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$(X_k)_{k \in \mathbb{N}}$ is iid with $\mathbf{P} [X_1 = \frac{1}{2}] = \mathbf{P} [X_1 = \frac{3}{2}] = \frac{1}{2}$. $M = (M_n)_{n \in \mathbb{N}_0}$ with $M_0 = 1$ and $M_n = \prod_{k = 1}^n X_k$ for $n > 0$ is a martingale.

For $0 < a < \frac{1}{2}$ and $b > 2$, $\tau := \inf_{n \in \mathbb{N}_0} \{ M_n \ge b \vee M_n \le a \} = \tau_a \wedge \tau_b$ is a stopping time with $\mathbf{P} [\tau < \infty] = 1$. Show that

$\frac{2 - 2 a}{3b - 2a} < \mathbf{P} [M_\tau \ge b] < \frac{2 - a}{2 b - a}$.

Thoughts: With the canonical symmetric random walk, I would use $\mathbf{P} [\tau < \infty] = 1$ (i.e., $M$ stops almost surely by hitting $a$ or $b$) to argue $\mathbf{E} [M_0] = \lim_{n \rightarrow \infty} \mathbf{E} [M_{\tau \wedge n}] = \mathbf{E} [M_\tau] = a \mathbf{P} [\tau = \tau_a] + b \mathbf{P} [\tau = \tau_b]$ and solve for $\mathbf{P} [\tau = \tau_b]$. In fact, it seems to me that $\mathbf{P} [\tau = \tau_b] = \mathbf{P} [M_\tau \ge b]$. However, $M$ can take all kinds of values (greater than $b$) at the time when it crosses $b$, so this doesn't work here.

Andy
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1 Answers1

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You are right that there might be overshooting, but we have some bounds on how much. First, since there might be overshooting now, we need to replace $a$ and $b$ with $E(M_\tau\mid \tau = \tau_a)$ and $E(M_\tau\mid \tau=\tau_b),$ and then we get $$ P(M_\tau\ge b) = P(\tau=\tau_b) = \frac{1-E(M_\tau\mid \tau=\tau_a)}{E(M_\tau\mid \tau=\tau_b)-E(M_\tau\mid \tau=\tau_a)}.$$

Then for bounds, we have $$ b\le E(M_\tau\mid \tau=\tau_b) \le \frac{3}{2}b$$ and $$ \frac{a}{2}\le E(M_\tau\mid \tau=\tau_a) \le a.$$

Taking the possibilities that result in the smallest and largest value for $P(M_\tau\ge b)$ gives the bounds in the problem.

spaceisdarkgreen
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  • Perfect, thank you! For the estimates of the conditional expectations, would I argue as follows? For $\mathbf{E} [M_{\tau \wedge n} | \tau = \tau_a]$, let $a$ be small, so that $M_{\tau \wedge n}$, given $\tau = \tau_a$, can only be $a$; then $a$ is an upper bound. If $a$ is larger, the expected value of the uniform distribution, $\frac{a}{2}$, is a lower bound, because $M_{\tau \wedge n} \sim b_{n + 1, \frac{1}{2}}$ and therefore the distribution of $M_{\tau \wedge n}$, given $\tau = \tau_a$, is left-skewed. The claim follows with $n \rightarrow \infty$. – Andy Jun 30 '19 at 09:44
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    @Andy What I had in mind was much simpler: If $\tau=\tau_b$ then $M_{\tau-1}< b$ and $X_\tau=3/2,$ so $M_{\tau}<\frac{3}{2}b.$ Similarly for $a.$ – spaceisdarkgreen Jun 30 '19 at 22:42
  • If my argument holds at all, yours is indeed much more intuitive and straightforward. Thank you again! – Andy Jul 01 '19 at 07:48