# 4.8. Optional stopping theorem


In this section, we generalize the bounded version of optional stopping. After that as an example we will cover theorem regarding assymetric random walk.

## Optional stopping theorem

Our first theorem will be the extension of theorem 4.2.9.

Let $(X_n)$ be a uniformly integrable submartingale and $N$ be a stopping time. Then $(X_{n\wedge N})$ is a uniformly integrable submartingale.

It is shown that $(X_{n\wedge N})$ is a submartingale in theorem 4.2.9. By Vitali's lemma $X_n$ converges almost surely and in $L^1$ to some $X_\infty.$ Since $x \mapsto x^+$ is convex and increasing, $X_n^+, X_{n\wedge N}^+$ are submartingales. Let $\tau=n, \sigma=n\wedge N$ then $\tau,\sigma$ are bounded stopping times. By Doob's inequality, $EX_{n\wedge N}^+ \le EX_n^+$ and $$\sup_n EX_{n\wedge N}^+ \le \sup_n EX_n^+ \le \sup_n E|X_n| < \infty.$$ By Submartingale convergence, $X_{n\wedge N} \to X_N$ a.s. and $E|X_N| < \infty.$ \begin{aligned} &E|X_{n\wedge N}| \mathbf{1}_{|X_{n\wedge N}| \ge a} \\ &\le E|X_{n\wedge N}| \mathbf{1}_{|X_{n\wedge N}| \ge a, N \le n} + E|X_{n\wedge N}| \mathbf{1}_{|X_{n\wedge N}| \ge a, N>n} \\ &= E|X_N| \mathbf{1}_{|X_N| \ge a} + E|X_n| \mathbf{1}_{|X_n| \ge a}. \end{aligned} Since both terms on the right-hand side goes to 0 as $a\to\infty,$ $X_{n\wedge N}$ is uniformly integrable.

Next theorem is the unbounded version of Doob’s inequality.

Let $(X_n)$ be a uniformly integrable submartingale, $N$ be a stopping time. Then $$EX_0 \le EX_N \le EX_\infty$$ where $X_\infty = \lim_n X_n$ a.s.

By the previous theorem $X_{n\wedge N}$ is a uniformly integrable submartingale. By Doob's inequality $$EX_0 \le EX_{n\wedge N} \le EX_n.$$ By Vitali's lemma, $EX_n \to EX_\infty$ and \begin{aligned} \lim_n X_{n\wedge N} = \begin{cases} X_N &,~ N<\infty \\ X_\infty = X_N &,~ N=\infty \end{cases} \end{aligned} Thus $X_{n\wedge N} \to X_N$ a.s. with $E|X_N| < \infty$ by Vitali's lemma and the desired result follows.

Finally we state and prove the main theorem.

Let $L \le M$ be stopping times and $(Y_{n\wedge M})$ be a uniformly integrable submartingale. Then $EY_L \le EY_M$ and $Y_L \le E(Y_M | \mathcal{F}_L)$ a.s.

Let $X_n = Y_{n\wedge M}$ then it directly follows that $EY_L \le EY_M.$ The rest of the proof is the same as the first proof of bounded stopping theorem.

Note that we do not need uniform integrability of $Y_n.$ The next theorem guarantees uniform integrability of stopped martingale of submartingale with uniformly bounded conditional increment.

Let $X_n$ be a submartingale with $E\left(\left|X_{n+1}-X_n\right| | \mathcal{F}_n\right) \le B$ a.s. and $N$ be a stopping time with $EN < \infty.$ Then $X_{n\wedge N}$ is uniformly integrable and $EX_0 \le EX_N.$

$$X_{n\wedge N} = X_0 + \sum\limits_{m=1}^n (X_m - X_{m-1}) \mathbf{1}_{m \le N}\\ |X_{n\wedge N}| \le |X_0| + \sum\limits_{m=1}^n |X_m - X_{m-1}| \mathbf{1}_{m \le N}$$ Let $Z$ be the right-hand side of the inequality. \begin{aligned} E|Z| &\le E|X_0| + \sum_m |X_m - X_{m-1}| \mathbf{1}_{m \le N} \\ &\le E|X_0| + \sum_m E\left( \mathbf{1}_{m\le N} E(\left|X_m - X_{m-1}\right| | \mathcal{F}_{m-1}) \right) \\ &\le E|X_0| + B \cdot \sum_m P(m \le N) \\ &= E|X_0| + B\cdot EN < \infty. \end{aligned} Thus $Z$ is integrable and $X_{n\wedge N}$ is uniformly integrable. $EX_0 \le EX_N$ follows directly.

## Assymetric random walk

As an application of optional stopping, we look into properties of assymetric random walk. We define assymetric random walk $S_n = \xi_1 + \cdots + \xi_n,$ $S_0 = 0$ where $\xi_i$’s are i.i.d. with $P(\xi_1=1)=p,$ $P(\xi_1=-1)=q,$ $p+q=1.$ Let $\text{Var}(\xi_1) = \sigma^2 < \infty$ and $\mathcal{F}_n = \sigma(\xi_1,\cdots,\xi_n)$ for $n\ge 1,$ $\mathcal{F}_0$ be a trivial $\sigma$-field. Let $\varphi(x) = (\frac{1-p}{p})^x.$

(a) $0 < p < 1 \implies \varphi(S_n)$ is a martingale.
(b) $T_x := \inf\{n: S_n=x\},$ $x\in\mathbb{Z}$ is a stopping time and $P(T_a < T_b) = \frac{\varphi(b)-\varphi(0)}{\varphi(b)-\varphi(a)}$ for $a<0<b.$
(c) $1/2<p<1$ and $a<0<b \implies$ $T_b < \infty$ a.s. and $P(T_a < \infty) < 1.$
(d) $1/2<p<1 \implies$ $ET_b = \frac{b}{2p-1},$ $b>0.$
expand proof
(b) Let $T_a \wedge T_b$ be a stopping time. By law of iterated logarithm, \begin{aligned} \limsup_n \frac{S_n - n(p-q)}{\sigma \sqrt{2n\log\log n}} &= 1 \text{ a.s.} \\ \liminf_n \frac{S_n - n(p-q)}{\sigma \sqrt{2n\log\log n}} &= -1 \text{ a.s.} \end{aligned} thus $S_n \approx n(p-q) \pm \sigma\sqrt{2n\log\log n}.$ If $p>q,$ $\lim_n S_n = \infty$ a.s. and $T_b < \infty$ a.s. Similarly $T_a$ or $T_b$ is almost surely finite in any cases, so $N < \infty$ a.s. If $N \ge n,$ $a \le S_{n\wedge N} = S_n \le b.$ If $N < n,$ $S_{n\wedge N} = S_N = a$ or $b.$ $\varphi(S_{n\wedge N})$ is a bounded, thus uniformly integrable and closable martingale. Note that $S_N = a\mathbf{1}_{T_a < T_b} + b\mathbb{1}_{T_a > T_b}.$ Also note that $E\varphi(S_N) = 1$ since $1 = E\varphi(S_0) = E\varphi(S_{n\wedge N}) \to E\varphi(S_N).$ \begin{aligned} 1 = E\varphi(S_N) &= \varphi(a) P(T_a < T_b) + \varphi(b) P(T_a > T_b) \\ &= (\varphi(a) - \varphi(b)) P(T_a < T_b) + \varphi(b). \end{aligned} Organizing both sides gives the result.
Observe that $T_\alpha < T_\beta$ for all $\beta<\alpha<0.$ Thus $\lim_{a\to-\infty} T_a = \infty.$ \begin{aligned} P(T_b < \infty) &= \lim_{a\to-\infty} P(T_b<T_a) \\ &= \lim_{a\to-\infty} \left(1 - \frac{\varphi(b) - 1}{\varphi(b) - \varphi(a)}\right) \\ &= \lim_{a\to-\infty} \frac{1-\varphi(a)}{\varphi(b) - \varphi(a)} = 1. \end{aligned} Similarly, $P(T_a < \infty) = 1/\varphi(a) < 1.$
Observe that if $a<0,$ $(\inf_n S_n \le a) = (T_a < \infty).$ Since $$P(\inf_n S_n \le a) = P(T_a < \infty) = \begin{cases} \left(\frac{1-p}{p}\right)^{-a} &,~ a<0 \\ 1 &,~ a\ge 0 \end{cases}$$ we get \begin{aligned} E|\inf_n S_n| &= \sum_{a=-\infty}^\infty |a| P(\inf_n S_n = a) \\ &= \sum_{a=-\infty}^\infty |a| \left( \left(\frac{1-p}{p}\right)^{-a} - \left(\frac{1-p}{p}\right)^{-(a-1)} \right) \\ &= \sum_{a=-\infty}^\infty |a| \left(\frac{1-p}{p}\right)^{-a} \left(1 - \frac{1-p}{p}\right) < \infty. \end{aligned} Thus $\inf_n S_n$ is integrable. Let $X_n = S_n - n(p-q)$ then $X_n$ is a martingale. Since $T_b < \infty$ a.s., $X_{n\wedge T_b}$ is also a martingale. \begin{aligned} ES_{n\wedge T_b} &= EX_{n\wedge T_b} + (p-q)E(T_b\wedge n) \\ &= \cancel{EX_0} + (p-q)E(T_b\wedge n). \end{aligned} Note that $\inf_n S_n \le S_{n\wedge T_b} \le b$ and $|S_{n\wedge T_b}| \le |\inf_n S_n| + b$ for all $n.$ By DCT, $ES_{n\wedge T_b} \to ES_{T_b} = b.$ By MCT, $E(T_b \wedge n) \uparrow ET_b.$ Thus the desired result follows.

Acknowledgement

This post series is based on the textbook Probability: Theory and Examples, 5th edition (Durrett, 2019) and the lecture at Seoul National University, Republic of Korea (instructor: Prof. Sangyeol Lee).