Chebyshevs Lemma Assignment Help

Limit Theorems - Chebyshevs Lemma

Chebyshev's Lemma:

Let X be a random variable with mean μ and variance σ2. Then Chebyshev's lemma states that for any k > 0

P{¦X-μ¦≥ k} ≤ σ2/k2

Proof:

Let X be a continuous random variable with pdf f ( x ). We have

σ2 = Var (X) = 1328_Chebyshevs Lemma.png(x-μ)2 f(x)dx

= 1464_Chebyshevs Lemma1.png(x-μ)2 f(x)dx + 721_Chebyshevs Lemma2.png(x-μ)2 f(x)dx

+949_Chebyshevs Lemma3.png (x-μ)2 f(x)dx ≥  1464_Chebyshevs Lemma1.png (x-μ)2 f(x)dx + 949_Chebyshevs Lemma3.png(x-μ)2 f(x)dx

≥ k2 {1464_Chebyshevs Lemma1.png f(x) dx +949_Chebyshevs Lemma3.png f(x) dx}

since in both the ranges of integration ( x - μ )2 ≥ k2.

Hence

σ 2 ≥ k2 P(X ≤ μ - k ) u X ≥ μ +k) =k2 p(¦X- μ¦≥ k)

or

P(¦X-μ¦≥ k) ≤ σ2/k2

The inequality can also be proved by taking X to be a discrete random variable.


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