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final09sol

# final09sol - STAT455/855 Fall 2009 Applied Stochastic...

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STAT455/855 Fall 2009 Applied Stochastic Processes Final Exam, Solutions 1. (15 marks) (a) 3 marks) True . Take the simple random walk with p = 1 / 2. (b) (3 marks) True . If state k were recurrent then the vector ρ ( k ) = ( ρ i ( k )) i S , where ρ i ( k ) is the mean number of visits to state i during a sojourn from k back to k , satisfies ρ ( k ) = ρ ( k ) P . (c) (3 marks) False . Take S to be the integers and for each i S suppose that p i,i +1 > 0, p i,i - 1 > 0, p i,i +2 > 0, and p i,i - 2 > 0. Then from state i one can make the transitions i i +1 i so that p ii (2) > 0. One can also make the transitions i i + 2 i + 1 i so that p ii (3) > 0. Since the greatest common divisor of 2 and 3 is 1, the period must be 1. (d) (3 marks) False . Take S to be the integers and for each i S suppose that p ii = r , where 0 < r < 1, p i,i +1 = (1 - r ) / 2, and p i,i - 1 = (1 - r ) / 2. If a stationary distribution existed then this chain is time reversible because for any states i and j and any path from i to j , the reverse path will have the same probability. So the stationary distribution must satisfy the local balance equations, which are π i 1 - r 2 = π i - 1 1 - r 2 for all i . These reduce to π i = π i - 1 for all i . Thus, all components of π must be the same. But this is impossible since S is infinite. (e) (3 marks) False . Take the simple, symmetric random walk ( p = 1 / 2), or the counterexample from part(d), where P is symmetric but no stationary distribution exists. (f) (2 bonus marks) True . If X were time reversible with stationary distribution π and P were symmetric then the local balance equations π i p ij = π j p ji would reduce to π i = π j for all i, j S , since p ij = p ji . That is, all components of π would have to be equal, which is impossible if S is infinite.

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STAT 455/855 -- Final Exam Solutions, 2009 Page 2 of 7 2. (15 marks) We check the local balance equations for all parts. (a) (5 marks) The local balance equations are π i n 2 - 1 = π j n 2 - 1 for all i, j such that j is equal to i except with two components interchanged. Thus, we get π i = π j for all i, j S . Since S has n ! elements, we get π i = 1 n ! for all i S.
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final09sol - STAT455/855 Fall 2009 Applied Stochastic...

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