A chance of being abnormally large is small proof

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Unformatted text preview: a ) ≤ E [X ] , for all a > 0. a Chance of being abnormally large is small. Proof: Define a random variable Y as (draw a figure to compare the distributions of Y and X ) Y = 0, a, if X < a; if X ≥ a. Clearly, E [Y ] ≤ E [X ] and P (Y = a) = P (X ≥ a). Thus E [Y ] ≤ E [X ] ⇒ 0 · P (Y = 0) + a · P (Y = a) = a · P (X ≥ a) ≤ E [X ]. M. Chen ([email protected]) ENGG2430C lecture 10 5 / 29 Markov Inequality If a random variable X takes only nonnegative values, then P (X ≥ a ) ≤ E [X ] , for all a > 0. a Example (Umbrella problem): Each of n students chooses one out of n umbrellas at random. What is the probability that at least 3 students get their own umbrellas? Define a r.v. Y to be the number of students who choose their own umbrellas. We already know E [Y ] = 1. Then P (Y ≥ 3) ≤ M. Chen ([email protected]) E [Y ] 1 =. 3 3 ENGG2430C lecture 10 6 / 29 Markov Inequality Markov inequality may give loose bound. [Example 7.1 in textbook] Let X be a r.v. uniformly distributed in [0, 4] E [X ] = 2 Applying Markov inequality, we have...
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