67 loose 3 e x 05 very loose 4 intuitively why markov

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Unformatted text preview: P (X ≥ 2) ≤ P (X ≥ 3) ≤ P (X ≥ 4) ≤ E [X ] = 1 (very loose) 2 E [X ] = 0.67 (loose) 3 E [X ] = 0.5 (very loose) 4 Intuitively, why? Markov inequality only utilizes mean. M. Chen (IE@CUHK) ENGG2430C lecture 10 7 / 29 Chebyshev Inequality For a random variable X with mean µ and variance σ 2 , P (|X − µ | ≥ c ) ≤ σ2 , for all c > 0. c2 Chance of being far away from the mean is small. Proof: Define a random variable Y as Y = (X − µ )2 Noticing E [Y ] = E [(X − µ )2 ] = Var(X ) = σ 2 , we apply Markov inequality and get P (|X − µ | ≥ c ) = P (Y ≥ c 2 ) ≤ M. Chen (IE@CUHK) ENGG2430C lecture 10 E [Y ] σ 2 = 2. c2 c 8 / 29 Chebyshev Inequality For a random variable X with mean µ and variance σ 2 , P (|X − µ | ≥ c ) ≤ σ2 , for all c > 0. c2 Example (Umbrella problem): Define a r.v. Y to be the number of students who choose their own umbrellas. We already know E [Y ] = 1 and Var(Y ) = 1. Then P (Y ≥ 3) = P (|Y − 1| ≥ 2) ≤ M. Chen (IE@CUHK) ENG...
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This document was uploaded on 03/31/2014.

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