X x f g is impossible because the r.v. X is never negative.

Therefore,

PfX xg¼ 0 for 1 x < 0 : (2.20)