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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 &gt; 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 &gt; 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...
View Full Document

## This document was uploaded on 03/31/2014.

Ask a homework question - tutors are online