From Bounded Increments to an Integrable Bound
Optional stopping at a bounded time follows by summing martingale increments up to that time. For an unbounded stopping time, the same finite-sum argument applies to its truncations, but an additional step is needed: one must justify passing from the truncated stopped values to the value at the stopping time. Uniformly bounded increments and a stopping time with finite expectation provide a direct integrable bound for that step.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a discrete-time filtration \((\mathcal{F}_n)_{n\ge0}\). A process \((M_n)_{n\ge0}\) is a martingale with respect to this filtration if each \(M_n\) is \(\mathcal{F}_n\)-measurable and integrable, and \(\mathbb{E}[M_n\mid\mathcal{F}_{n-1}]=M_{n-1}\) for \(n\ge1\). Let its state space be \(\mathbb{R}\). A stopping time \(\tau\) takes values in \(\{0,1,2,\ldots\}\cup\{\infty\}\) and satisfies \(\{\tau\le n\}\in\mathcal{F}_n\) for each \(n\). We assume \(M_0\) is integrable.
The important estimate is pathwise. Writing \(M_{n\wedge\tau}\) for the value at the earlier of \(n\) and \(\tau\), at most \(\tau\) increments are included. Thus every truncation is controlled by the same random variable \(|M_0|+C\tau\). Its integrability is exactly what allows the limiting argument.
Proof. On the event \(\{|Y_n|>K\}\), the bound \(|Y_n|\le Y\) implies \(Y>K\). Therefore, for every \(n\) and \(K\ge0\), \( |Y_n|\mathbf{1}_{\{|Y_n|>K\}}\le Y\mathbf{1}_{\{Y>K\}}\). Taking expectations and then the supremum over \(n\) gives an upper bound \(\mathbb{E}[Y\mathbf{1}_{\{Y>K\}}]\). Since \(Y\) is integrable, this tail expectation tends to zero as \(K\to\infty\), proving uniform integrability. \(\square\)
Proof by Truncation
For a finite deterministic horizon \(n\), the stopped-value identity is a finite sum. The indicator that the \(k\)th increment is included is \(\mathbf{1}_{\{\tau\ge k\}}\), and \(\{\tau\ge k\}=\{\tau>k-1\}\in\mathcal{F}_{k-1}\). This makes that indicator available before the \(k\)th increment.
Proof of the theorem. First, \(\mathbb{E}[\tau]<\infty\) implies \(\tau<\infty\) almost surely: if \(\mathbb{P}(\tau=\infty)>0\), then \(\mathbb{E}[\tau]\ge m\mathbb{P}(\tau=\infty)\) for every positive integer \(m\), contradicting finiteness. For each \(n\), telescoping gives \[ M_{n\wedge\tau}=M_0+\sum_{k=1}^{n}\mathbf{1}_{\{\tau\ge k\}}(M_k-M_{k-1}). \] The indicator in each summand is \(\mathcal{F}_{k-1}\)-measurable. The martingale property therefore gives \[ \mathbb{E}\!\left[\mathbf{1}_{\{\tau\ge k\}}(M_k-M_{k-1})\right] =\mathbb{E}\!\left[\mathbf{1}_{\{\tau\ge k\}}\mathbb{E}[M_k-M_{k-1}\mid\mathcal{F}_{k-1}]\right]=0. \] All terms are integrable: the increment is bounded and the indicator is at most one. Taking expectations in the finite sum yields \(\mathbb{E}[M_{n\wedge\tau}]=\mathbb{E}[M_0]\).
For every \(n\), the increment bound also gives \[ |M_{n\wedge\tau}|\le |M_0|+C(n\wedge\tau)\le |M_0|+C\tau. \] The right-hand side is integrable because \(\mathbb{E}[|M_0|]<\infty\) and \(\mathbb{E}[\tau]<\infty\). Since \(\tau<\infty\) almost surely, \(M_{n\wedge\tau}\to M_\tau\) almost surely. The dominated convergence theorem now gives \(\mathbb{E}[M_\tau]=\lim_{n\to\infty}\mathbb{E}[M_{n\wedge\tau}]=\mathbb{E}[M_0]\). The same bound, with \(n\wedge\tau\) replaced by \(\tau\), shows that \(M_\tau\) is integrable. \(\square\)
This proof has two distinct parts. Truncation gives an expectation identity at every finite horizon. The integrable envelope—not truncation by itself—justifies taking the limit of those identities.
Worked Example: Stop at the First Positive Increment
Let \(\varepsilon_1,\varepsilon_2,\ldots\) be independent fair signs, each equal to \(1\) or \(-1\) with probability \(\tfrac12\), on the sequence probability space with its natural filtration \(\mathcal{F}_n=\sigma(\varepsilon_1,\ldots,\varepsilon_n)\). Set \(M_0=0\), \(M_n=\sum_{k=1}^n\varepsilon_k\), and let \(\tau=\inf\{n\ge1:\varepsilon_n=1\}\). The martingale has state space \(\mathbb{Z}\), and its increments have absolute value \(1\). The event \(\{\tau\le n\}\) is determined by the first \(n\) signs, so \(\tau\) is a stopping time.
For \(k\ge1\), \(\mathbb{P}(\tau\ge k)=2^{-(k-1)}\): the first \(k-1\) signs must all be \(-1\). The tail-sum formula for a nonnegative integer-valued random variable gives \[ \mathbb{E}[\tau]=\sum_{k=1}^{\infty}\mathbb{P}(\tau\ge k) =\sum_{k=1}^{\infty}2^{-(k-1)}=2. \] The theorem applies with \(C=1\), so \(\mathbb{E}[M_\tau]=0\). In fact, if \(\tau=k\), the first \(k-1\) signs are \(-1\) and the \(k\)th is \(1\), giving \(M_\tau=1-(k-1)=2-k\). Thus \(\mathbb{E}[M_\tau]=2-\mathbb{E}[\tau]=2-2=0\), an independent check.
Check: \(\mathbb{P}(\tau=k)=2^{-k}\) for \(k\ge1\), and \(\sum_{k=1}^{\infty}2^{-k}=1\). Also, \(\sum_{k=1}^{\infty}k2^{-k}=2\), agreeing with the tail-sum calculation.
Worked Example: An Independent Geometric Horizon
On a probability space carrying two independent sequences, let \((\varepsilon_n)\) be independent fair signs and let \((A_n)\) be independent indicators with \(\mathbb{P}(A_n=1)=\mathbb{P}(A_n=0)=\tfrac12\); assume the two sequences are independent of each other. Use the filtration generated by both sequences through time \(n\). Define \(D_n=\varepsilon_n\) for odd \(n\), \(D_n=\tfrac12\varepsilon_n\) for even \(n\), and \(M_n=\sum_{k=1}^nD_k\), with \(M_0=0\). This is a martingale: each \(D_n\) is independent of \(\mathcal{F}_{n-1}\) and has conditional mean zero. Its state space is \(\mathbb{R}\), and \(|D_n|\le1\).
Let \(\tau=\inf\{n\ge1:A_n=1\}\). This is a stopping time for the stated filtration. Its tail probabilities are \(\mathbb{P}(\tau\ge k)=2^{-(k-1)}\), because the first \(k-1\) indicators must be zero. Consequently \(\mathbb{E}[\tau]=\sum_{k=1}^{\infty}2^{-(k-1)}=2\), so the theorem gives \(\mathbb{E}[M_\tau]=0\).
To check directly, condition on \(\tau=m\). Independence of the clock sequence from the signs gives \[ \mathbb{E}[M_\tau\mid\tau=m]=\sum_{k=1}^{m}\mathbb{E}[D_k]=0. \] Averaging over \(m\) therefore gives \(\mathbb{E}[M_\tau]=0\). The required integrability is also visible from \(|M_\tau|\le\tau\), so \(\mathbb{E}[|M_\tau|]\le2\).
Worked Example: A Martingale with Decreasing Increment Sizes
Let \((\varepsilon_n)\) again be independent fair signs on their sequence probability space, with the natural filtration. Define \(D_n=\varepsilon_n/n\) and \(M_n=\sum_{k=1}^nD_k\), with state space \(\mathbb{R}\) and \(M_0=0\). Each increment has conditional mean zero, so \((M_n)\) is a martingale, and \(|D_n|=1/n\le1\). Use the stopping time \(\tau=\inf\{n\ge1:\varepsilon_n=1\}\), whose finite mean was computed above: \(\mathbb{E}[\tau]=2\).
The theorem gives \(\mathbb{E}[M_\tau]=0\). The direct verification also shows the role of the stopping-time indicator. Since \(\tau\) is finite almost surely, \[ M_\tau=\sum_{k=1}^{\infty}\mathbf{1}_{\{\tau\ge k\}}\frac{\varepsilon_k}{k}, \] where only finitely many terms are nonzero on each outcome. The sum of the absolute values is bounded by \(\tau\), because each \(1/k\le1\); its expectation is at most \(2\). Thus the series can be integrated term by term. For each \(k\), the event \(\{\tau\ge k\}\) depends only on the first \(k-1\) signs and is independent of \(\varepsilon_k\). Hence \[ \mathbb{E}\!\left[\mathbf{1}_{\{\tau\ge k\}}\frac{\varepsilon_k}{k}\right] =\frac{1}{k}\mathbb{P}(\tau\ge k)\mathbb{E}[\varepsilon_k]=0. \] Summing these zero expectations proves \(\mathbb{E}[M_\tau]=0\), as required.
Check: for \(k=1\), the contribution has expectation \(\mathbb{E}[\varepsilon_1]=0\). For \(k=2\), it is \(\tfrac12\mathbb{P}(\varepsilon_1=-1)\mathbb{E}[\varepsilon_2]=\tfrac12\cdot\tfrac12\cdot0=0\). The same independence calculation applies at every \(k\).
Why the Finite-Mean Condition Matters
A common mistake is to prove \(\mathbb{E}[M_{n\wedge\tau}]=\mathbb{E}[M_0]\) for every \(n\), observe that \(M_{n\wedge\tau}\to M_\tau\) almost surely, and conclude that the expectations converge. Almost-sure convergence alone does not justify convergence of expectations. The proof above supplies the missing condition: a common integrable bound.
For a concrete failure, let \(M_n\) be the simple symmetric random walk started at \(0\), with independent increments equal to \(1\) and \(-1\) with probability \(\tfrac12\), and let \(\tau=\inf\{n\ge0:M_n=1\}\). The standard recurrence calculation for this walk gives \(\tau<\infty\) almost surely but \(\mathbb{E}[\tau]=\infty\). Its increments are bounded, yet \(M_\tau=1\) almost surely, so \(\mathbb{E}[M_\tau]=1\ne0=\mathbb{E}[M_0]\). Here the finite-mean hypothesis fails, and the bound \(|M_{n\wedge\tau}|\le\tau\) is not an integrable envelope. Indeed, the truncated expectations remain zero while the stopped values converge almost surely to \(1\); that convergence cannot be passed through expectation.
When applying the theorem, check each condition separately: the process is a martingale for the stated filtration; its increments have a common deterministic bound; the horizon is a stopping time; and its expectation is finite. The conclusion is an equality of expectations, not a claim that the stopped value equals the initial value on every outcome.
Check Your Understanding
For each question, identify the hypothesis or calculation that supports the conclusion.
- Why does \(\mathbb{E}[\tau]<\infty\) imply \(\tau<\infty\) almost surely?
- If \(|M_n-M_{n-1}|\le C\), derive a pathwise bound for \(|M_{n\wedge\tau}|\) that does not depend on \(n\).
- In the independent geometric-horizon example, compute \(\mathbb{P}(\tau\ge4)\) and \(\mathbb{E}[\tau]\).
- Why is \(\{\tau\ge k\}\) measurable with respect to \(\mathcal{F}_{k-1}\) for a discrete-time stopping time?
- What step fails if one uses almost-sure convergence of \(M_{n\wedge\tau}\) without an integrable dominating random variable?