Why a Common Bound Matters
When a stopping time is unbounded, its stopped value need not be controlled by any one fixed deterministic time. Uniform integrability provides a way to control a whole family of random variables through their tails. A particularly useful sufficient condition is simple: if every member of the family is bounded in absolute value by the same integrable random variable, then the family is uniformly integrable. The bound may itself be unbounded; its integrability is what matters.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). A family \(\mathcal{X}\) of integrable random variables is uniformly integrable if its expected tails vanish uniformly:
The tail event uses the strict inequality \(|X|>K\). Uniformity means that one threshold \(K\) controls every member of the family at once. It is stronger than saying that each \(X\) is integrable separately, or that their expectations are bounded.
Proof. For each \(X\in\mathcal{X}\), the bound \(|X|\le Y\) and \(\mathbb{E}[Y]<\infty\) imply \(\mathbb{E}[|X|]\le\mathbb{E}[Y]<\infty\). Fix \(K\ge0\). On the event \(\{|X|>K\}\), the domination implies \(Y>K\), and \(|X|\le Y\). Therefore, almost surely, \(|X|\mathbf{1}_{\{|X|>K\}}\le Y\mathbf{1}_{\{Y>K\}}\). Taking expectations and then the supremum over \(X\in\mathcal{X}\) gives \(\sup_{X\in\mathcal{X}}\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\le\mathbb{E}[Y\mathbf{1}_{\{Y>K\}}]\). Since \(Y\) is integrable, dominated convergence applied as \(K\to\infty\) shows that the right side tends to zero. This is exactly uniform integrability. \(\square\)
The proof gives a useful quantitative bound at the same threshold \(K\): every member’s tail expectation is at most the tail expectation of \(Y\). No boundedness of \(Y\) is required. In applications, the main task is often to find one integrable bound that works for all the stopped values under consideration.
Worked Example: An Unbounded Integrable Dominator
Let \(J\) take values \(0,1,2,\ldots\) with \(\mathbb{P}(J=j)=2^{-(j+1)}\). Set \(Y=J+1\), and for each integer \(n\ge0\) define \(X_n=Y\mathbf{1}_{\{J\le n\}}\). Show directly that the family \(\{X_n:n\ge0\}\) is uniformly integrable.
The probabilities sum to one because \(\sum_{j=0}^{\infty}2^{-(j+1)}=1\). The mean of the proposed dominator is
For every \(n\), \(0\le X_n\le Y\). To check the tail bound explicitly, take an integer \(K\ge1\). The event \(X_n>K\) implies \(Y>K\), which is the event \(J\ge K\). Thus
For the last equality, write \(j=K+r\); the sum becomes \(2^{-(K+1)}\sum_{r=0}^{\infty}(K+r+1)2^{-r}=2^{-(K+1)}(2K+4)=(K+2)/2^K\). This bound tends to zero as \(K\to\infty\), and it does not depend on \(n\), proving uniform integrability. At \(K=2\), the bound is \((2+2)/2^2=1\), which also agrees with the tail sum \(\sum_{j=2}^{\infty}(j+1)/2^{j+1}=1\).
Check: \(X_n\le Y\) holds also when \(J>n\), since then \(X_n=0\). The computed bound is valid for the strict tail event \(X_n>K\), including when \(X_n=K\), which is not counted in that event.
Stopped Values of Martingales
Let \((\Omega,\mathcal{F},\mathbb{P})\) carry a filtration \((\mathcal{F}_n)_{n\ge0}\), and let \((M_n)_{n\ge0}\) be an adapted, integrable discrete-time martingale. A stopping time \(\tau\) takes values in \(\{0,1,2,\ldots\}\cup\{\infty\}\) and satisfies \(\{\tau\le n\}\in\mathcal{F}_n\) for each \(n\). The stopped value \(M_\tau\) is defined on \(\{\tau<\infty\}\); statements about it require \(\tau<\infty\) almost surely unless another definition is given.
If an integrable \(Y\) satisfies \(|M_n|\le Y\) almost surely for every \(n\), then for any collection of stopping times that are finite almost surely, \(|M_\tau|\le Y\) almost surely for each \(\tau\) in that collection. The domination theorem therefore gives uniform integrability of the stopped values. In particular, if \(|M_n|\le C\) almost surely for every \(n\), where \(C\) is a fixed finite constant, then \(Y=C\) is an integrable common bound. The martingale property is not needed for this domination argument; it matters when we use optional stopping to identify expectations.
Worked Example: A Bounded Martingale and a Data-Dependent Stop
Let \(\xi\) and \(\eta\) be independent random variables, each taking \(1\) and \(-1\) with probability \(1/2\), on a probability space with four equally likely outcomes. Take \(\mathcal{F}_0=\{\varnothing,\Omega\}\), \(\mathcal{F}_1=\sigma(\xi)\), and \(\mathcal{F}_n=\sigma(\xi,\eta)\) for \(n\ge2\). Define \(M_0=0\), \(M_1=\xi/2\), and \(M_n=(\xi+\eta)/2\) for \(n\ge2\).
This is a martingale: \(\mathbb{E}[M_1\mid\mathcal{F}_0]=0=M_0\), and independence with \(\mathbb{E}[\eta]=0\) gives \(\mathbb{E}[M_2\mid\mathcal{F}_1]=\xi/2=M_1\). After time \(2\), the process is constant. Its possible values are \(0\), \(\pm\tfrac12\), and \(-1,0,1\), so \(|M_n|\le1\) for every \(n\).
Define \(\tau=1\) if \(\xi=1\), and \(\tau=2\) if \(\xi=-1\). It is a stopping time because \(\{\tau\le1\}=\{\xi=1\}\in\mathcal{F}_1\) and \(\{\tau\le2\}=\Omega\). The stopped value is \(M_\tau=\tfrac12\) when \(\xi=1\). When \(\xi=-1\), it is \((\eta-1)/2\), which is \(0\) if \(\eta=1\) and \(-1\) if \(\eta=-1\). Consequently, \(M_\tau\) equals \(\tfrac12,0,-1\) with probabilities \(\tfrac12,\tfrac14,\tfrac14\), respectively, and
Check: The probabilities sum to \(\tfrac12+\tfrac14+\tfrac14=1\), and the largest possible absolute stopped value is \(1\). Hence \(|M_\tau|\le1\), as required by the common bound; the expectation also agrees with \(M_0=0\).
A second useful source of domination is a common deterministic bound on the stopping horizons, even when the martingale itself is not bounded over all time.
Proof. Define \(Y_N=\max_{0\le k\le N}|M_k|\). This is a nonnegative random variable and \(Y_N\le\sum_{k=0}^{N}|M_k|\), so \(\mathbb{E}[Y_N]\le\sum_{k=0}^{N}\mathbb{E}[|M_k|]<\infty\). For each stopping time in the family, \(\tau\le N\) implies \(|M_\tau|\le Y_N\) almost surely. The integrable-domination theorem, applied with the same \(Y_N\) for every \(\tau\), proves uniform integrability. \(\square\)
Worked Example: A Two-Step Horizon for a Random Walk
Let \(\xi_1,\xi_2\) be independent fair signs on the four equally likely outcomes, and let \(\mathcal{F}_0\) be trivial, \(\mathcal{F}_1=\sigma(\xi_1)\), and \(\mathcal{F}_n=\sigma(\xi_1,\xi_2)\) for \(n\ge2\). Set \(M_0=0\), \(M_1=\xi_1\), and \(M_n=\xi_1+\xi_2\) for \(n\ge2\). The conditional mean of \(M_1\) given \(\mathcal{F}_0\) is zero, and \(\mathbb{E}[M_2\mid\mathcal{F}_1]=\xi_1=M_1\); thus this process is a martingale.
For any stopping time \(\tau\le2\), its stopped value is dominated by \(Z=\max(|M_0|,|M_1|,|M_2|)\). If the two signs agree, which occurs with probability \(\tfrac12\), then \(Z=2\). If they disagree, which also occurs with probability \(\tfrac12\), then \(Z=1\). Therefore \(\mathbb{E}[Z]=\tfrac12\cdot2+\tfrac12\cdot1=\tfrac32\), so \(Z\) is an integrable common bound for all these stopped values.
For a particular stopping time, take \(\tau=1\) if \(\xi_1=1\), and \(\tau=2\) otherwise. When \(\xi_1=1\), \(M_\tau=1\). When \(\xi_1=-1\), \(M_\tau=-1+\xi_2\), which is \(0\) if \(\xi_2=1\) and \(-2\) if \(\xi_2=-1\). Hence \(M_\tau\) equals \(1,0,-2\) with probabilities \(\tfrac12,\tfrac14,\tfrac14\), and its expectation is
Check: The three probabilities sum to \(1\), and each listed stopped value has absolute value at most \(2\). The family-wide bound \(Z\) is sharper on half the outcomes: there \(Z=1\), while on the other half \(Z=2\). At every threshold \(K\ge2\), the strict tail event \(Z>K\) is empty, so the domination bound’s tail expectation is zero.
Using Domination at a Finite Unbounded Stop
Domination also explains why a bounded martingale can be stopped at a stopping time that is finite almost surely but has no deterministic upper bound. The truncations \(\tau\wedge n=\min(\tau,n)\) are bounded stopping times. Bounded optional stopping applies to each truncation. If the martingale is bounded by one integrable random variable, its truncated stopped values are dominated by that same variable, allowing expectations to pass to the limit.
Proof. For each \(n\), \(\tau\wedge n\) is a stopping time bounded by \(n\), so bounded optional stopping gives \(\mathbb{E}[M_{\tau\wedge n}]=\mathbb{E}[M_0]\). Since \(\tau<\infty\) almost surely, for almost every outcome there is an \(n\) after which \(\tau\wedge n=\tau\). Thus \(M_{\tau\wedge n}\to M_\tau\) almost surely. Also \(|M_{\tau\wedge n}|\le Y\) for every \(n\). The dominated convergence theorem therefore gives \(\mathbb{E}[M_\tau]=\lim_{n\to\infty}\mathbb{E}[M_{\tau\wedge n}]=\mathbb{E}[M_0]\). \(\square\)
For a martingale with \(|M_n|\le C\) for every \(n\), the corollary applies with \(Y=C\), so any almost surely finite stopping time has the same expected stopped value as the initial value. The assumption that the stopping time is finite almost surely matters: without it, \(M_\tau\) may not be defined on the event \(\{\tau=\infty\}\).
A Common Mistake: Bounded Expectations Are Not Domination
A common error is to treat a uniform bound on expectations as if it supplied a common integrable random-variable bound. Here is a concrete failure. On \(\Omega=[0,1]\) with uniform probability, define \(X_n=n\mathbf{1}_{[0,1/n]}\) for integers \(n\ge1\). Each \(X_n\) is integrable and \(\mathbb{E}[|X_n|]=n\cdot(1/n)=1\). Nevertheless, for any fixed \(K>0\), choose an integer \(n>K\). Since \(X_n=n\) on an event of probability \(1/n\),
Thus the supremum of the tail expectations is at least \(1\) for every \(K\), so the family is not uniformly integrable. The calculation \(\sup_n\mathbb{E}[|X_n|]=1\) is correct, but it does not prove uniform integrability. What fails is the existence of an integrable common dominator, not the integrability of each member.
For stopped martingale arguments, first identify the family being controlled and then find a bound that holds almost surely for every member of that family. A deterministic bound on all martingale values works; a finite common horizon works through the maximum of finitely many values. Once the bound is established, the tail criterion—not merely a bound on expectations—supplies uniform integrability.
Check Your Understanding
For each question, pay attention to whether the bound is common to the whole family and whether the tail event is strict.
- If \(|X|\le Y\) for every \(X\) in a family and \(Y\) is nonnegative and integrable, what tail expectation bounds the uniform tail at threshold \(K\)?
- Why does the condition \(\sup_X\mathbb{E}[|X|]<\infty\) not by itself imply uniform integrability?
- For a martingale bounded by a fixed constant at every time, explain why its stopped values over all almost surely finite stopping times are uniformly integrable.
- For stopping times \(\tau\le N\), give an integrable common bound for \(|M_\tau|\) and justify its integrability.
- In the corollary, why does almost-sure finiteness of \(\tau\) imply \(M_{\tau\wedge n}\to M_\tau\) almost surely?