SECTION 6.1 Discrete Random Variables 323
Lemma. (Markov’s Inequality)Let Xbe a non-negative discrete ran-
dom variable. Then for any numberd > 0 , we have
P(X≥d)≤d^1 E(X).Proof. We define a new random variableY by setting
Y =
d ifX≥d
0 otherwise..
SinceY ≤X, it follows thatE(X)≥E(Y). Also note thatY has two
possible values: 0 andd; furthermore,
E(Y) = dP(Y =d) = dP(X≥d).SinceE(X)≥E(Y) =dP(X≥d), the result following immediately.
Lemma.(Chebyshev’s Inequality)LetXbe a discrete random variable
with meanμand varianceσ^2. Then for anyd > 0 we have
P(|X−μ|≥d) ≤σ^2
d^2.
Proof. Define the random variable Y = (X −μ)^2 ; it follows that
E(Y) =σ^2. Applying Markov’s inequality toY we have
P(|X−μ|≥d) =P(Y ≥d^2 )≤d^12 E(Y) =σ^2
d^2,
as required.
We now assume thatX 1 , X 2 , ...,Xnare random variables with the
same meanμ; we denote the average of these random variables thus:
X =
X 1 +X 2 +···+Xn
n