Gambler's Ruin Problem. A gambler starts playing a two-outcome betting game, starting with an initial wealth x ∈ R. Each time the gambler bets $1, he wins with probability p ∈ (0,1) and will stop when the total wealth reaches either 0 (he goes bankrupt) or a fixed amount M > x. Of course, the resulting process is a Random Walk with absorbing states at 0 or M. Let {X_n}_{n≥1} be the result of the n-th game. Then the sequence {X_1, X_2, …} is i.i.d., each taking values +1 or −1 with probability p and q := 1 − p, respectively. The wealth at time n ≥ 0 is given by:
W_n = ∑_{i=0}^n X_i, X_0 = x.
It is easy to see that the “profit” process S_n = W_n − x is an equivalent process, starting at 0 and stopping when reached either −x or M − x.
a) Assume that the game is fair, i.e. p = 1/2. Show that S_n is a martingale.
b) Argue that τ = min{n ≥ 0 : S_n = −x or S_n = M − x} is a stopping time.
c) Assume that the game is fair, and let p_0 be the probability of ruin, or, put it differently, the probability that S_n reaches −x before reaching M − x. Find p_0 using the optional sampling theorem.
d) Let us further calculate the expected time to ruin for a fair game. First, show that Z_n = S_n^2 − n is a martingale. Then, apply the optional sampling theorem for Z_n to obtain E[τ].
e) Assume now that the game is not fair, i.e. p ≠ 1/2. In order to apply the same reasoning, we need to construct a martingale. Check if S_n is in this case a martingale, a supermartingale, or a submartingale. Prove that
Y_n = (q/p)^{S_n}
is a martingale. Further, use the optional sampling theorem to calculate the probability of ruin p_0.