Digital Signal Processing Reference
In-Depth Information
Example 3.2.2 Find the joint distribution in Example 3.1.2.
Solution All outcomes have the same probability of occurrence, resulting in:
PfX 1 ¼ 1
; X 2 ¼ 1 g¼PfX 1 ¼ 0
; X 2 ¼ 0 g¼PfX 1 ¼ 1
; X 2 ¼ 0 g
¼ PfX 1 ¼ 0
; X 2 ¼ 1 1
=
4
:
(3.19)
From ( 3.19 ), we have:
F X 1 X 2 ðx 1 ; x 2 Þ¼þ 1
4 ½uðx 1 1 Þuðx 2 1 Þþuðx 1 1 Þuðx 2 Þþuðx 1 Þuðx 2 1 Þ
þ uðx 1 Þuðx 2 Þ:
=
(3.20)
Example 3.2.3 Find the joint distribution in Example 3.1.3.
Solution All outcomes have the same probability of occurrence. However, the
probabilities of the values of the random variables are:
Pfx 1 ¼ 1
; x 2 ¼ 1 g¼Pf H
;
T gþPf T
;
H 1
=
4 þ 1
=
4 ¼ 1
=
;
2
Pfx 1 ¼ 1 ; x 2 ¼ 0 g¼Pf H ; H 1 = 4 ;
Pfx 1 ¼ 0 ; x 2 ¼ 1 g¼Pf T ; T 1 = 4 :
(3.21)
The joint distribution is:
F X 1 X 2 ðx 1 ; x 2 Þ¼ 1
=
2 uðx 1 1 Þuðx 2 1 Þþ 1
=
4 uðx 1 1 Þuðx 2 Þþ 1
=
4 uðx 1 Þuðx 2 1 Þ:
(3.22)
Example 3.2.4 Find the probability that a random point ( x 1 , x 2 ) will fall in the area
A , as shown in Fig. 3.6 .
Solution Area A is defined as:
fx 1 a; x 2 cgfx 1 a; x 2 bg:
(3.23)
Fig. 3.6 Area A in Example 3.2.4
Search WWH ::




Custom Search