Solved Problems on Optional Stopping and Localization

How to Organize an Optional-Stopping Problem

Optional stopping is not a rule that preserves expectations at every stopping time. Its use depends on the stopping time and on integrability properties of the stopped process. A reliable solution therefore identifies the martingale, checks the stopping-time and integrability hypotheses, and only then applies an optional-sampling result. Localization adds a further caution: martingales obtained by stopping a local martingale need not give a valid expectation identity at an unbounded time unless an appropriate convergence condition is available.

Throughout, work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. The examples below use independent coin tosses as well; they may be placed on a product space with the Brownian motion, with the filtration generated by the Brownian motion and the coin tosses revealed so far. No stochastic integral is needed here. A stopping time $\tau$ satisfies $\{\tau\le t\}\in\mathcal{F}_t$ for every $t$. A martingale $M_t$ is an adapted integrable process satisfying $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ almost surely for $s\le t$.

Problem-solving rule: For bounded stopping times, the bounded optional-sampling theorem applies to a martingale under its stated hypotheses. For an almost surely finite unbounded stopping time, a standard sufficient condition is uniform integrability of the stopped family. For a local martingale, first check that localization yields true martingales; do not infer an expectation identity at the limiting stopping time without a justified convergence argument.

A Bounded Stopping Time: A Fair Betting Martingale

Let $\xi_1,\xi_2,\ldots$ be independent fair coin tosses, with $\xi_j=1$ for heads and $\xi_j=0$ for tails. Let $\mathcal{G}_n=\sigma(\xi_1,\ldots,\xi_n)$, with $\mathcal{G}_0$ trivial, and set $M_0=1$ and $M_n=2^n\mathbf{1}_{\{\xi_1=\cdots=\xi_n=1\}}$ for $n\ge1$. Between integer times define $M_t=M_{\lfloor t\rfloor}$, and reveal coin $n$ at time $n$. This is a nonnegative martingale: conditionally on the first $n$ tosses, $M_{n+1}$ is $2^{n+1}$ with probability $1/2$ if all $n$ tosses were heads, and is zero otherwise. Thus $\mathbb{E}[M_{n+1}\mid\mathcal{G}_n]=M_n$. At noninteger times the process is constant, so the martingale identity holds there as well.

Worked Example: Stop at the First Tail, but No Later Than Time Three

Let $\tau=\min\{n\ge1:\xi_n=0\}\wedge3$. This is a stopping time, since $\{\tau\le t\}$ is determined by the tosses revealed by time $t$; also $0\le\tau\le3$. Compute $\mathbb{E}[M_\tau]$ directly and compare it with $M_0$.

If the first tail occurs at toss 1, 2, or 3, then $M_\tau=0$. The respective probabilities are $\tfrac12,\tfrac14,\tfrac18$. If the first three tosses are heads, which has probability $\tfrac18$, then $\tau=3$ and $M_\tau=2^3=8$. Therefore

\mathbb{E}[M_\tau] =0\cdot\frac12+0\cdot\frac14+0\cdot\frac18+8\cdot\frac18 =1=M_0.

Check: the bounded optional-sampling theorem applies because $M$ is a martingale and $\tau\le3$. The direct calculation agrees with its conclusion. The stopped variable is integrable; indeed it takes only the values 0 and 8.

The example illustrates why checking boundedness is useful: the theorem applies without requiring the stopping time to be deterministic. In more general applications, the process may have unbounded values, so integrability of the stopped variables must also be addressed through the theorem’s hypotheses.

An Unbounded Time: Where Expectation Can Be Lost

Now use the same martingale, but let $\tau=\inf\{n\ge1:\xi_n=0\}$, the first toss showing tails. The event $\{\tau\le n\}$ depends only on the first $n$ tosses, so $\tau$ is a stopping time. It is finite almost surely because $\mathbb{P}(\tau>n)=2^{-n}\to0$. Nevertheless, stopping at $\tau$ does not preserve the initial expectation.

Worked Example: A Finite Stopping Time That Does Not Preserve Expectation

At the first tail the betting process becomes zero, so $M_\tau=0$ almost surely and $\mathbb{E}[M_\tau]=0$, whereas $M_0=1$. To see exactly how the discrepancy arises, stop first at $\tau\wedge n$, where $n$ is a fixed positive integer. If a tail occurs by time $n$, the stopped value is zero. If the first $n$ tosses are heads, the stopped value is $2^n$. Hence

M_{\tau\wedge n}=2^n\mathbf{1}_{\{\tau>n\}}, \qquad \mathbb{E}[M_{\tau\wedge n}] =2^n\mathbb{P}(\tau>n) =2^n\cdot2^{-n}=1.

For each fixed $n$, this is a bounded stopping time, and the expectation is preserved. But $M_{\tau\wedge n}\to0=M_\tau$ almost surely, while its expectation remains 1 for every $n$. There is no contradiction: almost-sure convergence alone does not imply convergence of expectations.

Check: the family $\{M_{\tau\wedge n}:n\ge1\}$ is not uniformly integrable. For any $K>0$, choose $n$ with $2^n>K$. Then

\mathbb{E}\!\left[M_{\tau\wedge n}\mathbf{1}_{\{M_{\tau\wedge n}>K\}}\right] =2^n\mathbb{P}(\tau>n)=1.

Thus the supremum of these tail expectations cannot tend to zero as $K\to\infty$, which is required for uniform integrability.

Proposition: Let $M_t$ be a right-continuous martingale and let $\tau$ be an almost surely finite stopping time. If the family $\{M_{t\wedge\tau}:t\ge0\}$ is uniformly integrable, then $\mathbb{E}[M_\tau]=\mathbb{E}[M_0]$.

Proof. For every positive integer $n$, the stopping time $\tau\wedge n$ is bounded. The bounded optional-sampling theorem gives $\mathbb{E}[M_{\tau\wedge n}]=\mathbb{E}[M_0]$. Since $\tau<\infty$ almost surely, for each such outcome $M_{\tau\wedge n}=M_\tau$ whenever $n\ge\tau$; hence $M_{\tau\wedge n}\to M_\tau$ almost surely. The assumed uniform integrability implies convergence in $L^1$, so $\mathbb{E}[M_{\tau\wedge n}]\to\mathbb{E}[M_\tau]$. Taking limits in the expectation identity proves the claim. $\square$

The proposition is a sufficient condition, not a claim that every unbounded stopping time requires precisely this condition. Its point is the passage from bounded stopping times to an unbounded one: the finite-time identities alone do not justify taking the limit. In the betting example, the stopped family fails the stated condition, and its expectations do not converge to the expectation of its almost-sure limit.

A Doob Maximal Bound with the Tail Event Checked

For a nonnegative submartingale $X_t$, Doob’s weak maximal inequality bounds the probability of a large maximum by the expected terminal value. The event convention matters: the version proved here uses the strict event $\{\sup_{0\le t\le T}X_t>\lambda\}$, with $\lambda>0$. The process is assumed right-continuous, adapted, and integrable at each time, and the submartingale property means $\mathbb{E}[X_t\mid\mathcal{F}_s]\ge X_s$ almost surely for $s\le t$.

Theorem: If $X_t\ge0$ is a right-continuous submartingale on $[0,T]$ and $\lambda>0$, then $$ \mathbb{P}\!\left(\sup_{0\le t\le T}X_t>\lambda\right)\le\frac{\mathbb{E}[X_T]}{\lambda}. $$

Proof. First take a finite grid $0=t_0<t_1<\cdots<t_m=T$. Let $A_j$ be the event that $j$ is the first grid index with $X_{t_j}>\lambda$. The events $A_j$ are disjoint and $A_j\in\mathcal{F}_{t_j}$. By the submartingale property and nonnegativity,

\mathbb{E}[X_T\mathbf{1}_{A_j}] =\mathbb{E}\!\left[\mathbb{E}[X_T\mid\mathcal{F}_{t_j}]\mathbf{1}_{A_j}\right] \ge\mathbb{E}[X_{t_j}\mathbf{1}_{A_j}] \ge\lambda\mathbb{P}(A_j).

Summing over $j$ and using $X_T\ge0$ gives $\lambda\mathbb{P}(\max_jX_{t_j}>\lambda)\le\mathbb{E}[X_T]$. Now take nested finite grids whose union is dense in $[0,T]$ and which all contain $T$. Right-continuity implies that if $X_t>\lambda$ at some $t<T$, then $X_s>\lambda$ for some grid point $s$ just to the right of $t$; the endpoint $T$ is included in every grid. Thus the grid events increase to the strict continuous-time event. Continuity of probability from below gives the asserted inequality. $\square$

Worked Example: Compute a Maximum Probability on a Three-Toss Grid

For the betting martingale above, compute the probability that $\max_{0\le t\le3}M_t>4$, and compare it with Doob’s bound at $\lambda=4$. The maximum is taken over all times, and the event uses a strict $>$.

The process takes its successive grid values $M_0=1,M_1,M_2,M_3$. It exceeds 4 only if all three tosses are heads: if a tail occurs on toss 1 or 2, the process becomes zero; if the first two tosses are heads and the third is a tail, the maximum is exactly $M_2=4$, which does not exceed 4. Therefore

\mathbb{P}\!\left(\max_{0\le t\le3}M_t>4\right) =\mathbb{P}(\xi_1=\xi_2=\xi_3=1) =\left(\frac12\right)^3 =\frac18.

Also, $\mathbb{E}[M_3]=8\cdot\tfrac18=1$. Doob’s inequality gives the upper bound $\mathbb{E}[M_3]/4=1/4$, so $\tfrac18\le\tfrac14$. Check: with the non-strict event $\{\max M_t\ge4\}$, the path heads-heads-tails would also count, so the probability would be $\tfrac14$. This distinction is why the event convention must be stated.

Localization: What the Stopped Martingales Do and Do Not Give

A local martingale $N_t$ is an adapted process for which there exists a nondecreasing sequence of stopping times $\tau_n$ tending to infinity almost surely such that each stopped process $N_{t\wedge\tau_n}$ is a martingale. Such a sequence is called a localizing sequence. The definition gives true martingales after stopping, but it does not by itself say that expectations of the original process, or of its value at an unbounded stopping time, are preserved.

Localization criterion: Suppose $N_t$ is a right-continuous local martingale with $N_0$ integrable, and $\tau_n$ is a localizing sequence. If for every fixed $t$ the family $\{N_{t\wedge\tau_n}:n\ge1\}$ is uniformly integrable, then $N_t$ is a martingale.

Proof. Fix $0\le s\le t$. For each $n$, the stopped process is a martingale, so $\mathbb{E}[N_{t\wedge\tau_n}\mid\mathcal{F}_s]=N_{s\wedge\tau_n}$. Since $\tau_n\to\infty$ almost surely, $N_{t\wedge\tau_n}\to N_t$ and $N_{s\wedge\tau_n}\to N_s$ almost surely. The assumed uniform integrability at $t$ gives $L^1$ convergence of the left-hand variables; uniform integrability at $s$ gives $L^1$ convergence on the right. Conditional expectation is continuous in $L^1$, because $\mathbb{E}[|\mathbb{E}[Y\mid\mathcal{F}_s]|]\le\mathbb{E}[|Y|]$. Passing to the limit in the conditional identity yields $\mathbb{E}[N_t\mid\mathcal{F}_s]=N_s$. The limits are integrable by $L^1$ convergence, so $N$ is a martingale. $\square$

In the betting example, $\tau_n=n$ is a localizing sequence: $M_{t\wedge n}$ is a martingale, bounded on each fixed stopped horizon. For any fixed $t$, when $n\ge t$ the stopped value is just $M_t$, so the family over $n$ is eventually constant and therefore uniformly integrable. This is consistent with $M_t$ being a martingale. It does not rescue the expectation identity at the separate unbounded time $\tau$: the relevant family there is $\{M_{\tau\wedge n}\}$, which was shown not to be uniformly integrable.

Common Pitfalls

Applying bounded optional sampling at an unbounded time. The invalid calculation is “$\tau<\infty$ almost surely, so $\mathbb{E}[M_\tau]=\mathbb{E}[M_0]$.” In the betting example this would give $0=1$. Almost-sure finiteness is not a substitute for the hypotheses that justify passage from $\tau\wedge n$ to $\tau$.

Confusing localization with uniform integrability. Knowing that each $N_{t\wedge\tau_n}$ is a martingale does not establish uniform integrability of the stopped family at a limiting time. In the example, each bounded-time calculation gives expectation 1, but the almost-sure limit at $\tau$ is zero.

Changing the maximal event silently. The bound above was stated for $>\lambda$. In the three-toss example, the strict event at level 4 has probability $\tfrac18$, while the event with $\ge4$ has probability $\tfrac14$. Check whether a problem asks for a strict or non-strict crossing before counting paths.

Check Your Understanding

For each problem, state the hypothesis that permits the conclusion, or identify precisely which step is not justified.

For the betting martingale, compute $\mathbb{E}[M_{\tau\wedge2}]$ when $\tau$ is the first-tail time.
Why does $\mathbb{P}(\tau<\infty)=1$ fail to imply $\mathbb{E}[M_{\tau\wedge n}]\to\mathbb{E}[M_\tau]$ in this example?
State the uniform-integrability condition used to pass from bounded stopping times to an almost surely finite unbounded stopping time.
For the three-toss example, compute the probability of $\max_{0\le t\le3}M_t\ge4$, and explain why it differs from the strict-crossing probability.
What additional fixed-time condition allows a localizing sequence to establish that a local martingale is a true martingale?