Removing Stopping Times with Uniform Integrability

Why Stopping Times Must Be Removed Carefully

Localization lets us prove statements about a process after stopping it at increasingly distant cutoffs. The stopped processes may be true martingales even when the original process has only been shown to be a local martingale. But if the stopping times tend to the time horizon, almost-sure convergence of the stopped variables does not by itself allow us to pass expectations or conditional expectations through the limit. Uniform integrability is the additional control that makes this passage valid.

Throughout, work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ with respect to the filtration: for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$, and has distribution $N(0,t-s)$. In particular, adaptedness alone is not enough. A process that reveals future Brownian values in $\mathcal{F}_0$ need not be Brownian with respect to that filtration, and its stochastic integrals need not have the martingale properties used below.

Definition: A family of integrable random variables $\{X_\alpha\}$ is uniformly integrable if $$ \lim_{K\to\infty}\sup_\alpha\mathbb{E}\!\left[|X_\alpha|\mathbf{1}_{\{|X_\alpha|>K\}}\right]=0. $$ The tail expectation must become small uniformly over the whole family, not just separately for each member.

A useful sufficient condition is a uniform $L^p$ bound for some $p>1$. Indeed, on $\{|X|>K\}$, $|X|\le |X|^p/K^{p-1}$, so $$ \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right]\le \frac{\mathbb{E}|X|^p}{K^{p-1}}. $$ Thus $\sup_\alpha\mathbb{E}|X_\alpha|^p<\infty$ implies uniform integrability.

The Limit Principle

Lemma: If $(X_n)$ is uniformly integrable and $X_n\to X$ almost surely, then $X$ is integrable and $\mathbb{E}|X_n-X|\to0$.

Proof. Fix $K>0$, and let $c_K(x)=\max(-K,\min(x,K))$ be the truncation of $x$ to $[-K,K]$. Fatou's lemma gives $$ \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right] \le \liminf_{n\to\infty}\mathbb{E}\!\left[|X_n|\mathbf{1}_{\{|X_n|>K\}}\right], $$ because $X_n\to X$ almost surely implies $|X_n|\mathbf{1}_{\{|X_n|>K\}}$ has lower limit at least $|X|\mathbf{1}_{\{|X|>K\}}$. Uniform integrability therefore makes the left-hand side tend to zero as $K\to\infty$. Also $|c_K(X)|\le K$, so $X$ is integrable: its truncated part is integrable, and its tail expectation is finite for sufficiently large $K$.

For fixed $K$, $c_K(X_n)\to c_K(X)$ almost surely and $|c_K(X_n)-c_K(X)|\le2K$. Dominated convergence gives $\mathbb{E}|c_K(X_n)-c_K(X)|\to0$. The truncation error satisfies $|x-c_K(x)|\le |x|\mathbf{1}_{\{|x|>K\}}$. Consequently, the triangle inequality bounds $\mathbb{E}|X_n-X|$ by the truncated difference and the two tail expectations. First choose $K$ so that both tail terms are uniformly small, using uniform integrability and the Fatou bound for $X$; then let $n\to\infty$. This proves convergence in $L^1$. $\square$

Theorem: Let $\tau_n$ be stopping times increasing to a terminal time $T$ almost surely. Suppose that for every $n$, the stopped process $M_{t\wedge\tau_n}$, $0\le t\le T$, is a martingale, and that for every fixed $t\in[0,T]$, the family $\{M_{t\wedge\tau_n}:n\ge1\}$ is uniformly integrable. If $M_{t\wedge\tau_n}\to M_t$ almost surely for each $t$, then $M$ is an integrable martingale on $[0,T]$.

Proof. Fix $0\le s\le t\le T$. The lemma applied at time $t$ gives $M_{t\wedge\tau_n}\to M_t$ in $L^1$; applied at time $s$, it gives $M_{s\wedge\tau_n}\to M_s$ in $L^1$. For each $n$, the stopped martingale property says $$ \mathbb{E}[M_{t\wedge\tau_n}\mid\mathcal{F}_s]=M_{s\wedge\tau_n}. $$ Conditional expectation is an $L^1$ contraction: $\mathbb{E}|\mathbb{E}[Y\mid\mathcal{F}_s]|\le\mathbb{E}|Y|$. Therefore the left sides converge in $L^1$ to $\mathbb{E}[M_t\mid\mathcal{F}_s]$, while the right sides converge in $L^1$ to $M_s$. Hence $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ almost surely. Integrability of $M_t$ follows from the lemma, so this is precisely the martingale property. $\square$

In applications, the stopped martingale identity is obtained from a localizing sequence. The key extra check is uniform integrability at each fixed time. A uniform second-moment bound is a common way to verify it; bounds that grow with $n$ do not suffice.

Worked Examples: Convergence and Stochastic Integrals

Worked Example: Expectations Can Fail Without Uniform Integrability

Let $U$ be uniformly distributed on $(0,1)$, and set $X_n=n\mathbf{1}_{\{U\le1/n\}}$. Determine the almost-sure limit and the expectations, and test uniform integrability.

For every outcome with $U>0$, there is an integer $N>1/U$. For all $n\ge N$, $U>1/n$, so $X_n=0$. Thus $X_n\to0$ almost surely. Since $\mathbb{P}(U\le1/n)=1/n$, $$ \mathbb{E}[X_n]=n\mathbb{P}(U\le1/n)=n\cdot\frac1n=1. $$ The expectations do not converge to the expectation of the limit, which is $0$.

For any $K>0$, choose an integer $n>K$. On $\{U\le1/n\}$, $X_n=n>K$, and otherwise $X_n=0$. Hence $$ \mathbb{E}\!\left[|X_n|\mathbf{1}_{\{|X_n|>K\}}\right]=n\cdot\frac1n=1. $$ Taking the supremum over $n$ leaves the value $1$ for every $K$, so the family is not uniformly integrable. This exhibits exactly the missing control in the limit principle.

Check: $\mathbb{E}X_n=1$ for every $n$, while $X_n\to0$ almost surely. The tail computation also shows why the hypothesis of the lemma cannot be omitted.

For the stochastic-integral applications, suppose $H$ is progressively measurable and belongs to $\mathcal{L}^2$ on $[0,T]$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. With $B$ Brownian with respect to the filtration, the Ito integral is square-integrable, and its isometry gives $\mathbb{E}|\int_0^t H_s\,dB_s|^2=\mathbb{E}\int_0^t H_s^2\,ds$. These facts verify both integrability and the martingale property for such an integral. The point here is how the same conclusion follows by first stopping and then removing the stops.

Worked Example: Removing Energy Cutoffs for an Ito Integral

On $[0,2]$, take $H_s=B_s$. This is progressively measurable, and $$ \mathbb{E}\int_0^2 B_s^2\,ds=\int_0^2\mathbb{E}[B_s^2]\,ds =\int_0^2s\,ds=\left.\frac{s^2}{2}\right|_0^2=2. $$ Thus $H\in\mathcal{L}^2$. Write $A_t=\int_0^t B_s^2\,ds$ and $\tau_n=\inf\{t\ge0:A_t\ge n\}\wedge2$, with the infimum of the empty set taken as infinity. Since $A_2<\infty$ almost surely, for almost every outcome $\tau_n=2$ once $n>A_2$. Therefore $\tau_n\uparrow2$ and $M_{t\wedge\tau_n}\to M_t$ almost surely, where $M_t=\int_0^t B_s\,dB_s$.

The stopped integrand $B_s\mathbf{1}_{\{s\le\tau_n\}}$ is progressively measurable and square-integrable. Its integral is a martingale. By the Ito isometry, $$ \mathbb{E}|M_{t\wedge\tau_n}|^2 =\mathbb{E}\int_0^{t\wedge\tau_n}B_s^2\,ds \le\mathbb{E}\int_0^2B_s^2\,ds=2. $$ The $L^2$ bound implies uniform integrability, so the theorem removes the stopping times and shows that $M$ is a martingale. In particular, $\mathbb{E}M_2=0$.

Independent check: Ito's formula for $f(x)=x^2$ has $f'(x)=2x$ and $f''(x)=2$, so its correction term is $\tfrac12 f''(B_t)\,dt=dt$. It gives $B_2^2=2\int_0^2B_s\,dB_s+2$, or $M_2=(B_2^2-2)/2$. Since $\mathbb{E}B_2^2=2$, this again gives $\mathbb{E}M_2=0$.

Worked Example: A Deterministic Integrand and an Explicit Uniform Bound

On $[0,1]$, let $H_s=s$, and define $N_t=\int_0^t s\,dB_s$. This integrand is progressively measurable and belongs to $\mathcal{L}^2$, since $$ \mathbb{E}\int_0^1s^2\,ds=\int_0^1s^2\,ds=\left.\frac{s^3}{3}\right|_0^1=\frac13. $$ Use the energy cutoffs $\tau_n=\inf\{t:\int_0^t s^2\,ds\ge n\}\wedge1$. Because $\int_0^1s^2\,ds=1/3$, every integer $n\ge1$ gives $\tau_n=1$; the stopped variables equal $N_t$. More generally, regardless of these particular cutoffs, the Ito isometry gives $$ \mathbb{E}|N_{t\wedge\tau_n}|^2 =\mathbb{E}\int_0^{t\wedge\tau_n}s^2\,ds \le\int_0^1s^2\,ds=\frac13. $$ Thus, for $K>0$, $$ \mathbb{E}\!\left[|N_{t\wedge\tau_n}|\mathbf{1}_{\{|N_{t\wedge\tau_n}|>K\}}\right] \le\frac{\mathbb{E}|N_{t\wedge\tau_n}|^2}{K} \le\frac{1}{3K}. $$ This is uniform in $n$, so the stopped family is uniformly integrable. The limiting process is a martingale by the theorem, and $\mathbb{E}N_1=0$.

Check: the Ito isometry gives $\mathbb{E}N_1^2=1/3$, a finite second moment consistent with the displayed uniform bound. The bound tends to zero as $K\to\infty$, as required by the definition of uniform integrability.

When Localization Does Not Supply the Needed Bound

For a progressively measurable integrand with only pathwise finite energy, $\int_0^T H_s^2\,ds<\infty$ almost surely, the energy cutoffs still make the stopped integrals square-integrable: their accumulated energy is bounded by the cutoff. But a bound of the form $\mathbb{E}|M_{T\wedge\tau_n}|^2\le n$ grows with $n$. It does not prove uniform integrability. One must find a bound independent of $n$, or use another argument that establishes uniform integrability. In the examples above, the stronger assumption $H\in\mathcal{L}^2$ provided the uniform bound $\mathbb{E}\int_0^T H_s^2\,ds$.

Common mistake. The incorrect inference is: “$M_{t\wedge\tau_n}\to M_t$ almost surely and every stopped variable has expectation zero, so $\mathbb{E}M_t=0$.” Almost-sure convergence does not justify exchanging limit and expectation. In the concrete calculation above, $X_n=n\mathbf{1}_{\{U\le1/n\}}\to0$ almost surely, but $\mathbb{E}X_n=1$ for every $n$. The correction is to verify uniform integrability, apply the limit lemma to obtain $L^1$ convergence, and only then pass expectations. For conditional identities, the same $L^1$ convergence and the $L^1$-contraction property of conditional expectation justify passage to the limit.

Takeaway: To remove stopping times, establish three separate facts: the stopped processes satisfy the required martingale identity; they converge to the unstopped process; and their values at each fixed time are uniformly integrable. The last condition turns almost-sure convergence into $L^1$ convergence, which is what allows expectations and conditional expectations to pass through the limit.

Check Your Understanding

Use the definition of uniform integrability and the stopped-martingale theorem to justify each answer.

Why does a uniform bound on $\mathbb{E}|X_n|^2$ imply uniform integrability? Give the tail estimate.
For $X_n=n\mathbf{1}_{\{U\le1/n\}}$, compute the tail expectation at a fixed threshold $K>0$. What does it show?
In the Brownian-integrand example, identify the $L^2$ bound that is uniform in $n$, and explain why it implies uniform integrability.
Why does the stopped-integral bound $\mathbb{E}|M_{T\wedge\tau_n}|^2\le n$ alone fail to establish uniform integrability as $n$ varies?
Which contraction property of conditional expectation is used to pass the stopped martingale identities to their limit?

Removing Stopping Times with Uniform Integrability

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