STAT
Test%201-Solutions

# Test%201-Solutions - 3(a Assume a probability distribution...

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3.(a) Assume a probability distribution p satisfies the detailed balance condition. Then for any j S , X i S p i P i,j = X i S p j P j,i = p j X i S P j,i = p j · 1 = p j . Thus, p is a stationary distribution of the MC. (b) Method 1: Let p be a stationary distribution for the MC, and take p to be the initial distribution of the MC. For any i = 0 , 1 , 2 , ... , P ( X 0 i ) = P ( X 1 i ) = i j =0 p j . Notice that since the transition probability matrix is tridiagonal, X 0 i but X 1 i if and only if X 0 = i and X 1 = i +1. Similarly, X 0 i but X 1 i if and only if X 0 = i + 1 and X 1 = i . Thus, it follows that 0 = P ( X 0 i ) - P ( X 1 i ) = P ( X 0 i, X 1 i ) - P ( X 0 i, X 1 i ) = P ( X 0 = i, X 1 = i + 1) - P ( X 0 = i + 1 , X 1 = i ) = P ( X 1 = i + 1 | X 0 = i ) P ( X 0 = i ) - P ( X 1 = i | X 0 = i + 1) P ( X 0 = i + 1) = p i P i,i +1 - p i +1 P i +1 ,i . Therefore, the detailed balance condition is satisfied for all adjacent states i and j . For i, j such that | i - j | > 1, P i,j = P j,i = 0, and so the detailed balance condition is also satisfied in this case.
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• Fall '13
• Inductive Reasoning, Markov chain, Necessary and sufficient condition, stationary distribution, Πi, balance condition

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