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STAT_333_Assignment_3_Solutions

STAT_333_Assignment_3_Solutions - STAT 333 Assignment 3...

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STAT 333 Assignment 3 SOLUTIONS 1. Consider a discrete- time Markov Chain with state space S = {0, …, 7} and transition matrix P = 1/2 1/4 0 0 0 0 1/4 0 1/4 0 0 0 0 0 3/4 0 0 0 1/3 0 0 0 0 2/3 0 1/5 1/5 1/5 1/5 1/5 0 0 1/6 0 0 0 1/3 1/6 1/6 1/6 0 0 0 0 0 1 0 0 1/4 0 0 0 0 0 3/4 0 0 0 2/3 0 0 0 0 1/3 a. Determine the classes of this chain, which are open or closed, write P in simplified form, and find the period of each closed class. C 1 = {0, 1, 6} closed so positive recurrent aperiodic C 2 = {2, 7} closed so positive recurrent aperiodic C 3 = {3} open so transient C 4 = {4} open so transient C 5 = {5} closed so positive recurrent aperiodic P sim = 1/2 1/4 1/4 0 0 0 0 0 1/4 0 3/4 0 0 0 0 0 1/4 0 3/4 0 0 0 0 0 0 0 0 1/3 2/3 0 0 0 0 0 0 2/3 1/3 0 0 0 0 0 0 0 0 1 0 0 0 1/5 0 1/5 0 1/5 1/5 1/5 1/6 0 1/6 0 1/6 1/6 0 1/3 with new order of states {0, 1, 6, 2, 7, 5, 3, 4} b. Find the equilibrium distribution corresponding to each closed class. Write down the general form of all stationary distributions for this chain. Within C 2 , we can see the submatrix is doubly stochastic so equilibrium is (½ , ½) Within C 5 , there is only one state, so equilibrium is (1) Within C 1 , we get ½ π 0 + ¼ π 1 + ¼ π 6 = π 0 ¼ π 0 = π 1 π 0 + π 1 + π 6 = 1 solving gives (1/3, 1/12, 7/12) The general form is: α (1/3, 1/1 2 , 0, 0, 0, 0, 7/12, 0) + β (0, 0, ½, 0, 0, 0, 0, ½) + (1– α – β)(0, 0, 0, 0, 0, 1, 0, 0) where α, β, and (1 – α – β) are between 0 and 1.

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c. Find the absorption probabilities from each transient state into each closed class. A 4, C1 = P 4, C1 + P 4, 4 *A 4, C1 = (1/6 + 1/6) + 1/3 *A 4, C1 So A 4, C1 = ½ A 4, C2 = P 4, C2 + P 4, 4 *A 4, C2 = 1/6 + 1/3 *A 4, C2 So A 4, C1 = ¼ So A 4, C5 = 1 A 4, C1 A 4, C2 = ¼ A 3, C1 = P 3, C1 + P 3, 3 *A 3, C1 + P 3, 4 *A 4, C1 = 1/5 + 1/5 *A 3, C1 + 1/5*1/2 So A 3, C1 = 3/8 A 3, C2 = P 3, C2 + P 3, 3 *A 3, C2 + P 3, 4 *A 4, C2 = 1/5 + 1/5 *A 3, C1 + 1/5*1/4 So A 3, C1 = 5/16 So A 3, C5 = 1 A 3, C1 A 3, C2 = 5/16 d. Describe the long-run behaviour of the chain if X 0 = 0. Do the same if X 0 = 3. If X0 = 0, we are starting in closed class C1, so the long-run behaviour is (1/2, 1/12, 0, 0, 0, 0, 7/12, 0) If X0 = 3, the long-run behavior is (3/8)(1/3, 1/12, 0, 0, 0, 0, 7/12, 0)+(5/16)(0, 0, ½, 0, 0, 0, 0, ½)+(5/16)(0, 0, 0, 0, 0, 1,0,0) 2. Consider a discrete-time Markov Chain with five transient states (1, 2, 3, 4, and 5) and two recurrent states (6 and 7). The chain is equally likely to start in any of the transient states. The transition matrix is P = 1/2 0 1/4 0 0 0 1/4 0 1/3 0 1/3 0 1/3 0 1/5 1/5 1/5 1/5 1/5 0 0 0 0 0 1/4 1/4 1/4 1/4 2/3 0 1/3 0 0 0 0 0 0 0 0 0 1/2 1/2 0 0 0 0 0 1/3 2/3 Note: The R code used is available in a separate document.
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STAT_333_Assignment_3_Solutions - STAT 333 Assignment 3...

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