Local Behaviour of First Passage Probabilities - MIMS - The ...

p 361 of [6]), so we conclude thatlim lim sup c n P 1 = 0:!0 n!1Exactly the same argument applies to P 3 , and since q u (v) is clearly boundedbelow by a positive constant for u; v 2 [D 1 ; D] we have shown that (26) holds,except that the RHS is multiplied by some constant C: However if C 6= 1, bysumming over y we easily get a contradiction, and this …nishes the proof.5 Proof of Theorem 25.1 Proof when x=c n ! 0:Proof. As already indicated, the proof involves applying the estimates in Proposition12 to the representation (16), which we recall isP (T x = n + 1) = X y0P (S n = x y; T x > n)F (y); x 0; n > 0: (41)In the case < 1, given " > 0 we can …nd K " and n " such that nF (K " c n ) "for n n " and, using (23) from Proposition 12,P (S n = xy; T x > n) 2U(x)f(0)V (y)nc nfor all x_y "c n and n n " : (42)We can then use (24) of Proposition 12 to show that we can also assume, increasingthe value of n " if necessary, that for x "c n and y 2 ("c n ; K " c n );1 " c nP (S n = x y; T x > n)U(x)P ( > n)~p(y n ) 1 + " for all n n " : (43)For …xed " it is clear that as n ! 1XZ K"Z K"~p(y n )F (y)=c n v ~p(z)F (c n z)dz v n 1 ~p(z)z dz: (44)"c""n n)F (y) = 0;"#0 n!1 U(x)P ( = n)y2[0;"c n][[K "c n;1)(45)it will follow from (43) that P (T x = n) v U(x)k 11 1 P ( = n): Since this holdsin particular for x = 0; we see that k 11 = ; so it remains only to verify (45).Note …rst thatXP (S n = x y; T x > n)F (y) F (K " c n )P (T x > n)y2[K "c n;1)"n 1 P (T x > n);14