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It Xhas a varianceσ2, then σ, the positive square root of the variance, is called the standard deviation. It Xhas to mean μand standard deviationσ, to show thatP{|X-μ|≥kσ}≤1k2.

Short Answer

Expert verified

Since|X-μ|≥0,

|X-μ|kσ2≥|X-μ|kσ≥I{|X-μ|≥kσ}⇓1k2=Eδ|X-μ|kσ2≥EI{|X-μ|>kσ}=P{|X-μ|≥kσ}

On the other hand, this result can be proven directly using Chebyshev's inequality. Sincekσ>0,

P{|X-μ|≥kσ}≤σ2(kσ)2=1k2.

Step by step solution

01

Given Information.

It Xhas a varianceσ2, thenσ, the positive square root of the variance is called the standard deviation. It Xhas to mean μand standard deviationσ,

02

Explanation.

Assume that the random variable Xhas meant μand standard deviationσ, and letk>0. Thenkσ>0.

Let

I{|X-μ|≥kσ}=1if|X-μ|≥kσ0,otherwise

Then

P{|X-μ|≥kσ}=EI{|X-μ|≥kσ}

Since|X-μ|>0note that.

|X-μ|kσ2≥|X-μ|kσ≥I{|X-μ|>kσ}E|X-μ|kσ2≥EI{|X-μ|≥kσ}1k2σ2E|X-μ|2Áåž=σ2≥P{|X-μ|≥kσ}â„•1k2≥P{|X-μ|≥kσ}

03

Explanation.

On the other hand, this result can be proven directly using Chebyshev's inequality. Sincekσ>0,

P{|X-μ|≥kσ}≤σ2(kσ)2=1k2.

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