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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?

BCLC
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1 Answers1

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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 $E[T] \le \frac{b}{p-q} < \infty$.

Nate Eldredge
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  • Seriously? I think somewhere in your argument you assumed $E[T] < \infty$. Am I wrong? – BCLC Dec 14 '15 at 06:46
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    @BCLC: I'm pretty sure I didn't. Note that until the very last line, we are only working with the stopping time $T \wedge k$, which is bounded. And monotone convergence doesn't require assuming $E[T]<\infty$. – Nate Eldredge Dec 14 '15 at 06:48
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    And if you're worried about $X_{T \wedge k}$, note that in addition to being bounded above by $b$, it's bounded below by $-k$. So its expectation is also finite. – Nate Eldredge Dec 14 '15 at 06:54
  • MCT myb not but what about after MCT? How do you know the equality in the last line is true? Because $T < \infty$? I think I read a proof on that. Didn't think it was important – BCLC Dec 14 '15 at 06:59
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    @BCLC: The equality in the last line *is* the monotone convergence theorem. – Nate Eldredge Dec 14 '15 at 07:31
  • Sorry for the confusion. How do you know $T \wedge k \to T$ ? – BCLC Dec 14 '15 at 20:33
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    It's basically just the definition of "limit". Fix $\omega$. If $T(\omega) < \infty$, then there exists an integer $N$ so large that $N \ge T(\omega)$. For all $k \ge N$ we have $T(\omega) \wedge k = T(\omega)$. Hence $T(\omega) \wedge k \to T(\omega)$. If $T(\omega) = \infty$ then $T(\omega) \wedge k = k \to \infty = T(\omega)$. In either case $T(\omega) \wedge k \to T(\omega)$, and $\omega$ was arbitrary. – Nate Eldredge Dec 14 '15 at 21:18
  • Wow so like you seem better than Williams :O Thanks Nate Eldredge! ^-^ I think you didn't use optional stopping, but rather this thm based on a probability principle ? – BCLC Dec 15 '15 at 14:52
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    @BCLC: You can call it what you like. It is a special case of the optional stopping theorem, applied to the bounded stopping time $T \wedge k$. – Nate Eldredge Dec 15 '15 at 16:49