B.2 Chebyshev’s Inequality 423Therefore,P[Z > 1 −a]≥ 1 −1 −μ
a =a+μ− 1
a.B.2 Chebyshev’s Inequality
Applying Markov’s inequality on the random variable (Z−E[Z])^2 we obtain
Chebyshev’s inequality:∀a > 0 , P[|Z−E[Z]|≥a] =P[(Z−E[Z])^2 ≥a^2 ]≤Var[Z]
a^2, (B.4)
where Var[Z] =E[(Z−E[Z])^2 ] is the variance ofZ.
Consider the random variablem^1∑m
i=1Zi. SinceZ^1 ,...,Zmare i.i.d. it is easy
to verify thatVar[
1
m∑mi=1Zi]
= Var[Z^1 ]
m.
Applying Chebyshev’s inequality, we obtain the following:lemmaB.2 LetZ 1 ,...,Zmbe a sequence of i.i.d. random variables and assume
thatE[Z 1 ] =μandVar[Z 1 ]≤ 1. Then, for anyδ∈(0,1), with probability of at
least 1 −δwe have
∣∣
∣∣
∣1
m∑mi=1Zi−μ∣∣
∣∣
∣
≤
√
1
δ m.
Proof Applying Chebyshev’s inequality we obtain that for alla > 0P
[∣∣
∣∣
∣
1
m∑mi=1Zi−μ∣∣
∣∣
∣
> a]
≤
Var[Z 1 ]
ma^2≤
1
ma^2.
The proof follows by denoting the right-hand sideδand solving fora.The deviation between the empirical average and the mean given previously
decreases polynomially withm. It is possible to obtain a significantly faster
decrease. In the sections that follow we derive bounds that decrease exponentially
fast.B.3 Chernoff’s Bounds
LetZ 1 ,...,Zmbe independent Bernoulli variables where for everyi,P[Zi= 1] =
piandP[Zi= 0] = 1−pi. Letp=∑m
i=1piand letZ=∑m
i=1Zi. Using the