1 Random Walks Conditioned to Stay Positive Bob Keener Let S n be a random walk formed by summing i.i.d. integer valued random variables X i, i 1: S n = X 1 + + X n. If the drift EX i is negative, then S n as n. If A n is the event that S k 0 for k = 1,..., n, then P (A n ) 0 as n. In this talk we will consider conditional distributions for the random walk given A n. The main result will show that finite dimensional distributions for the random walk given A n converge to those for a time homogeneous Markov chain on {0, 1,...}.
2 Exponential Families Let h denote the mass function for an integer valued random variable X under P 0. Assume that E 0 X = xh(x) = 0, and define e ψ(ω) = E 0 e ωx = x e ωx h(x). (1) Then f ω (x) = h(x) exp[ωx ψ(ω)], is a probability mass function whenever ω Ω = {ω : ψ(ω) < }.. Let X, X 1,... be i.i.d. under P ω with marginal mass function f ω. Differentiating (1), ψ (ω)e ψ(ω) = x xe ωx h(x), and so E ω X = ψ (ω). Similarly, Var ω (X) = ψ (ω). Note that since ψ > 0, ψ is increasing, and since ψ (0) = E 0 X = 0, ψ (ω) < 0 when ω < 0.
3 Classical Large Deviation Theory Exponential Tilting: Let S n = X 1 + + X n and X n = S n /n. Since X 1,..., X n have joint mass function n ( h(xi ) exp[ωx i ψ(ω)] ) [ n ] = h(x i ) exp[ωs n nψ(ω)], and i=1 In particular, i=1 E ω f(x 1,..., X n ) = E 0 f(x 1,..., X n ) exp[ωs n nψ(ω)]. P ω (S n 0) = e nψ(ω) E 0 [e ωs n ; S n 0]. For notation, E[Y ; A] def = E[Y 1 A ]. Using this, it is easy to argue that if ω < 0, 1 n log P ω(s n 0) ψ(ω), or P ω (S n 0) = e nψ(ω) e o(n). Also, for any ɛ > 0, P ω (X n > ɛ X n 0) 0, as n. [For regularity, need 0 Ω o.]
4 Refinements Local Central Limit Theorem: If the distribution for X is lattice with span 1, then as n, P 0 [S n = k] 1/ 2πnψ (0). Using this, for ω < 0, P ω (S n 0) = e nψ(ω) E 0 [e ωs n ; S n 0] e nψ(ω) = k=0 e ωk 2πnψ (0) e nψ(ω) (1 e ω ) 2πnψ (0), and P (S n = k S n 0) (1 e ω )e ωk, the mass function for a geometric distribution with success probability e ω.
5 Goal, Sufficiency, and Notation Let τ = inf{n : S n < 0} and note that τ > n if and only if S j 0, j = 1,..., n. Goal: Study the behavior of the random walk given τ > n for large n. Sufficiency: Under P ω, the conditional distribution for X 1,..., X n given S n does not depend on ω. New Measures: Under P ω (a), the summands X 1, X 2,... are still i.i.d. with common mass function f ω, but S n = a + X 1 + + X n. Finally, P n (a,b) denotes conditional probability under P ω (a) this measure, S k goes from a to b in n steps. given S n = b. Under
6 Positive Drift If ω > 0, then E ω X > 0. In this case, P ω (a) (τ = ) > 0, and conditioning on τ = is simple. For x 0, P (a) ω (S 1 = x τ = ) = P (a) ω (S 1 = x, τ = ) P (a) ω (τ = ) = P (a) ω (S 1 = x) P ω (x) (τ = ) P ω (a) (τ = ). Given τ =, the process S n, n 0, is a random walk, and the conditional transition kernel is an h-transform of the original transition kernel for S n.
7 Simple Random Walk If X i = ±1 with p = P ω (X i = 1), q = 1 p = P ω (X i = 1), and ω = 1 2 log(p/q), then S n, n 0, is a simple random walk. Theorem: For a simple random walk, as n, P (a,b) n (τ > n) 2(a + 1)(b + 1). n Proof. By the reflection principle, P (a) 0 (τ < n, S n = b) = P (a) 0 (S n = b 2). Dividing by P 0 (S n = b) (a binomial probability), P (a,b) n (τ < n) = Result follows from Stirling s formula. ( n+b a 2 ( n a b 2 2 ) (! n b+a )! ).! 2 ) (! n+a+b+2 2
8 Result for Simple Random Walks Let Y n, n 0, be a Markov chain on {0, 1,...} with Y 0 = 0 and transition matrix 0 1 0 0 0 1/4 0 3/4 0 0 0 2/6 0 4/6 0 0 0 3/8 0 5/8........... Theorem: For a simple random walk with p < 1/2, P ω (S 1 = y 1,..., S j = y j τ > n) P (Y 1 = y 1,..., Y j = y j ) as n.
9 Proof: Let B = {S 1 = y 1,..., S j = y j = y} with y 1 = 1, y i 0, and y i+1 y i = 1. Then Use this in P (0,b) n (B, τ > n) = P 0(B, τ > n, S n = b) P 0 (S n = b) = (1/2)j P (y,b) n j (τ > n j)p 0(S n j = b y) P 0 (S n = b) 2(y + 1)(b + 1) 2 j. n P ω (B τ > n) = b P (0,b) n (B, τ > n)p ω (S n = b) b P (0,b) n (τ > n)p ω (S n = b).
10 General Results For the general case, let Y n, n 0, be a stationary Markov chain with Y 0 = 0 and P (Y n+1 = z Y n = y) = P 0 (X = z y) E(z) 0 (z S τ ) E (y) 0 (y S τ ). Remark: the transition kernel for Y is an h-transform of the kernel for the random walk under P 0. Theorem: For ω < 0, as n. P ω (S 1 = y 1,..., S k = y k τ > n) P (Y 1 = y 1,..., Y k = y k ) Theorem: Let τ + (b) = inf{n : S n > b}. For ω < 0, P ω (S n = b τ > n) ebω E 0 S τ + (b). k=0 ekω E 0 S τ + (k)
11 Approximate Reflection: P (a,b) n (τ > n), b of order n. Proposition: If 0 b = O( n), P (a,b) n (τ > n) = 2b nψ (0) E(a) 0 (a S τ ) + o ( 1/ n ). Proof: Let g k denote the P 0 mass function for S k, and define L k (x) = P n (a,b) (S k = x) P n (a, b) (S k = x) = g n k(b x)g n ( b a) g n k ( b x)g n (b a). Then L k (S k ) is dp n (a,b) /dp n (a, b) restricted to σ(s k ) or σ(x 1,..., X k ), and P n (a,b) (τ n) = E (a, b) n L τ (S τ ) = E (a, b) n [ 1 2b nψ (0) (a S τ ) + o ( 1/ n ) Finish by arguing that E n (a, b) S τ E (a) 0 S τ. ].
12 Approximate Reflection: P n (a,b) (τ > n), b of order one. Corollary: As n, P (a,b) n (τ > n) = 2 nψ (0) E(a) 0 (a S τ )E (b) 0 (b S τ ) + o(1/n). Proof: Take m = n/2. Then P (a,b) n (τ > n) = c=0 P (a,b) n (S m = c)p m (a,c) (τ > m)p n m(τ (b,c) > n m). Result follows using the prior result since S m under P n (a,b) normal. is approximately
13 References Keener (1992). Limit theorems for random walks conditioned to stay positive. Ann. Probab. Iglehart (1974). Functional central limit theorems for random walks conditioned to stay positive. Ann. Probab. Durrett (1980). Conditioned limit theorems for random walks with negative drift. Z. Wahrsch. verw. Gebiete Bertoin and Doney (1992). On conditioning a random walk to stay nonnegative. Ann. Probab.