Why Bounded Stopping Preserves Expectation
A martingale has no predictable average gain when viewed one step at a time. Bounded optional stopping says that this balance remains true when the time at which we stop depends on the observations so far, provided the stopping time has a deterministic finite upper bound. This gives a rigorous way to calculate expectations at random horizons without assuming that the stopping rule is independent of the process.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a filtration \((\mathcal{F}_n)_{n\ge0}\), an increasing sequence of information sets. Let \((M_n)_{n\ge0}\) be a real-valued, integrable, adapted process: each \(M_n\) is \(\mathcal{F}_n\)-measurable and \(\mathbb{E}|M_n|<\infty\). It is a martingale if \(\mathbb{E}[M_n\mid\mathcal{F}_{n-1}]=M_{n-1}\) almost surely for every \(n\ge1\). A random time \(\tau\) taking values in \(\{0,1,2,\ldots\}\) is a stopping time if \(\{\tau\le n\}\in\mathcal{F}_n\) for every \(n\). It is bounded if some deterministic integer \(N\) satisfies \(\mathbb{P}(\tau\le N)=1\).
The key is to write the stopped value as a finite sum of increments. Define \(\Delta M_k=M_k-M_{k-1}\). The event that the process is still running at step \(k\) is \(\{\tau\ge k\}\), which is determined by time \(k-1\): its complement is \(\{\tau\le k-1\}\). Thus the decision to include the \(k\)th increment is made before that increment is observed.
Proof. Since \(\tau\le N\) almost surely, the following equality holds pathwise: $$ M_\tau=M_0+\sum_{k=1}^{N}\mathbf{1}_{\{\tau\ge k\}}\Delta M_k. $$ Indeed, for a sample point with \(\tau=r\), the indicators equal \(1\) for \(k=1,\ldots,r\) and \(0\) for \(k>r\); the sum then telescopes to \(M_r-M_0\). Each term is integrable because \(|\mathbf{1}_{\{\tau\ge k\}}\Delta M_k|\le|\Delta M_k|\), and \(\Delta M_k\) is integrable. Also, \(\{\tau\ge k\}\in\mathcal{F}_{k-1}\), so the martingale property gives $$ \mathbb{E}\!\left[\mathbf{1}_{\{\tau\ge k\}}\Delta M_k\right] =\mathbb{E}\!\left[\mathbf{1}_{\{\tau\ge k\}}\mathbb{E}[\Delta M_k\mid\mathcal{F}_{k-1}]\right]=0. $$ Taking expectations in the finite sum therefore yields \(\mathbb{E}[M_\tau]=\mathbb{E}[M_0]\). The finite sum also establishes the asserted integrability of \(M_\tau\). \(\square\)
Proof. The sum is \(M_\tau-M_0\) by telescoping, including when \(\tau=0\), where it is an empty sum equal to zero. Both terms are integrable by the theorem and the martingale assumptions. Hence its expectation is \(\mathbb{E}[M_\tau]-\mathbb{E}[M_0]=0\). \(\square\)
Finite-Horizon Applications
Worked Example: A Fair Walk Stopped at a Head or at Time Three
Let \((\xi_n)_{n\ge1}\) be independent fair signs, each equal to \(1\) or \(-1\) with probability \(\tfrac12\), on their product probability space. Let \(\mathcal{F}_n\) be the information from the first \(n\) signs, set \(\mathcal{F}_0=\{\varnothing,\Omega\}\), and define \(S_0=0\), \(S_n=\sum_{k=1}^n\xi_k\). This is a real-valued martingale: \(\mathbb{E}[\xi_n\mid\mathcal{F}_{n-1}]=0\), so \(\mathbb{E}[S_n\mid\mathcal{F}_{n-1}]=S_{n-1}\). Stop at the first positive sign or at time three, whichever comes first.
In symbols, \(\tau=\min(\inf\{n\ge1:\xi_n=1\},3)\). This is a stopping time because whether a positive sign has appeared by time \(n\) is known from the first \(n\) signs. It is bounded by \(3\). The possible stopped outcomes are: a positive first sign, giving \(S_\tau=1\) with probability \(\tfrac12\); a negative first sign followed by a positive second sign, giving \(S_\tau=0\) with probability \(\tfrac14\); and two initial negative signs, followed by either sign at time three, giving \(-1\) or \(-3\), each with probability \(\tfrac18\).
Therefore $$ \mathbb{E}[S_\tau] =\tfrac12(1)+\tfrac14(0)+\tfrac18(-1)+\tfrac18(-3) =\tfrac12-\tfrac18-\tfrac38=0. $$ This agrees with \(\mathbb{E}[S_0]=0\). As a check, the four outcome probabilities sum to \(\tfrac12+\tfrac14+\tfrac18+\tfrac18=1\), and direct substitution gives the same expectation.
Worked Example: Stopping a Conditional-Probability Martingale
Let \(\Omega=\{H,T\}^3\) with each of its eight outcomes equally likely, and let \(\mathcal{F}_n\) record the first \(n\) fair-coin tosses. Let \(A\) be the event that at least two of the three tosses are heads, and set \(M_n=\mathbb{P}(A\mid\mathcal{F}_n)\) for \(0\le n\le3\). This is a martingale, since conditional expectations of a fixed event indicator satisfy the tower property. Its state space is \([0,1]\).
Initially, four of the eight outcomes belong to \(A\), so \(M_0=\tfrac12\). After one toss, \(M_1=\tfrac34\) if the toss is \(H\), and \(M_1=\tfrac14\) if it is \(T\). After two tosses, the values are \(1\) for \(HH\), \(\tfrac12\) for \(HT\) or \(TH\), and \(0\) for \(TT\). At time three, \(M_3=\mathbf{1}_A\).
Define \(\tau\) as the first \(n\in\{1,2\}\) for which \(M_n\) equals \(0\) or \(1\), and set \(\tau=3\) if there is no such \(n\). This is a stopping time: at each time \(n\le2\), whether the value has reached \(0\) or \(1\) is known from the tosses already observed. It is bounded by \(3\). The stopped value is \(1\) on \(A\) and \(0\) off \(A\), because the event is resolved by the third toss. Thus $$ \mathbb{E}[M_\tau]=1\cdot\mathbb{P}(A)+0\cdot\mathbb{P}(A^c) =1\cdot\tfrac12+0\cdot\tfrac12=\tfrac12=M_0. $$ For a path check, \(HH\) gives value \(1\) with probability \(\tfrac14\), and \(TT\) gives \(0\) with probability \(\tfrac14\). The two mixed prefixes, \(HT\) and \(TH\), each have probability \(\tfrac14\); the last toss makes the event true for one of its two possibilities and false for the other. Together they add probability \(\tfrac14\) to each stopped value. Hence the total probabilities of stopped values \(1\) and \(0\) are both \(\tfrac12\), as used above.
Worked Example: A Quadratic Martingale Gives a Stopping-Time Identity
Use the fair-sign walk \(S_n\) with its natural filtration from the previous example. The process \(M_n=S_n^2-n\) is an integrable martingale. To verify this, write \(S_{n+1}=S_n+\xi_{n+1}\), where \(\xi_{n+1}^2=1\). Then $$ M_{n+1}-M_n=2S_n\xi_{n+1}, $$ whose conditional expectation given \(\mathcal{F}_n\) is \(2S_n\mathbb{E}[\xi_{n+1}\mid\mathcal{F}_n]=0\).
Let \(\tau\) be the first \(n\ge1\) for which \(|S_n|=2\), or \(3\) if no such time occurs by then. The event of hitting either boundary by a given time depends only on signs observed by that time, so \(\tau\) is a stopping time; it is bounded by \(3\). After two steps, the paths \(++\) and \(--\) hit a boundary, with total probability \(\tfrac12\), and have \(\tau=2\), \(S_\tau^2=4\). The paths \(+-\) and \(-+\), also with total probability \(\tfrac12\), have \(S_2=0\); at time three they end at \(1\) or \(-1\), so \(\tau=3\) and \(S_\tau^2=1\).
Consequently, $$ \mathbb{E}[\tau]=\tfrac12(2)+\tfrac12(3)=\tfrac52,\qquad \mathbb{E}[S_\tau^2]=\tfrac12(4)+\tfrac12(1)=\tfrac52. $$ Optional stopping applied to \(M_n=S_n^2-n\) gives \(\mathbb{E}[S_\tau^2-\tau]=\mathbb{E}[M_0]=0\), in agreement with these computations. More directly, \(M_\tau\) equals \(4-2=2\) with probability \(\tfrac12\), and \(1-3=-2\) with probability \(\tfrac12\), so \(\mathbb{E}[M_\tau]=\tfrac12(2)+\tfrac12(-2)=0\). This checks both the stopping identity and the two separate expectations.
Why the Bound Cannot Be Dropped
Bounded means bounded by one deterministic integer almost surely; it does not mean merely that the stopping time is finite almost surely. The finite sum in the proof is essential. Without a deterministic bound, an infinite sum of stopped increments may require additional conditions before expectations can be interchanged with summation.
For a concrete failure, take an infinite sequence of independent fair coin tosses, with \(\mathcal{F}_n\) recording the first \(n\). Define \(M_0=1\) and \(M_n=2^n\mathbf{1}_{\{\text{the first }n\text{ tosses are all heads}\}}\) for \(n\ge1\). This is an integrable martingale: if the first \(n\) tosses are not all heads, both \(M_n\) and \(M_{n+1}\) are zero; otherwise the next value is \(2^{n+1}\) with probability \(\tfrac12\) and zero with probability \(\tfrac12\), whose conditional mean is \(2^n=M_n\). In particular, \(\mathbb{E}[M_n]=1\) for every \(n\).
Let \(\tau\) be the time of the first tail. It is a stopping time and \(\mathbb{P}(\tau>n)=2^{-n}\), so it is finite almost surely. But \(\mathbb{P}(\tau>n)>0\) for every \(n\), so it has no deterministic finite bound. At time \(\tau\), a tail has occurred and \(M_\tau=0\). Thus \(\mathbb{E}[M_\tau]=0\ne1=\mathbb{E}[M_0]\). This example pinpoints the limitation: almost-sure finiteness alone does not give the bounded optional stopping conclusion.
Common mistake. A tempting but invalid argument is to use the bounded-time identity for a stopping time that is only known to be finite almost surely, on the grounds that each realized path stops after finitely many steps. In the example above that argument would give \(\mathbb{E}[M_\tau]=\mathbb{E}[M_0]=1\), contradicting the direct calculation \(M_\tau=0\) almost surely. The correction is to check for a single deterministic \(N\) with \(\mathbb{P}(\tau\le N)=1\) before applying this theorem. If no such bound exists, this result does not apply; other hypotheses and results are needed.
Check Your Understanding
For each question, identify the relevant hypotheses before using the expectation identity.
- Why is \(\{\tau\ge k\}\) measurable with respect to \(\mathcal{F}_{k-1}\) when \(\tau\) is a stopping time?
- For a martingale bounded stopping time \(\tau\le N\), write the finite increment sum for \(M_\tau-M_0\). What is the expected value of its \(k\)th term?
- In the fair-sign walk example stopped at a positive sign or time three, list the possible values of \(S_\tau\) and their probabilities, then recompute its expectation.
- For the quadratic martingale example, verify directly that \(S_\tau^2-\tau\) has expectation zero using the two possible stopping cases.
- In the first-tail example, explain why \(\tau\) is finite almost surely but is not bounded by any deterministic integer.