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MIT6_042JF10_rec23_sol

# MIT6_042JF10_rec23_sol - 6.042/18.062J Mathematics for...

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6.042/18.062J Mathematics for Computer Science December 8, 2010 Tom Leighton and Marten van Dijk Notes for Recitation 23 Theorem 1. Let E 1 , . . . , E n be events, and let X be the number of these events that occur. Then: Ex ( X ) = Pr { E 1 } + Pr { E 2 } + . . . + Pr { E n } Theorem 2 (Markov’s Inequality) . Let X be a nonnegative random variable. If c > 0 , then: Ex ( X ) Pr { X c } ≤ c Theorem 3 (Chebyshev’s Inequality) . For all x > 0 and any random variable R , Var [ R ] Pr {| R Ex ( R ) | ≥ x } ≤ x 2 Theorem 4 (Union Bound) . For events E 1 , . . . , E n : Pr { E 1 . . . E n } ≤ Pr { E 1 } + . . . + Pr { E n } Theorem 5 (“Murphy’s Law”) . If events E 1 , . . . E n are mutually independent and X is the number of these events that occur, then: Pr { E 1 . . . E n } ≥ 1 e Ex( X ) Theorem 6 (Chernoff Bounds) . Let E 1 , . . . , E n be a collection of mutually independent events, and let X be the number of these events that occur. Then: Pr { X c Ex ( X ) } ≤ e ( c ln c c + 1) Ex ( X ) when c > 1

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Recitation 23 2 1 Getting Dressed Sometimes I forget a few items when I leave the house in the morning. For example, here are probabilities that I forget various pieces of footwear: left sock 0 . 2 right sock 0 . 1 left shoe 0 . 1 right shoe 0 . 3 a. Let X be the number of these that I forget. What is Ex ( X )? Solution. By Theorem 1 , the expected number of events that happen is the sum of the event probabilities. So, Ex ( X ) = 0 . 2 + 0 . 1 + 0 . 1 + 0 . 3 = 0 . 7 b. Upper bound the probability that I forget one or more items. Make no independence assumptions. Solution. By the Union Bound, the probability that I forget something is at most: 0 . 1 + 0 . 1 + 0 . 2 + 0 . 3 = 0 . 7 c. Use the Markov Inequality to upper bound the probability that I forget 3 or more items. Solution. Ex ( X ) 7 = Pr { X 3 } ≤ 3 30 d. Now suppose that I forget each item of footwear independently. Use Chebyshev’s Inequality to upper bound the probability that I forget two or more items. Solution. Let X 1 be the event I bring my left sock, X 2 my right sock, X 3 my left shoe, and X 4 my right shoe. Then Ex ( X 1 ) = . 2, Ex ( X 2 ) = . 1, Ex ( X 3 ) = . 1, and Ex ( X 4 ) = . 3. Moreover, since the X i are Bernoulli random variables (binomial with n = 1), we have Var [ X 1 ] = . 2(1 . 2) = . 16 , Var [ X 2 ] = . 1(1
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