Stochastic

Chapter 2 poisson processes 21 exponential

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Unformatted text preview: be the time of the first visit to y and let hN (x) = Px (VN < V0 ). By considering what happens on the first step, we can write hN (x) = px hN (x + 1) + rx hN (x) + qx hN (x 1) Set hN (1) = cN and solve this equation to conclude that 0 is recurrent if and P1 Qy 1 Q0 only if y=1 x=1 qx /px = 1 where by convention x=1 = 1. 1.71. To see what the conditions in the last problem say we will now consider ↵ some concrete examples. Let px = 1/2, qx = e cx /2, rx = 1/2 qx for x 1 and p0 = 1. For large x, qx ⇡ (1 cx ↵ )/2, but the exponential formulation keeps the probabilities nonnegative and makes the problem easier to solve. Show that the chain is recurrent if ↵ > 1 or if ↵ = 1 and c 1 but is transient otherwise. 1.72. Consider the Markov chain with state space {0, 1, 2, . . .} and transition probability ✓ ◆ 1 1 1 for m 0 p(m, m + 1) = 2 m+2 ✓ ◆ 1 1 p(m, m 1) = 1+ for m 1 2 m+2 and p(0, 0) = 1 p(0, 1) = 3/4. Find the stationary distribution ⇡ . 1.73. Consider the Markov chain with state space {1, 2, . . .} and transition probability p(m, m + 1) = m/(2m + 2) f...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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