80 Probability and DistributionsProof.In Theorem 1.10.2 takeu(X)=(X−μ)^2 andc=k^2 σ^2 .ThenwehaveP[(X−μ)^2 ≥k^2 σ^2 ]≤E[(X−μ)^2 ]
k^2 σ^2.Since the numerator of the right-hand member of the preceding inequality isσ^2 ,the
inequality may be written
P(|X−μ|≥kσ)≤1
k^2
,which is the desired result. Naturally, we would take the positive numberkto be
greater than 1 to have an inequality of interest.
Hence, the number 1/k^2 is an upper bound for the probabilityP(|X−μ|≥kσ).
In the following example this upper bound and the exact value of the probability
are compared in special instances.
Example 1.10.1.LetXhave the uniform pdf
f(x)={ 1
2
√
3 −√
3 <x<√
3
0elsewhere.Based on Example 1.9.1, for this uniform distribution, we haveμ=0andσ^2 =1.
Ifk=^32 , we have the exact probability
P(|X−μ|≥kσ)=P(
|X|≥3
2)
=1−∫ 3 / 2− 3 / 21
2√
3dx=1−√
3
2.By Chebyshev’s inequality, this probability has the upper bound 1/k^2 =^49 .Since
1 −
√
3 /2=0.134, approximately, the exact probability in this case is considerably
less than the upper bound^49 .Ifwetakek= 2, we have the exact probability
P(|X−μ|≥ 2 σ)=P(|X|≥2) = 0. This again is considerably less than the upper
bound 1/k^2 =^14 provided by Chebyshev’s inequality.
In each of the instances in Example 1.10.1, the probabilityP(|X−μ|≥kσ)and
its upper bound 1/k^2 differ considerably. This suggests that this inequality might
be made sharper. However, if we want an inequality that holds for everyk> 0
and holds for all random variables having a finite variance, such an improvement is
impossible, as is shown by the following example.
Example 1.10.2.Let the random variableXof the discrete type have probabilities
1
8 ,
6
8 ,1
8 at the pointsx=−^1 ,^0 ,1, respectively. Hereμ=0andσ(^2) = 1
4 .Ifk=2,
then 1/k^2 =^14 andP(|X−μ|≥kσ)=P(|X|≥1) =^14. That is, the probability
P(|X−μ|≥kσ) here attains the upper bound 1/k^2 =^14. Hence the inequality
cannot be improved without further assumptions about the distribution ofX.
A convenient form of Chebyshev’s Inequality is found by takingkσ= for >0.
Then Equation (1.10.2) becomes
P(|X−μ|≥ )≤
σ^2
2
, for all > 0. (1.10.3)
The second inequality of this section involves convex functions.