Download Chebyshev`s inequality Let X be a random variable taking

Probability and Statistics
Grinshpan
Chebyshev’s inequality
Let X be a random variable taking nonnegative values and having a finite expectation.
For simplicity, we will assume that X is discrete (the general case is treated similarly).
Let xi ≥ 0 be the values of X, and consider the mean value
∑
E[X] =
xi P (X = xi ).
i
Every term in the above sum is nonnegative. If we omit the terms with xi below a fixed bound t > 0,
the net value of the sum would not be increased:
∑
∑
xi P (X = xi ) ≥
xi P (X = xi ).
i: xi ≥t
i
Observe that
∑
xi P (X = xi ) ≥
i: xi ≥t
∑
tP (X = xi ) = tP (X ≥ t).
i: xi ≥t
Therefore, putting the chain of inequalities together,
E[X]
,
t > 0.
P (X ≥ t) ≤
t
Actually, this estimate is only meaningful for t > E[X]. We may also write it in the form
E[X]
.
t
Let now Y be a (real) random variable without value restrictions, and let µ be its mean.
P (X < t) ≥ 1 −
Then X = (Y − µ)2 is nonnegative. So, according to the derived estimate,
E[(Y − µ)2 ]
,
t > 0.
t
√
Letting s = t, and noting that (Y − µ)2 ≥ t is equivalent to |Y − µ| ≥ s, we obtain
P ((Y − µ)2 ≥ t) ≤
P (|Y − µ| ≥ s) ≤
Var(Y )
.
s2
This is Chebyshev’s inequality. It is meaningful for any Y with finite mean and variance and s > σ,
where σ is the standard deviation of Y.
Var(Y )
For the complementary event |Y − µ| < s, we have P (|Y − µ| < s) ≥ 1 −
.
s2
A restatement of Chebyshev’s inequality is often in use. If s = kσ, the inequality becomes
1
P (|Y − µ| ≥ kσ) ≤ 2 .
k
For instance, the probability that Y falls two standard deviations away from its mean is under 1/4.