Given random variables $Y_1, Y_2, \ldots \stackrel{iid}{\sim} P(Y_i = 1) = p = 1 - q = 1 - P(Y_i = -1)$ where $p > q$ in a filtered probability space $(\Omega, \mathscr F, \{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)$ where $\mathscr F_n = \mathscr F_n^Y$,
define $X = (X_n)_{n \ge 0}$ where $X_0 = 0$ and $X_n = \sum_{i=1}^{n} Y_i$.
It can be shown that the stochastic process $M = (M_n)_{n \ge 0}$ where $M_n = X_n - n(p-q)$ is a $(\{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)$-martingale.
Let $b$ be a positive integer and $T:= \inf\{n: X_n = b\}$.
It can be shown that $T$ is a $\{\mathscr F_n\}_{n \in \mathbb N}$-stopping time.
Prove that $E[T] < \infty$.
One proposition to use is 'What always stands a reasonable chance of happening will (almost surely) happen - sooner rather than later' (or here)
So let us show either that $\exists N \in \mathbb N, \epsilon > 0$ s.t. $\forall n \in \mathbb N$,
$$P(T \le n + N \mid \mathscr F_n) > \epsilon$$
or the weaker condition that $\exists N \in \mathbb N, \epsilon > 0$ s.t. $\forall n \in \mathbb N$,
$$P(T > kN) \le (1 - \epsilon)^k$$
I tried the first one:
$$P(T \le \infty \mid \mathscr F_n) = E(1_{T \le \infty} \mid \mathscr F_n) = \sum_{i=1}^{n} 1_{T=i} + \sum_{i=n+1}^{\infty} E[1_{T=i} \mid \mathscr F_n]$$
Hence, we must find and integer $N$ and a positive number $\epsilon$ s.t.
$$P(T \le n + N \mid \mathscr F_n) = E(1_{T \le n + N} \mid \mathscr F_n) = \sum_{i=1}^{n} 1_{T=i} + \sum_{i=n+1}^{n+N} E[1_{T=i} \mid \mathscr F_n] > \epsilon$$
where $\forall i > n$,
$$E[1_{T=i} \mid \mathscr F_n] = P(T=i \mid \mathscr F_n)$$
$$= P(X_i = b, X_1 \ne b, X_2 \ne b, \ldots, X_n \ne b, \ldots, X_{i-1} \ne b \mid \mathscr F_n)$$
$$= \prod_{j=1}^{n} 1_{X_j \ne b} E[1_{X_i = b} \prod_{j=n+1}^{i-1} 1_{X_j \ne b} \mid \mathscr F_n]$$
That's all I got. How can I use the proposition? Might there be another way I can approach this problem?
Quick and easy: since $M_n$ is a martingale, by optional stopping, for any $k$ we have $E[M_{T \wedge k}] = E[M_0] = 0$. Rearranging, this says $$E[T \wedge k] = \frac{1}{p-q} E[X_{T \wedge k}] \le \frac{b}{p-q}$$ since $X_{T \wedge k} \le b$. By monotone convergence, $E[T] = \lim_{k \to \infty} E[T \wedge k]$, so $E[T] \le \frac{b}{p-q} < \infty$.