6.262.PS1.sol

6.262.PS1.sol - 6.262 Discrete Stochastic Processes, Spring...

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6.262 Discrete Stochastic Processes, Spring 2010 Problem Set 1 — Solutions due: Friday, February 12, 2010 Problem 1 (Exercise 1.11) a We have two IID random variables X 1 and X 2 . One is strictly smaller than the other with probability 1 because P( X 1 = X 2 ) = 0. By symmetry, each variable is equally likely to be the smallest, so P ( X 1 < X 2 ) = 1 / 2 . b The answer is very similar to that of the previous problem, but let’s be a little more sys- tematic. With probability 1, the random variables X 1 ,X 2 ,...,X n can be ordered to form a strictly increasing finite sequence since the random variables are continuous. It follows that ( X 1 ,X 2 ,...,X n ) can be put in correspondence with all possible orderings of the set { 1 , 2 ,...,n } . For instance, for n = 3, the event { X 3 > X 1 > X 2 } then corresponds to (2 , 1 , 3). Now it’s easy: since the random variables are IID, all the n ! orderings of { 1 , 2 ,...,n } are equally likely. Since the events of interest correspond to having n in the n th position, it follows that the desired probability is ( n - 1)! /n ! = 1 /n . c Again, let’s be systematic. Let I n denote the indicator associated with the event that X n is the record to date. Then, E[ m n =1 I n ] yields the expected number of records in the sequences X 1 ,X 2 ,...,X m . Since expectation is linear, the quantity of interest becomes m n =1 E[ I n ] = m n =1 n - 1 . As m → ∞ , we obtain the (divergent) harmonic series, which tells us that the expected number of records-to-date in an infinite sequence is itself infinite. Problem 2 (Exercise 1.23) (a) At a suitably abstracted level, this problem is identical to the previous one. Can you see why the answer is ( n - 1)! /n ! = 1 /n ? (b) Let I k denote the indicator associated with the event that user k receives the correct pass- word. Then, E[ n k =1 I k ] yields the expected number of users who are able to continue using the system. Again, since expectation is linear, the quantity of interest becomes n k =1 E[ I k ] = n k =1 n - 1 = n/n = 1. Challenge question: can you compute the probability that exactly k users receive correct passwords? Problem 3 (Exercise 1.31) (a) We are interested in calculating the probability that S n is contained within an interval of fixed width 2 m + 1 centered at the mean, . As n increases, the standard deviation of S n increases as well (as n , in fact), and we expect the probability mass associated with the interval to vanish in the limit. 1
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Intuition is useful, but intuition can be wrong, so let’s do this precisely. Let Y n denote the zero-mean, unit-variance version of S n , that is, Y n = ( S n - ) / . (Note that σ = p var( X 1 ) = p δ (1 - δ ) does not vary with n .) Then, X - m i + m P ( S n = i ) = F S n ( + m ) - F S n ( - m ) - = F Y n ± m ² - F Y n ± - m ² - Note 1: To clarify some notation, F S n ( a ) = P( S n a ), while F S n ( a ) - = P( S n < a ). Can you see why
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6.262.PS1.sol - 6.262 Discrete Stochastic Processes, Spring...

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