From an Unbounded Time to Bounded Times
Optional stopping at an unbounded time is not obtained merely by writing down the bounded-time identity and letting the time horizon increase. Two separate facts are needed: the truncated stopped values must converge to the value at the stopping time, and their expectations must converge as well. Almost-sure finiteness gives the first fact. Uniform integrability supplies the second.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a discrete-time filtration \((\mathcal{F}_n)_{n\ge0}\). Let \((M_n)_{n\ge0}\) be an integrable, real-valued martingale adapted to this filtration: for each \(n\), \(M_n\) is \(\mathcal{F}_n\)-measurable and integrable, and \(\mathbb{E}[M_n\mid\mathcal{F}_{n-1}]=M_{n-1}\) for \(n\ge1\). A stopping time \(\tau\) takes values in \(\{0,1,2,\ldots\}\cup\{\infty\}\) and satisfies \(\{\tau\le n\}\in\mathcal{F}_n\) for every \(n\). For an integer \(n\ge0\), its truncation is \(\tau\wedge n=\min(\tau,n)\).
First, truncation really does produce a bounded stopping time. For integers \(k,n\ge0\),
When \(k<n\), this event belongs to \(\mathcal{F}_k\) because \(\tau\) is a stopping time. When \(k\ge n\), it is all of \(\Omega\), which belongs to \(\mathcal{F}_k\). Thus \(\tau\wedge n\) is a stopping time and is bounded by \(n\).
The Limit Step: Uniform Integrability and Convergence
The precise passage from truncated values to the terminal value is an instance of a general fact. A sequence converging almost surely need not have expectations converging; uniform integrability rules out expectation being carried by increasingly rare, increasingly large values.
Proof. Choose \(K_0\) so large that \(\sup_n\mathbb{E}[|Y_n|\mathbf{1}_{\{|Y_n|>K_0\}}]\le1\). For every \(n\), splitting according to \(|Y_n|\le K_0\) or \(|Y_n|>K_0\) gives \(\mathbb{E}|Y_n|\le K_0+1\). Fatou's lemma and almost-sure convergence then give \(\mathbb{E}|Y|\le\liminf_n\mathbb{E}|Y_n|\le K_0+1\), so \(Y\) is integrable. For any \(K>0\), almost-sure convergence implies \(|Y|\mathbf{1}_{\{|Y|>K\}}\le\liminf_n |Y_n|\mathbf{1}_{\{|Y_n|>K\}}\): if \(|Y|>K\), then eventually \(|Y_n|>K\), and otherwise the left side is zero. Fatou's lemma therefore yields \(\mathbb{E}[|Y|\mathbf{1}_{\{|Y|>K\}}]\le\sup_n\mathbb{E}[|Y_n|\mathbf{1}_{\{|Y_n|>K\}}]\). Define \(c_K(x)=\max(-K,\min(x,K))\). This is continuous and bounded, so dominated convergence gives \(\mathbb{E}|c_K(Y_n)-c_K(Y)|\to0\) for each fixed \(K\). Also \(|x-c_K(x)|\le |x|\mathbf{1}_{\{|x|>K\}}\). Consequently, \(\mathbb{E}|Y_n-Y|\le \mathbb{E}|Y_n-c_K(Y_n)|+\mathbb{E}|c_K(Y_n)-c_K(Y)|+\mathbb{E}|c_K(Y)-Y|\). The first and third terms are bounded by the corresponding uniform-integrability tail bounds, which can be made arbitrarily small by choosing \(K\) large. For that fixed \(K\), the middle term tends to zero. Hence \(\mathbb{E}|Y_n-Y|\to0\). \(\square\)
Convergence in mean also gives convergence of expectations: \(|\mathbb{E}Y_n-\mathbb{E}Y|\le\mathbb{E}|Y_n-Y|\). This is the exact implication needed for optional stopping.
Optional Stopping by Truncation
Proof. We first verify the expectation identity for each truncation. For \(n\ge1\), the stopped value has the finite-sum representation \(M_{\tau\wedge n}=M_0+\sum_{k=1}^n\mathbf{1}_{\{\tau\ge k\}}(M_k-M_{k-1})\). Indeed, on each outcome the sum includes exactly the increments up to time \(\tau\wedge n\). For every \(k\), \(\{\tau\ge k\}=\{\tau>k-1\}\in\mathcal{F}_{k-1}\), so the martingale property gives \(\mathbb{E}[\mathbf{1}_{\{\tau\ge k\}}(M_k-M_{k-1})]=\mathbb{E}[\mathbf{1}_{\{\tau\ge k\}}\mathbb{E}[M_k-M_{k-1}\mid\mathcal{F}_{k-1}]]=0\). Taking expectations in the finite sum proves \(\mathbb{E}M_{\tau\wedge n}=\mathbb{E}M_0\). For \(n=0\), the same identity holds because \(\tau\wedge0=0\).
On the event \(\{\tau<\infty\}\), for every \(n\ge\tau\) we have \(M_{\tau\wedge n}=M_\tau\). Since this event has probability one, \(M_{\tau\wedge n}\to M_\tau\) almost surely. Apply the proposition to the uniformly integrable family \(Y_n=M_{\tau\wedge n}\). It gives integrability of \(M_\tau\) and \(\mathbb{E}M_{\tau\wedge n}\to\mathbb{E}M_\tau\). Each expectation on the left equals \(\mathbb{E}M_0\), so \(\mathbb{E}M_\tau=\mathbb{E}M_0\). \(\square\)
This proof separates the roles of the hypotheses. Almost-sure finiteness ensures that the truncations eventually equal the value at the finite stopping time. Uniform integrability justifies convergence of their expectations. The identity for each truncation comes from bounded stopping, here shown directly through a finite sum of martingale increments.
Worked Examples
Worked Example: A Geometric Stopping Time with Uniformly Integrable Truncations
Let \((\Omega,\mathcal{F},\mathbb{P})\) carry two independent sequences: fair coin tosses determining a time \(\tau\), and independent increments \(\varepsilon_k\) with \(\mathbb{P}(\varepsilon_k=1)=\mathbb{P}(\varepsilon_k=-1)=\tfrac12\). Let \(\tau\) be the index of the first tails in the first sequence, and set \(S_0=0\), \(S_n=\sum_{k=1}^n\varepsilon_k\). Use the filtration that reveals both sequences up to time \(n\). The process \((S_n)\) has state space \(\mathbb{Z}\), is a martingale, and \(\tau\) is a stopping time for this filtration.
The probability of no tails in the first \(n\) tosses is \(\mathbb{P}(\tau>n)=2^{-n}\), which tends to zero, so \(\tau<\infty\) almost surely. For \(k\ge1\), \(\mathbb{P}(\tau\ge k)=2^{-(k-1)}\). The number of increments in \(S_{\tau\wedge n}\) is \(\tau\wedge n\), and, conditional on this number, the independent symmetric increments have mean zero and variance equal to their number. Therefore
For any \(K>0\), the strict tail event satisfies \(\mathbb{E}[|S_{\tau\wedge n}|\mathbf{1}_{\{|S_{\tau\wedge n}|>K\}}]\le\mathbb{E}[S_{\tau\wedge n}^{\,2}]/K\le2/K\), since \(|x|\le x^2/K\) whenever \(|x|>K\). Thus the stopped truncations are uniformly integrable. The theorem gives \(\mathbb{E}S_\tau=0\). As a further check, conditional on \(\tau=k\), \(\mathbb{E}S_k=0\) and \(\mathbb{E}S_k^2=k\); moreover, \(\mathbb{E}\tau=\sum_{k\ge1}2^{-(k-1)}=2\). Hence \(\mathbb{E}S_\tau=0\) and \(\mathbb{E}S_\tau^2=2\), consistent with the truncation calculation.
Worked Example: Rare Large Values Can Defeat the Limit
Let \((M_n)_{n\ge0}\) be the martingale on fair coin tosses defined by \(M_0=1\) and \(M_n=2^n\) if the first \(n\) tosses are all heads, and \(M_n=0\) otherwise. Given that the first \(n-1\) tosses were heads, the conditional expectation of \(M_n\) is \(\tfrac12 2^n+\tfrac12 0=2^{n-1}=M_{n-1}\); after a tail, both the current and all later values are zero. Thus this is a martingale. Let \(\tau\) be the index of the first tails. Then \(\mathbb{P}(\tau>n)=2^{-n}\to0\), so \(\tau<\infty\) almost surely, and \(M_\tau=0\).
For each fixed \(n\), \(M_{\tau\wedge n}=0\) if a tail occurs by time \(n\), and equals \(2^n\) if the first \(n\) tosses are heads. The latter event has probability \(2^{-n}\). Therefore \(\mathbb{E}M_{\tau\wedge n}=2^n\cdot2^{-n}=1\), while \(M_{\tau\wedge n}\to0\) almost surely. For any fixed \(K>0\), choose \(n\) with \(2^n>K\). Then \(\mathbb{E}[|M_{\tau\wedge n}|\mathbf{1}_{\{|M_{\tau\wedge n}|>K\}}]=2^n\cdot2^{-n}=1\). The tail expectations therefore cannot tend uniformly to zero, so the family is not uniformly integrable. This check explains why the limit step cannot be inferred from almost-sure convergence alone.
Worked Example: What Changes When the Stopping Time May Be Infinite
Let \(Z\) be a fair Bernoulli random variable, take \(\mathcal{F}_0\) to be the trivial sigma-field and \(\mathcal{F}_n=\sigma(Z)\) for \(n\ge1\), and define \(M_0=\tfrac12\) and \(M_n=Z\) for \(n\ge1\). This is a martingale because \(\mathbb{E}Z=\tfrac12=M_0\), and after time \(1\) the process is constant. Its state space is \(\{0,1\}\). Define \(\tau=1\) when \(Z=1\), and \(\tau=\infty\) when \(Z=0\). For \(n=0\), \(\{\tau\le n\}=\varnothing\); for \(n\ge1\), \(\{\tau\le n\}=\{Z=1\}\in\mathcal{F}_n\). Thus \(\tau\) is a stopping time, but \(\mathbb{P}(\tau<\infty)=\tfrac12\), not one.
For every \(n\ge1\), \(M_{\tau\wedge n}=Z\), so the truncations are uniformly bounded and converge almost surely to \(Z\). However, on \(\{Z=0\}\), which has probability \(\tfrac12\), there is no finite index \(\tau\) at which to evaluate \(M_\tau\). The limit of the truncations is a value at infinite time on that event, not a finite-time stopped value. This example isolates the role of almost-sure finiteness: it is what identifies the almost-sure limit with \(M_\tau\) as a finite-time random variable.
Common Pitfalls and the Main Distinction
Common mistake. In the rare-large-values example, one might write \(\mathbb{E}M_\tau=\lim_n\mathbb{E}M_{\tau\wedge n}=1\). The calculation immediately before the limit shows why this is wrong: \(\mathbb{E}M_{\tau\wedge n}=1\) for every \(n\), but \(M_{\tau\wedge n}\to M_\tau=0\) almost surely. The expectations do not converge to the expectation of the limit because uniform integrability fails; for every fixed \(K\), some truncation has strict tail expectation exactly \(1\).
In an application, keep the two limit checks distinct. First establish \(\tau<\infty\) almost surely to obtain \(M_{\tau\wedge n}\to M_\tau\) almost surely. Then establish uniform integrability of the stopped truncations to obtain convergence in mean. Neither condition should be silently replaced by the other.
| Requirement | What it provides |
|---|---|
| \(\tau\wedge n\) is bounded | The expectation identity for each truncation |
| \(\mathbb{P}(\tau<\infty)=1\) | \(M_{\tau\wedge n}\to M_\tau\) almost surely |
| \(\{M_{\tau\wedge n}\}\) is uniformly integrable | Convergence in mean and convergence of expectations |
Check Your Understanding
For each question, identify which part of the truncation argument is being used.
- For a stopping time \(\tau\), write \(\{\tau\wedge n\le k\}\) separately for \(k<n\) and \(k\ge n\), and explain why both events belong to \(\mathcal{F}_k\).
- In the geometric-stopping example, compute \(\mathbb{E}[S_{\tau\wedge 2}^{\,2}]\) directly from the probabilities of \(\tau\wedge2=1\) and \(\tau\wedge2=2\). Does it agree with \(2(1-2^{-2})\)?
- For the rare-large-values example, fix \(K=3\). Find an \(n\) for which the strict tail event has positive probability, and compute its tail expectation.
- Which hypothesis identifies the almost-sure limit of \(M_{\tau\wedge n}\) as \(M_\tau\), and which hypothesis ensures convergence of expectations?
- Why is the expression \(M_\tau\) not a finite-time stopped value on \(\{\tau=\infty\}\)?