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# Chebyshevs Lemma Assignment Help

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Limit Theorems - Chebyshevs Lemma
*Chebyshev's Lemma:*

Let X be a random variable with mean μ and variance σ^{2}. Then Chebyshev's lemma states that for any k > 0

P{¦X-μ¦≥ k} ≤ σ^{2}/k^{2}

**Proof:**

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

σ^{2} = Var (X) = (x-μ)^{2} f(x)dx

= (x-μ)^{2} f(x)dx + (x-μ)^{2} f(x)dx

+ (x-μ)^{2} f(x)dx ≥ (x-μ)^{2} f(x)dx + (x-μ)^{2} f(x)dx

≥ k^{2 }{ f(x) dx + f(x) dx}

since in both the ranges of integration ( x - μ )^{2} ≥ k^{2}.

Hence

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

or

P(¦X-μ¦≥ k) ≤ σ^{2}/k^{2}

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

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*Chebyshev's Lemma:*^{2}. Then Chebyshev's lemma states that for any k > 0

^{2}/k

^{2}

**Proof:**

^{2}= Var (X) = (x-μ)

^{2}f(x)dx

^{2}f(x)dx + (x-μ)

^{2}f(x)dx

^{2}f(x)dx ≥ (x-μ)

^{2}f(x)dx + (x-μ)

^{2}f(x)dx

^{2 }{ f(x) dx + f(x) dx}

^{2}≥ k

^{2}.

^{2}≥ k

^{2}P(X ≤ μ - k ) u X ≥ μ +k) =k

^{2}p(¦X- μ¦≥ k)

^{2}/k

^{2}