6.262.PS2.sol

6.262.PS2.sol - 6.262 Discrete Stochastic Processes Spring 2010 Problem Set 2 Solutions due Friday Problem 1 a We have 1 1 and P(Xn = 0 = 1 n n

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6.262 Discrete Stochastic Processes, Spring 2010 Problem Set 2 — Solutions due: Friday, February 19, 2010 Problem 1 a We have P( X n = 1) = 1 n and P( X n = 0) = 1 - 1 n . Equivalently, looking at the cumulative distribution function, P( X n a ) = 0 a < 0 1 - 1 n 0 a < 1 1 1 a Consider the random variable Y distributed as P( Y = 0) = 1. The corresponding CDF is given by P( Y a ) = ± 0 a < 0 1 0 a Then, clearly, lim n →∞ P( X n a ) = P( Y a ) for all a R . Therefore, X n Y in distribution. Now let Z n = | X n - Y n | and note that P( Z n = 1) = 1 n and P( Z n = 0) = 1 - 1 n . Thus, E( | X n - Y | 2 ) = E( Z 2 n ) = 1 n , and therefore lim n →∞ E( | X n - Y | 2 ) = 0, from which we conclude that X n converges to Y in mean square. b We now have P( X n a ) = 0 a < 0 1 - 1 n 0 a < n 1 n a Letting Y be as above, we again obtain lim n →∞ P( X n a ) = P( Y a ) for all a R . However, letting Z n = | X n - Y n | , we obtain P( Z n = n ) = 1 n and P( Z n = 0) = 1 - 1 n , and therefore E( | X n - Y | 2 ) = E( Z 2 n ) = n 2 n = n . It follows that lim n →∞ E( | X n - Y | 2 ) 6 = 0, from which we conclude that X n does not converge to Y in mean square. 1
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c Recall that in the double-or-quarter game, P( X k = 2) = P( X i = 1 / 4) = 1 / 2 and W n = X 1 X 2 ...X n , assuming an initial bet of $1. First we claim that W n converges in distribution to Y , where Y is a random variable distributed as P( Y = 0) = 1. Here, it suffices to show that lim n →∞ P ( W n a ) = 1 for all a > 0. (Why are we not concerned about the case where a < 0? How about a = 0? Think carefully.) Taking complements, it is equivalent to show that lim n →∞ P ( W n > a ) = 0 for all a > 0. Fix a > 0 and let k be some positive integer such that a > (1 / 2) k . Then P( W n > a ) P( W n (1 / 2) k ). Now let n be even and large enough so that 1 k < n/ 2. Consider n coin tosses and note that in order to have W n > (1 / 2) k , the number of heads must at least equal to the number of tails, that is, { W n (1 / 2) k } ⊂ { H n n/ 2 } , (1) where H n denotes the number of heads in n tosses. One way to see this is to show the contrapositive, that is, taking complements, we wish to show that { W n < (1 / 2) k } ⊃ { H n < n/ 2 } . (2) This is easy: since W n = 2 H n (1 / 4) n - H n and 1 k < n/ 2, we have that H n < n/ 2 = W n < 2 n/ 2 (1 / 4) n - n/ 2 = (1 / 2) n/ 2 < (1 / 2) k , which is exactly what we need to show to prove (2). Taking probabilities in (1) yields that P( W n (1 / 2) k ) P( H n n/ 2) . Now by Chebyshev inequality, P( H n n/ 2) V ar ( H n ) ( n/ 2) 2 = n/ 4 ( n/ 2) 2 . The right-hand side of the above equation gets arbitrarily small as
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This note was uploaded on 10/21/2010 for the course EE 5581 taught by Professor Moon,j during the Spring '08 term at Minnesota.

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6.262.PS2.sol - 6.262 Discrete Stochastic Processes Spring 2010 Problem Set 2 Solutions due Friday Problem 1 a We have 1 1 and P(Xn = 0 = 1 n n

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