By Richard Durrett
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Extra resources for Essentials of Stochastic Processes, 2nd ed. (draft)
Let Nn (y) be the number of visits to y at times ≤ n. As n → ∞ 1 Nn (y) → n Ey Ty Why is this true? Suppose first that we start at y. The times between returns, t1 , t2 , . . 15) k If we do not start at y then t1 < ∞ and t2 , t3 , . . 15). Writing ak ∼ bk when ak /bk → 1 we have R(k) ∼ kEy Ty . Taking k = n/Ey Ty we see that there are about n/Ey Ty returns by time n. Proof. 15). To turn this into the desired result, we note that from the definition of R(k) it follows that R(Nn (y)) ≤ n < R(Nn (y) + 1).
Tx }. Since X(Tx ) = X0 = x it follows that µ = µp. 2: Picture of the cycle trick. Proof. To formalize this intuition, let p¯n (x, y) = Px (Xn = y, Tx > n) and interchange sums to get ∞ µ(y)p(y, z) = y p¯n (x, y)p(y, z) n=0 y 56 CHAPTER 1. MARKOV CHAINS Case 1. Consider the generic case first: z = x. p¯n (x, y)p(y, z) = y Px (Xn = y, Tx > n, Xn+1 = z) y = Px (Tx > n + 1, Xn+1 = z) = p¯n+1 (x, z) Here the second equality holds since the chain must be somewhere at time n, and the third is just the definition of p¯n+1 .
Proof. Let T = Va ∧ Vb . 3 that Px (T < ∞) = 1 for all x ∈ C. 16) implies that h(x) = Ex h(X1 ) when x = a, b. The Markov property implies h(x) = Ex h(XT ∧n ).