The Question is,
For the gambling policies model, consider the "triple 'til you win" policy defined by,
$W_1 \equiv 1$; and for $n \ge 2$, $ W_n=3^{n-1}$ if $Z_1=...=Z_{n-1}=-1$ otherwise $W_n=0$.
($\{W_i\}$ are the fortune at each time i; and $\{Z_i\}$ are assumed to be i.i.d random variables indicating success or failure at each time i, i.e., $P(Z_i=1)=p$ and $P(Z_i=-1)=1-p$.)
(a) Prove that, with probability 1, the limit $\lim_{n \to \infty}X_n$ exists.
(b) Describe precisely the distribution of $\lim_{n \to \infty}X_n$.
I solved (a) like below.
Let, $\tau=inf\{n \ge 1; Z_n=+1 \}$, the first time game successes, and $X_n=a+W_1Z_1+...+W_nZ_n$ when $a$ is gambler's starting budget.
Then, $$X_n=a+({3^\tau}-\sum_{k=0}^{\tau-1}3^k)=a+{3^\tau +1 \over 2}$$ if $\tau \le n$. And $$P(\tau>n)=(1-p)^n$$ $$P(\tau<\infty)=1-P(\tau>\infty)=1-\lim_{n \to \infty}P(\tau>n)=1$$ Therefore, $P(\lim_{n \to \infty}X_n=a+{3^\tau +1 \over 2})=1$, implying that with probability 1, $\lim_{n \to \infty}X_n$ exists.
But I have no idea how to describe precisely the distribution of $\lim_{n \to \infty}X_n$. Thanks for any help.
If $X=\lim_{n \to \infty}X_n$ then I would have thought you had
$$P\left(X=a+{3^t +1 \over 2}\right) = p(1-p)^{t-1}$$ for positive integers $t$ as the distriution of the limit