Page 74 - Probability, Random Variables and Random Processes

RANDOM VARIABLES [CHAP 2

Since the integrand is an even function, we have

Then

Next,

Let y = x2/(2a2). Then dy = x dx/a2, and so

Hence, by Eq. (2.31),

2.35. Consider a continuous r.v. X with pdf f,(x). If fx(x) = 0 for x < 0, then show that, for any a > 0,
Clx
a
P(X 2 a) 5 -
where px = E(X). This is known as the Markov inequality.
From Eq. (2.23),

Since fx(x) = 0 for x < 0,

2.36. For any a > 0, show that

where px and ax2 are the mean and variance of X, respectively. This is known as the Chebyshev
inequality.
From Eq. (2.23),

By Eq. (2.29),