Why Uniform Integrability Extends Optional Stopping
For a bounded stopping time, optional stopping follows by adding a finite number of martingale increments. If the stopping time can be arbitrarily large, that finite-sum argument no longer directly applies. A useful route is to stop first at a deterministic time \(n\), apply the bounded result, and then let \(n\) increase. Almost-sure finiteness ensures that the truncated stopped values eventually equal the value at the stopping time. Uniform integrability supplies the additional control needed to pass from that almost-sure convergence to convergence of expectations.
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 integrable and \(\mathcal{F}_n\)-measurable, and \(\mathbb{E}[M_n\mid\mathcal{F}_{n-1}]=M_{n-1}\) for \(n\ge1\). A stopping time \(\tau\) takes values in \(\{0,1,2,\ldots\}\cup\{\infty\}\) and satisfies \(\{\tau\le n\}\in\mathcal{F}_n\) for each \(n\). Assume \(\tau<\infty\) almost surely when forming \(M_\tau\). For each deterministic \(n\), \(n\wedge\tau=\min(n,\tau)\) is bounded.
Proof of the Theorem
We first give the convergence fact that connects uniform integrability to the limiting stopped value. This also verifies that the limit can be passed through expectation rather than merely through probability.
Proof. Uniform integrability implies \(\sup_n\mathbb{E}|Z_n|<\infty\): choose \(K\) so that all tail expectations above \(K\) are at most \(1\), and use \(\mathbb{E}|Z_n|\le K+\mathbb{E}[|Z_n|\mathbf{1}_{\{|Z_n|>K\}}]\). Fatou's lemma then gives \(\mathbb{E}|Z|\le\liminf_n\mathbb{E}|Z_n|<\infty\). For \(K>0\), on \(\{|Z|>2K\}\), almost-sure convergence implies that eventually \(|Z_n|>K\). Fatou's lemma therefore gives $$ \mathbb{E}\big[|Z|\mathbf{1}_{\{|Z|>2K\}}\big] \le \liminf_{n\to\infty}\mathbb{E}\big[|Z_n|\mathbf{1}_{\{|Z_n|>K\}}\big]. $$ The right side becomes uniformly small as \(K\to\infty\), so the tails of \(Z\) also become small. Define the clipping function \(c_K(x)=\max(-K,\min(x,K))\). For fixed \(K\), \(c_K(Z_n)\to c_K(Z)\) almost surely and \(|c_K(Z_n)-c_K(Z)|\le2K\), so bounded convergence gives \(\mathbb{E}|c_K(Z_n)-c_K(Z)|\to0\). Also, $$ |Z_n-Z|\le |c_K(Z_n)-c_K(Z)| +|Z_n|\mathbf{1}_{\{|Z_n|>K\}} +|Z|\mathbf{1}_{\{|Z|>K\}}. $$ Taking the limit superior in expectation, then letting \(K\to\infty\), makes the last two terms vanish by uniform integrability and the tail bound for \(Z\). Thus \(\mathbb{E}|Z_n-Z|\to0\). \(\square\)
Proof. Set \(Z_n=M_{n\wedge\tau}\). First, for every deterministic \(n\), bounded optional stopping gives \(\mathbb{E}[Z_n\mid\mathcal{F}_0]=M_0\). In discrete time this identity can also be checked directly: for \(k\ge1\), \(\{\tau\ge k\}\in\mathcal{F}_{k-1}\), and $$ Z_n=M_0+\sum_{k=1}^n\mathbf{1}_{\{\tau\ge k\}}(M_k-M_{k-1}). $$ Each summand has expectation zero by the martingale property. Multiplying by the indicator of any event in \(\mathcal{F}_0\) gives the conditional identity.
Because \(\tau<\infty\) almost surely, for almost every outcome there is a finite index after which \(n\wedge\tau=\tau\). Hence \(Z_n\to M_\tau\) almost surely. The assumed uniform integrability and the lemma imply \(\mathbb{E}|Z_n-M_\tau|\to0\), in particular that \(M_\tau\) is integrable. For any \(A\in\mathcal{F}_0\), convergence in mean gives $$ \mathbb{E}[\mathbf{1}_A M_\tau] =\lim_{n\to\infty}\mathbb{E}[\mathbf{1}_A Z_n] =\mathbb{E}[\mathbf{1}_A M_0]. $$ This is precisely \(\mathbb{E}[M_\tau\mid\mathcal{F}_0]=M_0\) almost surely. Taking expectations proves the unconditional identity. \(\square\)
The proof has two separate ingredients. The stopping-time identity at \(n\wedge\tau\) is a bounded-time calculation. Uniform integrability is what permits its limit to retain the same expectation. Almost-sure convergence alone is not enough.
Worked Example: A Two-Toss Conditional-Expectation Martingale
Let two independent fair coin tosses be observed in sequence, with \(\mathcal{F}_n\) the information from the first \(n\) tosses. For \(n\ge2\), let \(\mathcal{F}_n\) contain both tosses. Set \(Y=\mathbf{1}_{\{\text{the first two tosses are both heads}\}}\) and \(M_n=\mathbb{E}[Y\mid\mathcal{F}_n]\). Find \(\mathbb{E}[M_\tau]\) for the stopping time \(\tau=1\) if the first toss is heads and \(\tau=2\) otherwise.
There are four equally likely outcomes, so \(\mathbb{E}[M_0]=\mathbb{E}[Y]=\tfrac14\). After a first head, the conditional probability that both tosses are heads is \(\tfrac12\); after a first tail, it is \(0\). Thus \(M_1=\tfrac12\) on a first head and \(M_1=0\) on a first tail. At time \(2\), \(M_2=Y\). The rule is a stopping time: \(\{\tau\le1\}\) is the event that the first toss is heads, which belongs to \(\mathcal{F}_1\), and \(\tau\le2\) always.
This martingale is bounded between \(0\) and \(1\), so every family of its stopped values is uniformly integrable. The stopped value is \(\tfrac12\) on a first head and \(0\) on a first tail. Therefore $$ \mathbb{E}[M_\tau]=\tfrac12\cdot\tfrac12+\tfrac12\cdot0=\tfrac14=\mathbb{E}[M_0]. $$ Check: Directly, \(M_\tau\) equals \(M_1\) on a first head and \(M_2\) on a first tail. In the latter case \(Y=0\), so the same two contributions, \(\tfrac14\) and \(0\), give the stated expectation.
Worked Example: An Unbounded UI Family and a Geometric Stopping Time
Let \(Y\) have the geometric distribution \(\mathbb{P}(Y=k)=2^{-k}\), \(k=1,2,\ldots\). Let \(\mathcal{F}_n\) reveal the exact value of \(Y\) if \(Y\le n\), and otherwise reveal only that \(Y>n\); take \(\mathcal{F}_0\) to be trivial. Set \(M_n=\mathbb{E}[Y\mid\mathcal{F}_n]\) and \(\tau=Y\). We verify the theorem even though the stopped values are not bounded by one common constant.
If \(Y\le n\), then \(M_n=Y\). If \(Y>n\), then \(\mathbb{P}(Y=n+r\mid Y>n)=2^{-r}\) for \(r\ge1\), since \(\mathbb{P}(Y>n)=2^{-n}\). As \(\sum_{r\ge1}r2^{-r}=2\), this gives \(M_n=n+2\) on \(\{Y>n\}\). Also \(M_0=\mathbb{E}Y=\sum_{k\ge1}k2^{-k}=2\). The event \(\{Y\le n\}\) is known at time \(n\), so \(\tau=Y\) is a stopping time; and \(\mathbb{P}(\tau>n)=2^{-n}\to0\), so it is finite almost surely.
For \(Z_n=M_{n\wedge\tau}\), the explicit values are \(Z_n=Y\) when \(Y\le n\) and \(Z_n=n+2\) when \(Y>n\). For integer \(K\ge1\), using the strict tail event \(|Z_n|>K\), $$ \mathbb{E}\big[|Z_n|\mathbf{1}_{\{|Z_n|>K\}}\big] =\sum_{\substack{1\le k\le n\\k>K}}k2^{-k} +(n+2)2^{-n}\mathbf{1}_{\{n+2>K\}}. $$ The first term is at most \(\sum_{k>K}k2^{-k}=(K+2)2^{-K}\). In the second term, \(n+2>K\) implies \(n\ge K-1\); since \((n+2)2^{-n}\) decreases with \(n\ge0\), that term is at most \((K+1)2^{-(K-1)}\). Both bounds tend to zero as \(K\to\infty\), uniformly in \(n\). Thus the stopped values are uniformly integrable.
At the stopping time, \(M_\tau=M_Y=Y\), so \(\mathbb{E}[M_\tau]=\sum_{k\ge1}k2^{-k}=2=M_0\). Check: the truncated values converge to \(Y\): for each outcome with \(Y=k\), every \(n\ge k\) has \(Z_n=k\). This agrees with the almost-sure convergence used in the theorem.
Why the Uniform-Integrability Condition Matters
The next example keeps the stopping time almost surely finite but removes uniform integrability. It shows that this hypothesis cannot be replaced by almost-sure finiteness alone.
Worked Example: A Finite Stopping Time with a Failed Expectation Identity
Let \((H_n)_{n\ge1}\) be independent fair coin tosses and let \(\mathcal{F}_n\) record the first \(n\) tosses. Define \(M_0=1\), and for \(n\ge1\) set \(M_n=2^n\) if the first \(n\) tosses are all heads and \(M_n=0\) otherwise. Let \(\tau\) be the time of the first tail.
To check the martingale property, if the first \(n\) tosses are all heads, then \(M_{n+1}=2^{n+1}\) with probability \(\tfrac12\) and \(0\) with probability \(\tfrac12\), giving conditional expectation \(\tfrac12 2^{n+1}=2^n=M_n\). If a tail has already occurred, both \(M_n\) and \(M_{n+1}\) are zero. Also \(\mathbb{E}M_1=\tfrac12\cdot2+\tfrac12\cdot0=1=M_0\).
The event \(\{\tau>n\}\) means the first \(n\) tosses are heads, so \(\mathbb{P}(\tau>n)=2^{-n}\to0\). Thus \(\tau<\infty\) almost surely. But the toss at time \(\tau\) is a tail, so \(M_\tau=0\) almost surely and \(\mathbb{E}M_\tau=0\ne1=\mathbb{E}M_0\).
Indeed, for \(Z_n=M_{n\wedge\tau}\), \(Z_n=2^n\) on \(\{\tau>n\}\) and is \(0\) otherwise. For any \(K>0\), choose \(n\) with \(2^n>K\). Then $$ \mathbb{E}\big[|Z_n|\mathbf{1}_{\{|Z_n|>K\}}\big] =2^n\mathbb{P}(\tau>n)=2^n2^{-n}=1. $$ This holds for arbitrarily large \(K\), so the family is not uniformly integrable. Check: the stopped expectation at deterministic time \(n\) is \(2^n\cdot2^{-n}=1\), while the almost-sure limit is \(0\). The expectations do not converge to the expectation of the limit because the family is not uniformly integrable.
Common mistake. From \(M_{n\wedge\tau}\to M_\tau\) almost surely, it is tempting to conclude that \(\mathbb{E}[M_{n\wedge\tau}]\to\mathbb{E}[M_\tau]\). Almost-sure convergence by itself does not justify that step. In the last example, the stopped expectations are all \(1\), but the limit has expectation \(0\). The missing condition is uniform integrability, not merely the fact that \(\tau\) is finite almost surely.
The theorem is useful whenever a random stopping rule has no fixed deterministic bound, but the stopped martingale values cannot send a persistent amount of expectation into increasingly rare, increasingly large outcomes. Check the actual family \(\{M_{n\wedge\tau}\}\): uniform integrability of some related family, or a bound on the stopping time alone, is not a substitute unless it implies this condition.
Check Your Understanding
For each question, identify which part of the theorem or proof supplies the needed step.
- Why does \(\tau<\infty\) almost surely imply \(M_{n\wedge\tau}\to M_\tau\) almost surely?
- Which condition permits the passage from this almost-sure convergence to convergence in mean?
- In the geometric example, calculate \(\mathbb{P}(Y>3)\) and the value of \(M_3\) on that event.
- In the fair-toss counterexample, for a threshold \(K=8\), choose \(n\) with \(2^n>K\) and calculate the corresponding tail expectation.
- Does the theorem assume \(\mathbb{E}[\tau]<\infty\)? State the assumptions it does require.