Uniform Integrability and Class D Martingales

Why Uniform Integrability Matters

Convergence in probability does not by itself control expectations: rare outcomes can have small probability but very large magnitude. Uniform integrability rules out this loss of mass in the tails. For martingales, the related class D condition requires this control not just at deterministic times, but across values taken at stopping times.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. The results below concern integrable random variables and martingales; no stochastic integral is used. All equalities between random variables are understood to hold almost surely. An integrable random variable $X$ satisfies $\mathbb{E}|X|<\infty$.

Definition: A family $\mathcal{X}$ of integrable random variables is uniformly integrable (UI) if $$ \lim_{K\to\infty}\ \sup_{X\in\mathcal{X}}\mathbb{E}\big[|X|\mathbf{1}_{\{|X|>K\}}\big]=0. $$ The tail event is strict: it is $\{|X|>K\}$, not $\{|X|\ge K\}$. A process $M=(M_t)_{0\le t\le T}$ is of class D on $[0,T]$ if the family $\{M_\tau:\tau\text{ is a stopping time with }0\le\tau\le T\}$ is UI. On an unbounded time interval, class D means UI over all almost surely finite stopping times in that interval.

Here and below, a stopping time $\tau$ is a random time for which $\{\tau\le t\}\in\mathcal{F}_t$ for every $t$. Class D is stronger than requiring the deterministic-time family $\{M_t\}$ to be UI: it also tests random times chosen using the observed history.

Uniform Integrability and Convergence in Mean

Theorem: Suppose $(X_n)_{n\ge1}$ is a UI family and $X_n\to X$ in probability, where $X$ is finite almost surely. Then $X$ is integrable, $X_n\to X$ in $L^1$, and $\mathbb{E}[X_n]\to\mathbb{E}[X]$.

Proof. Write $u(K)=\sup_n\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]$, so $u(K)\to0$. We first control the tail of $X$. From any subsequence of $X_n$, convergence in probability permits extraction of a further subsequence $X_{n_j}$ converging to $X$ almost surely. For fixed $K>0$, the pointwise inequality $$ |X|\mathbf{1}_{\{|X|>K\}}\le\liminf_{j\to\infty}|X_{n_j}|\mathbf{1}_{\{|X_{n_j}|>K\}} $$ holds. Indeed, on $\{|X|>K\}$, convergence implies that $|X_{n_j}|>K$ eventually, so the right side is $|X|$; outside that event the left side is zero. Fatou's lemma gives $\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\le u(K)$. In particular, taking $K$ sufficiently large makes this tail expectation finite and small; the part where $|X|\le K$ is bounded by $K$, so $X$ is integrable.

For $K>0$, let $c_K(x)=\max(-K,\min(x,K))$. Then $|x-c_K(x)|\le |x|\mathbf{1}_{\{|x|>K\}}$, and $|c_K(x)|\le K$. Since $c_K(X_n)\to c_K(X)$ in probability, for every $\varepsilon>0$, $$ \mathbb{E}|c_K(X_n)-c_K(X)|\le \varepsilon+2K\,\mathbb{P}(|c_K(X_n)-c_K(X)|>\varepsilon). $$ For fixed $K$, the limit superior as $n\to\infty$ is at most $\varepsilon$, and then letting $\varepsilon\downarrow0$ shows that this expectation tends to zero. The triangle inequality therefore gives $$ \limsup_{n\to\infty}\mathbb{E}|X_n-X| \le u(K)+\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\le 2u(K). $$ Let $K\to\infty$; this proves $L^1$ convergence. Finally, $|\mathbb{E}[X_n]-\mathbb{E}[X]|\le\mathbb{E}|X_n-X|\to0$. $\square$

Worked Example: A Bounded Martingale Is Class D

Suppose $M=(M_t)_{0\le t\le T}$ is a martingale with right-continuous paths, and $|M_t|\le 3$ almost surely for every fixed $t$. Show it is class D on $[0,T]$.

First, the bound holds for all times at once. Let $N$ be the union of the null sets on which $|M_t|>3$, taken over rational $t\in[0,T]$ and $t=T$. Then $N$ is null, and by right continuity, $|M_t(\omega)|\le3$ for every $t\in[0,T]$ and every $\omega\notin N$.

For any stopping time $\tau\le T$, the stopped value satisfies $|M_\tau|\le3$ almost surely. Thus, for every $K\ge3$, the strict tail event $\{|M_\tau|>K\}$ is empty up to a null set. Consequently, $$ \sup_{\tau\le T}\mathbb{E}\big[|M_\tau|\mathbf{1}_{\{|M_\tau|>K\}}\big]=0\qquad(K\ge3). $$ This is uniform integrability, so $M$ is class D. Each $M_\tau$ is integrable because it is bounded. The argument uses boundedness of the process, not an optional-stopping identity.

Check: at $K=3$, equality $|M_\tau|=3$ does not enter the strict tail event. The conclusion remains valid because only values strictly greater than $K$ are counted.

Why right continuity (or some simultaneous bound) is needed. A bound that holds almost surely at each fixed time does not by itself control $M_\tau$ at a random time. Let $U$ be uniform on $(0,1)$, let $\mathcal{F}_t$ contain $\sigma(U)$ for every $t$, and set $M_0=0$ and $M_t=U^{-1}\mathbf{1}_{\{U=t\}}$ for $t>0$. For each fixed $t$, $\mathbb{P}(U=t)=0$, so $M_t=0$ almost surely. Thus $M$ is a martingale with $|M_t|\le3$ almost surely at every fixed time. But $\tau=U$ is a stopping time, and $M_\tau=U^{-1}$, which is not even integrable because $\mathbb{E}[U^{-1}]=\int_0^1u^{-1}\,du=\infty$. This process does not have right-continuous paths, and it is not class D on $[0,1]$.

Conditional Expectations Form a UI Family

Proposition: If $X$ is integrable, then the family $\{\mathbb{E}[X\mid\mathcal{G}]:\mathcal{G}\text{ is a sub-}\sigma\text{-field of }\mathcal{F}\}$ is uniformly integrable.

Proof. Let $Y=\mathbb{E}[X\mid\mathcal{G}]$ and $A=\{|Y|>K\}$, which belongs to $\mathcal{G}$. Conditional Jensen's inequality gives $|Y|\le\mathbb{E}[|X|\mid\mathcal{G}]$. Hence $$ \mathbb{E}[|Y|\mathbf{1}_A]\le\mathbb{E}[|X|\mathbf{1}_A]. $$ For any $L>0$, split the right side according to whether $|X|>L$: $$ \mathbb{E}[|X|\mathbf{1}_A]\le \mathbb{E}[|X|\mathbf{1}_{\{|X|>L\}}]+L\mathbb{P}(A). $$ If $\mathbb{E}|X|=0$, then $X=0$ almost surely and all the conditional expectations vanish. Otherwise Markov's inequality and $\mathbb{E}|Y|\le\mathbb{E}|X|$ give $\mathbb{P}(A)\le\mathbb{E}|X|/K$. Given $\varepsilon>0$, integrability permits choosing $L$ so the first term is less than $\varepsilon/2$; then choose $K$ so that $L\mathbb{E}|X|/K<\varepsilon/2$. Both choices are independent of $\mathcal{G}$. Thus the tail expectations are uniformly less than $\varepsilon$, proving the proposition. $\square$

This proposition provides a useful route to class D. Suppose $X$ is integrable and $M_t=\mathbb{E}[X\mid\mathcal{F}_t]$ has an adapted, right-continuous version on $[0,T]$, with the filtration satisfying the usual conditions. Then $M$ is a martingale: it is integrable at each time and, for $s\le t$, the tower property gives $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$. In fact, $M$ is class D.

To see why stopping times are covered, let $\tau\le T$ be a stopping time. Approximate it from above by finite-valued stopping times $\tau_n$ on nested dyadic grids of $[0,T]$, including $T$; then $\tau_n\downarrow\tau$. For a finite-valued stopping time $S$, partitioning over its possible values and using the deterministic-time conditional expectation identity gives $M_S=\mathbb{E}[X\mid\mathcal{F}_S]$. The proposition therefore makes $\{M_{\tau_n}:n\ge1\}$ UI. Right continuity gives $M_{\tau_n}\to M_\tau$ almost surely. The convergence theorem just proved gives convergence in $L^1$. For every $A\in\mathcal{F}_\tau$, the stopping-time sigma-field inclusion $\mathcal{F}_\tau\subseteq\mathcal{F}_{\tau_n}$ implies $$ \mathbb{E}[\mathbf{1}_A M_{\tau_n}]=\mathbb{E}[\mathbf{1}_A X]. $$ Passing to the limit in $L^1$ gives $\mathbb{E}[\mathbf{1}_A M_\tau]=\mathbb{E}[\mathbf{1}_A X]$. Since $M_\tau$ is $\mathcal{F}_\tau$-measurable, it is a version of $\mathbb{E}[X\mid\mathcal{F}_\tau]$. The proposition applied to this conditional expectation shows that the family $\{M_\tau:\tau\le T\}$ is UI. Thus $M$ is class D.

Worked Example: A Conditional-Expectation Martingale

Let $\Omega=\{\omega_1,\omega_2,\omega_3,\omega_4\}$, with probability $1/4$ on each outcome. Define $X$ to take the values $4,0,-2,2$, respectively. Let $\mathcal{F}_0$ be trivial, let $\mathcal{F}_1$ reveal whether the outcome is in $\{\omega_1,\omega_2\}$ or $\{\omega_3,\omega_4\}$, and let $\mathcal{F}_2$ reveal the full outcome. Compute $M_t=\mathbb{E}[X\mid\mathcal{F}_t]$ at these three times.

The unconditional mean is $(4+0-2+2)/4=1$, so $M_0=1$. On the first $\mathcal{F}_1$-atom the conditional mean is $(4+0)/2=2$; on the second it is $(-2+2)/2=0$. At time 2 the value is $X$. Thus $M_1$ takes values $2$ and $0$, each with probability $1/2$, and $M_2$ takes values $4,0,-2,2$, each with probability $1/4$. The tower property verifies the martingale identities: $\mathbb{E}[M_1\mid\mathcal{F}_0]=(2+0)/2=1=M_0$, and $\mathbb{E}[M_2\mid\mathcal{F}_1]$ is $2$ on the first atom and $0$ on the second, exactly $M_1$.

All values have absolute value at most $4$, so every stopping-time value on this finite horizon is bounded by $4$, and the martingale is class D. More generally, the conditional-expectation proposition explains the same conclusion for right-continuous versions of conditional-expectation martingales on a bounded horizon. Check: $\mathbb{E}[M_2]=(4+0-2+2)/4=1=\mathbb{E}[M_0]$, as required.

Class D Is a Genuine Restriction

Worked Example: A Martingale That Is Not Class D

Let $U$ be uniform on $(0,1]$, and for integers $n\ge0$ put $A_n=\{U\le2^{-n}\}$. Let $\mathcal{F}_n=\sigma(A_0,\ldots,A_n)$, and define $M_n=2^n\mathbf{1}_{A_n}$. Verify the martingale property and show that the deterministic-time values are not UI.

For $n\ge1$, $A_n\subseteq A_{n-1}$, and $A_{n-1}$ has probability $2^{-(n-1)}$. Conditional on $A_{n-1}$, the event $A_n$ has probability $2^{-n}/2^{-(n-1)}=1/2$. Outside $A_{n-1}$, both $M_n$ and $M_{n-1}$ vanish. Therefore $\mathbb{E}[M_n\mid\mathcal{F}_{n-1}]=2^n(1/2)\mathbf{1}_{A_{n-1}}=M_{n-1}$. Also, $\mathbb{E}|M_n|=2^n2^{-n}=1$, so each value is integrable.

For any fixed $K>0$, choose $n$ with $2^n>K$. Then $\{|M_n|>K\}=A_n$, and $$ \mathbb{E}\big[|M_n|\mathbf{1}_{\{|M_n|>K\}}\big]=2^n\mathbb{P}(A_n)=1. $$ Thus the supremum of the tail expectations is at least $1$ for every $K$; the family is not UI.

This example can be viewed in continuous time by setting $M_t=M_{\lfloor t\rfloor}$ and $\mathcal{F}_t=\mathcal{F}_{\lfloor t\rfloor}$, where $\lfloor t\rfloor$ is the greatest integer not exceeding $t$. The process is right-continuous and is a martingale: between integer times it is constant, and across successive integer intervals the conditional expectation identity is the one just checked. Each deterministic time is a stopping time, so the failure of UI among the $M_n$ already proves that this process is not class D on $[0,\infty)$. Check: the expectation stays exactly $1$ at every integer time, while all of that expectation lies on an event of probability $2^{-n}$ with value $2^n$. This is the tail concentration that uniform integrability excludes.

A Threshold Pitfall

When using Fatou's lemma to control the tail of a limit, do not assume that the tail indicators converge at the threshold. For example, take constant random variables $X_n=K+1/n$ and $X=K$, for $K>0$. Then $X_n\to X$, but $X_n\mathbf{1}_{\{X_n>K\}}\to K$, whereas $X\mathbf{1}_{\{X>K\}}=0$. The claimed pointwise convergence fails at $|X|=K$.

The correct argument uses a one-sided inequality: on $\{|X|>K\}$, convergence implies $|X_n|>K$ eventually, so $|X|\mathbf{1}_{\{|X|>K\}}\le\liminf_n |X_n|\mathbf{1}_{\{|X_n|>K\}}$; outside $\{|X|>K\}$, the left side is zero. Fatou's lemma applies to this inequality, as in the proof above. This distinction matters whenever a strict tail event is used.

Takeaway: Uniform integrability prevents expectation from escaping into increasingly rare, increasingly large values. It turns convergence in probability into convergence in $L^1$. Class D imposes this tail control on a martingale at every stopping time, while martingales bounded simultaneously at all times (for example, right-continuous martingales bounded at each fixed time) and right-continuous conditional-expectation martingales with an integrable terminal variable provide important sufficient examples.

Check Your Understanding

For each question, distinguish deterministic-time control from the stronger stopping-time requirement in class D.

Why does a bound $|M_t(\omega)|\le C$, valid for all $t\in[0,T]$ outside one null set, imply that $\{M_\tau:\tau\le T\}$ is UI?
In the conditional-expectation proposition, where is the fact that $A=\{|\mathbb{E}[X\mid\mathcal{G}]|>K\}$ belongs to $\mathcal{G}$ used?
What does the convergence theorem imply about $\mathbb{E}[X_n]$ when $X_n\to X$ in probability and the family is UI?
For the nested-event martingale, calculate $\mathbb{E}[|M_n|\mathbf{1}_{\{|M_n|>K\}}]$ when $2^n>K$, and explain why this rules out UI.
Why is pointwise convergence of the strict-tail indicators not guaranteed when $|X|=K$?