Page 57 - Schaum's Outlines - Probability, Random Variables And Random Processes

RANDOM VARIABLES [CHAP 2

Fig. 2-10

2.6. Verify (a) Eq. (2.1 0); (b) Eq. (2.1 1); (c) Eq. (2.1 2).
(a) Since (X _< b) = (X I a) u (a < X _< b) and (X I a) n (a < X 5 h) = @, we have
b)
P(X I h) = P(X 5 a) + P(u < X I
or F,y(b) = FX(a) + P(u < X I h)
Thus, P(u < X 5 b) = Fx(h) - FX(u)
n
(b) Since (X 5 a) u (X > a) = S and (X I (X > a) = a, we have
a)
P(X S a) + P(X > a) = P(S) = 1
Thus, P(X > a) = 1 - P(X 5 a) = 1 - Fx(u)

(c) Now P(X < h) = P[lim X 5 h - E] = lim P(X I b - E)
c-0 c+O
c>O E>O
= lim Fx(h - E) = Fx(b - )
8-0
8: > 0

2.7. Show that
(a) P(a i X i b) = P(X = a) + Fx(b) - Fx(a)
(b) P(a < X < b) = Fx(b) - F,(a) - P(X = h)
(c) P(a i X < b) = P(X = u) + Fx(b) - Fx(a) - P(X = b)
(a) Using Eqs. (1.23) and (2.10), we have
P(a I X I h) = P[(X = u) u (a < X I b)]
= P(X = u) + P(a < X 5 b)
= P(X = a) + F,y(h) - FX(a)
(b) We have
P(a < X 5 b) = P[(u < X c h) u (X = b)]
= P(u < X < h) + P(X = b)