Essentials of Stochastic Processes, 2nd ed. (draft) by Richard Durrett

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By Richard Durrett

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Then as n → ∞, pn (x, y) → π(y). Proof. Let S be the state space for p. Define a transition probability p¯ on S × S by p¯((x1 , y1 ), (x2 , y2 )) = p(x1 , x2 )p(y1 , y2 ) In words, each coordinate moves independently. Step 1. We will first show that if p is aperiodic and irreducible then p¯ is irreducible. Since p is irreducible, there are K, L, so that pK (x1 , x2 ) > 0 and pL (y1 , y2 ) > 0. Since x2 and y2 have period 1, it 54 CHAPTER 1. 9 that if M is large, then pL+M (x2 , x2 ) > 0 and pK+M (y2 , y2 ) > 0, so p¯K+L+M ((x1 , y1 ), (x2 , y2 )) > 0 Step 2.

In symbols, if n is odd then pn (x, x) = 0 for all x. 22. Renewal chain. We will explain the name later. ” Let fk be a distribution on the positive integers and let p(0, k − 1) = fk . For 34 CHAPTER 1. MARKOV CHAINS states i > 0 we let p(i, i − 1) = 1. In words the chain jumps from 0 to k − 1 with probability fk and then walks back to 0 one step at a time. If X0 = 0 and the jump is to k − 1 then it returns to 0 at time k. If say f5 = f15 = 1/2 then pn (0, 0) = 0 unless n is a multiple of 5. 8. The period of a state is the largest number that will divide all the n ≥ 1 for which pn (x, x) > 0.

5. 036, we lose exactly one of our sales. 623. Suppose we use a 0,3 inventory policy. 2 dollars per day. 11 dollars per day. 11 so the 1,3 inventory policy is optimal. 1 CHAPTER 1. 9. A transition matrix p is said to be doubly stochastic if its COLUMNS sum to 1, or in symbols x p(x, y) = 1. , y p(x, y) = 1. 12. If p is a doubly stochastic transition probability for a Markov chain with N states, then the uniform distribution, π(x) = 1/N for all x, is a stationary distribution. Proof. To check this claim we note that if π(x) = 1/N then π(x)p(x, y) = x 1 N p(x, y) = x 1 = π(y) N Looking at the second equality we see that conversely, if the stationary distribution is uniform then p is doubly stochastic.

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