9 We show next that the existence of π implies that all states are non null and

# 9 we show next that the existence of π implies that

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We show next that the existence of π implies that all states are non-null and that π i = 1 μ i for each i . Suppose that X 0 has distribution π , so that P { X 0 = i } = π i for each i . Then π j μ j = P { X 0 = j } X n =1 P { T j n | X 0 = j } = X n =1 P { T j n, X 0 = j } . 10
However, P { T j 1 , X 0 = j } = P { X 0 = j } , and for n 2 , P { T j n, X 0 = j } = P { X 0 = j, X m 6 = j for 1 m n - 1 } = P { X m 6 = j for 1 m n - 1 } - P { X m 6 = j for 0 m n - 1 } = P { X m 6 = j for 0 m n - 2 } - P { X m 6 = j for 0 m n - 1 } by stationarity = a n - 2 - a n - 1 where a n = P { X m 6 = j for 0 m n } . 11

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Sum over n to obtain π j μ j = P { X 0 = j } + P { X 0 6 = j } - lim n →∞ a n = 1 - lim n →∞ a n . However, a n P { X m 6 = j for all m } = 0 as n → ∞ , by the persistence of j . We have shown that π j μ j = 1 , so that μ j = 1 π j < if π j > 0 . To see that π j > 0 for all j , suppose on the contrary that π j = 0 for some j . 12
Then 0 = π j = X i π i p ij ( n ) π i p ij ( n ) for all i and n , yielding that π i = 0 whenever i j . The chain is assumed irreducible, so that π i = 0 for all i in contradiction of the fact that π i ’s sum to 1 . Hence μ j < and all states of the chain are non-null. Furthermore, we see that π j are specified uniquely as 1 μ j . 13

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Thus, if π exists then it is unique and all the states of the chain are non-null persistent. Conversely, if the states of the chain are non-null persistent then the chain has a stationary distribution given by Lemma (1.3). 14
Proposition 1.5. If i j then i is null persistent if and only if j is null persistent. Proof. Let C ( i ) be the irreducible closed equivalence class of states which contains the non-null persistent state i . Suppose that X 0 C ( i ) . Then X n C ( i ) for all n , and Lemma (1.3) and Proposition (1.2) combine to tell us that all the states in C ( i ) are non-null. 15

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Proposition 1.6. Let s S be any state of an irreducible chain. The chain is transient if and only if there exists a non-zero solution { y i : i 6 = s } , satisfying | y i | ≤ 1 for all i , to the equations y i = X j : j 6 = s p ij y j , i 6 = s. 16
1.0.1 Example: Random walk with retaining barrier A particle performs a random walk on the non-negative integers with a retaining barrier at 0 . The transition probabilities are p 0 , 0 = q, p i,i +1 = p for i 0 p i,i - 1 = q for i 1 , Let ρ = p/q . (a) If q < p , take s = 0 to see that y i = 1 - 1 ρ i satisfies the equation in Proposition (1.6), and so the chain is transient. 17

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(b) Solve the equation π = π P to find that there exists a stationary distribution, with π j = ρ j (1 - ρ ) , if and only if q > p . Thus the chain is non-null persistent if and only if q > p . (c) If q = p = 1 2 the chain is persistent since symmetric random walk is persistent (just reflect negative excursions of a symmetric random walk into the positive half-line). Solve the equation x = xP to find that x i = 1 for all i is the solution, unique up to a multiplicative constant. However, i x i = so that the chain is null by Proposition (1.4).
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