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SolutionsII

# SolutionsII - ECE 496 SOLUTIONS TO HOMEWORK ASSIGNMENT II...

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ECE 496 SOLUTIONS TO HOMEWORK ASSIGNMENT II Spring 2007 1. Since this Markov chain has finitely many states, it’s convenient to solve parts of the problem using the transition matrix P defined by [ P ] ij = P ( i, j ), where P ( i, j ) is the probability of making a one-step transition from i to j . For this example, P = 2 6 6 4 r p 0 q q r p 0 0 q r p p 0 q r 3 7 7 5 . (a) We could do this by computing P 2 , the square of the transition matrix. Then P (2) ( i, j ) = [ P 2 ] ij for all i and j . However, it’s easier to look at the picture. To make a two-step transition from 1 to 2, you can either go 1 to 1 to 2 or 1 to 2 to 2. These are mutually exclusive possibilities, so their probabilities add to give P (2) (1 , 2) = rp + pr = 2 pr . The same reasoning applies to two-step transitions from 2 to 3, 3 to 4, and 4 to 1, so P (2) (1 , 2) = P (2) (2 , 3) = P (2) (3 , 4) = P (2) (4 , 1) = 2 pr Similarly, using the q ’s instead of the p , we get P (2) (1 , 4) = P (2) (2 , 1) = P (2) (3 , 2) = P (2) (4 , 3) = 2 qr For the i -to- i transitions, the only way to make any of them in two steps is to stay put or move to an “adjacent” state and back. For each i , this yields P (2) ( i, i ) = r 2 + 2 pq . To make a 1-to-3 transition, you can go either 1 to 2 to 3 or 1 to 4 to 3. The 3-to-1 transitions follow those paths in reverse. Because of the symmetries in the probabilities, the results come out the same, namely P (2) (1 , 3) = P (2) (3 , 1) = p 2 + q 2 . You’ll find that the same holds for the remaining two possibilities: P (2) (2 , 4) = P (2) (4 , 2) = p 2 + q 2 . Note that P 4 j =1 P (2) ( i, j ) = 1 for all i . Just to check this for i = 1, the sum is r 2 + 2 pq + 2 pr + p 2 + q 2 + 2 pq = ( p + q + r ) 2 = 1 , (b) Here are the possible 4-step trips from 1 to 3 that don’t hit 3 until step 4, along with their probabilities: 1-1-1-2-3: r 2 p 2 1-1-2-2-3: rprp = r 2 p 2 1-2-2-2-3: pr 2 p = r 2 p 2 1-2-1-2-3: pqpp = p 3 q 1-1-1-4-3: r 2 q 2 1-1-4-4-3: r 2 q 2 1-4-4-4-3: r 2 q 2 1-4-1-4-3: q 3 p Add these together and you get f (4) (1 , 3). 1

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(c) The chain is “generic” in the sense that it has only one recurrence class, and that class is positively recurrent, so the chain has a unique stationary distribution π * , which satisfies 4 X i =1 π * i P ( i, j ) = π * j for all j . Thinking of π * as a row vector, that’s the same as π * P = π * , where P is the transition matrix above. The equations read ( r - 1) π * 1 + * 2 + * 4 = 0 * 1 + ( r - 1) π * 2 + * 3 = 0 * 2 + ( r - 1) π * 3 + * 4 = 0 * 1 + * 3 + ( r - 1) π * 4 = 0 The unique distribution solving these equations is π * i = 1 / 4 for all i . This might not seem too surprising considering how symmetric the chain is. (d) From the convergence theorems in class, 1 m j = π * j for all j , so the limiting fraction of the time the chain spends in each state is just 1/4 for every state. Again, not so surprising. 2. (a) States 3 and 4 are clearly transient, because you can leave the “3-4 cluster” (and therefore never return) with positive probability. States 1 and 2 are recurrent — starting from state 1, the only way you could leave 1 and never return is to go to 2 and stay there forever, which has probability (1 - δ ) = 0, where I’m assuming and δ are both positive. Similarly, 5, 6, and 7 are recurrent. The only
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