Question

Problem 1.0 For the gamblers ruin problem, let Ma denote the mean number of games that must be played until the game ends (either the gambler goes broke or wins all the money) given that the gamble starts with d dollars, d0,..N. Recall that N is the total amount of money in the game, and using the Section 1.3.3 notation, (a) Show that Mo = MN = 0 and Md = 1 + pMy+1 +gMd-1 for d=1,2, A-1. (b) Use (a) to show that when p = q = Note: Proof by substitution is not acceptable!! Also, when p the solution is (not necessary to show this) 4-р

Answer 1

Here, Solution :- Let, - Mo=MN=0 Md = 1 + PMd to +9 M-isl, --N-I © since, Ptq=1, o gives (Ptq) Md = 1+ PMd to tq PMd tqMd = 1+ PMd to tq Md-1 PMats -PMas qMo -q Mady- Mato - Md = (a/p) (Md - Ma-1) 11/26 - (ii) Mo = 0; d=in (ii) M2 - Mo = (culp) (Mo - Mo). + - Calp) Mi-4p M3-M2 = (a/p) (M2-M1)-1/P = (all) { (alp) Mo - Up }="/p = [alp? Mo-alp="/p My – m3 = [UL] [M3-Ma) - 1/p * {0] [coule)?m, -4122-Vp] -VP .. - Calp?m, - 9 //3 - 2/ , /p.

My - Ma-1-( alp) (Mod4 - Md -2) - Vp -(alpst mi- vp content content -- +1) where, :-2, 3, ...N. pasta From Ma-Ma-1 we have i Ma - Mon af model, q=P Carnesto M, [ci-colpson) q&P. wher, paq , 1-2,3, --N. =