Thus n j1 j1 j 2n 1 2 2 2n 1 e xn p p 2n 2n

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Unformatted text preview: Inequality) If X ∈ L2 , then P {|X | ≥ a} ≤ for every a > 0. 21–1 E(X 2 ) a2 Proof. Recall that X ∈ L1 if and only if |X | ∈ L1 . Since |X | ≥ 0, we can write |X | ≥ a1{|X |≥a} . Taking expectations of the previous expression implies E(|X |) ≥ aE(1{|X |≥a} ) = P {|X | ≥ a} and Markov’s inequality follows. Similarly, X 2 ≥ a2 1{X 2 ≥a2 } so that if X ∈ L2 , taking expectations implies ￿ ￿ E ( X 2 ) ≥ a2 P X 2 ≥ a2 . Hence, ￿ ￿ E(X 2 ) P {|X | ≥ a} ≤ P X 2 ≥ a ≤ a2 yielding Chebychev’s inequality. Definition. If X ∈ L2 we say that X has finite variance and define the variance of X to be Var(X ) = E[(X − E(X )]2 ) = E(X 2 ) − [E(X )]2 . Thus, Chebychev’s inequality sometimes takes the form P {|X − E(X )| ≥ a} ≤ Var(X ) . a2 Computing Expectations Having proved a number of the main expectation theorems, we will now take a small detour to discuss computing expectations. We will also show over the course of the next several lectures how the formulas fo...
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This document was uploaded on 04/07/2014.

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