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zz-33 - Utah State University ECE 6010 Stochastic Processes...

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Utah State University ECE 6010 Stochastic Processes Homework # 11 Solutions 1. Let M n denote the sequence of sample means from an iid random process X n : M n = X 1 + X 2 + · · · + X n n . (a) Is M n a Markov process? M n = 1 n n X i =1 X i = 1 n [ X n + ( n - 1) M n - 1 ] = 1 n X n + (1 - 1 n ) M n - 1 Clearly if M n - 1 is given then M n depends only on X n and is independent of M n - 2 , M n - 3 , . . . . There- fore, M n is a Markov process. (b) If the answer to part a is yes, find the following state transition pdf: f M n ( x | M n - 1 = y ). f M n ( x | M n - 1 = y ) = P [ x < M n x + dx | M n - 1 = y ] = P [ x < 1 n + (1 - 1 n ) y x + dx ] = P [ nx - ( n - 1) y < x n nx - ( n - 1) y + dx ] = f x ( nx - ( n - 1) y ) dx 2. An urn initially contains five black balls and five white balls. The following experiment is repeated indef- initely: A ball is drawn from the urn; if the ball is white it is put back in the urn, otherwise it is left out. Let X n be the number of black balls remaining in the urn after n draws from the urn. (a) Is X n a Markov process? If so, find the apropriate transition probabilities. (b) Do the transition probabilities depend on n ? The number X n of black balls in the urn completely specifies the probability of the outcomes of a trial; therefore X n is independent of its past values and X n is a Markov proces. P [ X n = 4 | X n - 1 = 5] = 5 10 = 1 - P [ X n = 5 | X n - 1 = 5] P [ X n = 3 | X n - 1 = 4] = 4 9 = 1 - P [ X n = 4 | X n - 1 = 4] P [ X n = 2 | X n - 1 = 3] = 3 8 = 1 - P [ X n = 3 | X n - 1 = 3] P [ X n = 1 | X n - 1 = 2] = 2 7 = 1 - P [ X n = 2 | X n - 1 = 2] P [ X n = 0 | X n - 1 = 1] = 1 6 = 1 - P [ X n = 1 | X n - 1 = 1] P [ X n = 0 | X n - 1 = 0] = 1 All the transition probability are independent of time.
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