Show that $P(T \le n + N \mid \mathscr F_n) > \epsilon$ where T is a stopping time

Question

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 $X_n = a + \sum_{i=1}^{n} Y_i$ where $0 < a$.

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 $b > a$ be a positive integer and $T:= \inf\{n: X_n = 0 \ or \ X_n = b\}$.

It can be shown that $T$ is a $\{\mathscr F_n\}_{n \in \mathbb N}$-stopping time.

Show that $\exists N \in \mathbb N, \epsilon > 0$ s.t. $\forall n \in \mathbb N$,

$$P(T \le n + N \mid \mathscr F_n) > \epsilon \ a.s.$$

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What I tried:

By using induction on n, we have for the base case:

Suppose $P(T \le N) > \epsilon$. Show that $P(T \le N+1 | \mathscr F_1) > \epsilon$.

I tried considering $P(T \le N+1 | \mathscr F_1)$ and then hopefully I could use the assumption somewhere:

$$P(T \le N+1 | \mathscr F_1) = E[1_{T \le N+1} | \mathscr F_1]$$

$$= E[1_{T \le N} 1_{T = N+1} | \mathscr F_1]$$

$$= E[E[1_{T \le N} 1_{T = N+1} | \mathscr F_N] | \mathscr F_1]$$

$$= E[1_{T \le N} E[ 1_{T = N+1} | \mathscr F_N] | \mathscr F_1]$$

Now what is $E[ 1_{T = N+1} | \mathscr F_N]$ exactly?

Well, up to time N we have already hit $X_n = b$, or we haven't. If we have, then $E[ 1_{T = N+1} | \mathscr F_N] = 0$. Otherwise, $E[ 1_{T = N+1} | \mathscr F_N] = p1_{X_{N} = b-1}$. Continuing:

$$= E[1_{T \le N} p1_{X_{N} = b-1} | \mathscr F_1]$$

$$= pE[1_{T \le N} 1_{X_{N} = b-1} | \mathscr F_1]$$

$$= pE[1_{T \le N-1} 1_{X_{N} = b-1} | \mathscr F_1]$$

Similarly, I got

$$= p^2 E[1_{T \le N-2} 1_{X_{N-1} = b-2} | \mathscr F_1]$$

$$= p^2 E[1_{T \le N-2} E[1_{X_{N-1} = b-2} | \mathscr F_{n-2}] | \mathscr F_1]$$

However, I'm not quite sure that

$$E[1_{X_{N-1} = b-2} | \mathscr F_{n-2}] = p1_{X_{N-2} = b-3}$$

I think we have that

$$E[1_{X_{N-1} = b-2} | \mathscr F_{n-2}] = p1_{X_{N-2} = b-3} + q1_{X_{N-2} = b-1}$$

Um, am I on the right track? Did I make a mistake somewhere?

Answer 1

Step 1. Define $T_x$ by

$$ T_x = \inf\{n \geq 0 : x+Y_1+\cdots+Y_n \in \{0, b\} \}. $$

For any given $x$, we know that $x+Y_1+\cdots+Y_n \to \infty$ almost surely as $n\to\infty$. (For instance, the SLLN is enough to justify this.) So we have $\Bbb{P}(T_x > N) < 1$ for any sufficiently large $N$. Then we can choose $N$ such that

$$ c := \max_{0 < x < b} \Bbb{P}(T_x > N) < 1. $$

Step 2. We claim that the inequality holds with this $N$ and $\epsilon = 1-c$. To this end, write

\begin{align*} \Bbb{P}(T > n+N \mid \mathscr{F}_n) &= \Bbb{E}[ \mathbf{1}_{\{T > n+N\}} \mid \mathscr{F}_n] \\ &= \Bbb{E}[ \mathbf{1}_{\{T-n > N\}}\mathbf{1}_{\{T > n\}} \mid \mathscr{F}_n] \\ &= \sum_{x : 0 < x < b } \Bbb{E}[ \mathbf{1}_{\{T-n > N\}} \mathbf{1}_{\{T > n, X_n = x \}} \mid \mathscr{F}_n] \end{align*}

Now define $\tilde{T}_x$ by

$$ \tilde{T}_x := \inf\{k \geq 0 : x + Y_{n+1} + \cdots + Y_{n+k} \in \{0, b\}\}. $$

Then given $\{T > n, X_n = x\}$, we have $T-n = \tilde{T}_x$. Also it is clear that $\tilde{T}_x$ is independent of $\mathscr{F}_n$ and has the same distribution as $T_x$. So $\Bbb{P}$-a.s., we have

\begin{align*} \Bbb{P}(T > n+N \mid \mathscr{F}_n) &= \sum_{x : 0 < x < b } \Bbb{P}(T_x > N) \mathbf{1}_{\{T > n, X_n = x \}} \\ &\leq \sum_{x : 0 < x < b } (1-\epsilon) \mathbf{1}_{\{T > n, X_n = x \}} \\ &\leq 1-\epsilon. \end{align*}

This is equivalent to the desired inequality.

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Remark. If you have acquaintance with Markov property, you will see that Step 2 is a typical Markov property argument. In this case, we can shorten Step 2 as follows: $\Bbb{P}^a$-almost surely,

\begin{align*} \Bbb{P}^a(T > n+N \mid \mathscr{F}_n) &= \Bbb{E}^a[ \mathbf{1}_{\{T-n > N\}}\mathbf{1}_{\{T > n\}} \mid \mathscr{F}_n] \\ &= \Bbb{P}^{X_n}(T > N) \mathbf{1}_{\{T > n\}} \\ &\leq 1-\epsilon. \end{align*}