HW2solns - HW2 Problem Solutions#2.1.6 First we check the...

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Unformatted text preview: February 24, 2007 HW2 Problem Solutions #2.1.6. First, we check the sufficiency of the condition X j ∈ A l p ij = q kl (not depending on i ) ∀ k, l ∀ i ∈ A k ( * ) for the HMC property of the ˆ X n chain. It suffices to calculate P ( ˆ X n +1 = l | ˆ X n = k, ˆ X m = k m , ≤ m < n ) = X i ∈ A k X j ∈ A l P ( X n = i | ˆ X n = k, ˆ X m = k m , ≤ m < n ) · · P ( X n +1 = j | X n = i, ˆ X m = k m , ≤ m < n ) and in this double-sum, the first conditional probability in the summand does not depend on j while the second conditional probability does not depend on the state-values ( k m , ≤ m < n ) because of the Markov property and its sum over j ∈ A l is equal to q kl by the assumption (*). We conclude that the double sum is equal to q kl X i ∈ A k P ( X n = i | ˆ X n = k, ˆ X m = k m , ≤ m < n ) = q kl which completes the proof. As mentioned in the assignment (after the correction was inserted), the necessity assertion in #2.1.6 is actually false. To see this rigorously, assume that A 1 = A contains at least 2 states, while all other partition classes A k , k ≥ 2 contain only one element each, and suppose that p bi = r i (not depending on b ) ∀ b ∈ S ∀ i ∈ A ( ] ) Then we note, whether (*) holds or not , that P ( ˆ X n +1 = l | ˆ X n = k, ˆ X m = k m , ≤ m < n ) is equal to ∑ j ∈ A l p ij ≡ q kl by the HMC property of X n whenever k ≥ 2 with A k = { i } , while if k = 1 , n ≥ 1 this displayed probability is equal to X i ∈ A X b ∈ A k n- 1 P ( X n- 1 = b, X n = i | ˆ X n = k, ˆ X m = k m , ≤ m < n ) · · P ( X n +1 ∈ A l | X n- 1 = b, X n = i, ˆ X m = k m , ≤ m < n- 1) = X i ∈ A X b ∈ A k n- 1 P ( X n +1 ∈ A l | X n = i ) p bi...
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This note was uploaded on 09/09/2009 for the course STAT 650 taught by Professor Slud during the Spring '09 term at Maryland.

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HW2solns - HW2 Problem Solutions#2.1.6 First we check the...

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