Uniform Integrability from an Integrable Envelope

Why a Common Envelope Controls the Tails

Uniform integrability asks whether the expected contribution from large values becomes small uniformly across a family of random variables. A useful way to guarantee this is to find one integrable random variable that dominates the absolute value of every member. The same envelope then controls every tail, even when the family has many members or a continuous parameter.

The key is that the event $\{|X|>K\}$ can occur only where the envelope itself exceeds $K$. Thus the tail of any family member is bounded by a tail of the single integrable envelope. We will make this comparison precise and then apply it in several settings.

Uniform Integrability and Integrable Envelopes

Let $\mathcal X$ be a family of real-valued random variables on a probability space. The family is uniformly integrable if every $X\in\mathcal X$ is integrable and its tail expectations vanish uniformly as the cutoff increases.

Definition: A family $\mathcal X$ of random variables is uniformly integrable if each $X\in\mathcal X$ satisfies $\mathbb E|X|<\infty$ and $$ \lim_{K\to\infty}\sup_{X\in\mathcal X} \mathbb E\!\left[|X|\mathbf 1_{\{|X|>K\}}\right]=0. $$

Here $\mathbf 1_A$ denotes the indicator of an event $A$. The supremum is taken over the whole family before the cutoff $K$ is sent to infinity. That order matters: the definition requires one cutoff to control the tails of every member at once.

An integrable envelope for $\mathcal X$ is a nonnegative integrable random variable $Y$ such that, for every $X\in\mathcal X$, $|X|\le Y$ almost surely. The almost-sure inequality may have a different null set for different $X$. No common null set for the entire family is needed, because the expectation comparison below is applied separately to each member.

Theorem: If a family $\mathcal X$ has an integrable envelope $Y$, then $\mathcal X$ is uniformly integrable. More precisely, for every $K\ge 0$, $$ \sup_{X\in\mathcal X} \mathbb E\!\left[|X|\mathbf 1_{\{|X|>K\}}\right] \le \mathbb E\!\left[Y\mathbf 1_{\{Y>K\}}\right]. $$

The Tail of an Integrable Envelope

The theorem relies on a basic fact: the expected tail of a nonnegative integrable random variable tends to zero. We state and prove the fact so the limiting step in the main argument is explicit.

Lemma: If $Y\ge 0$ and $\mathbb E Y<\infty$, then $$ \lim_{K\to\infty}\mathbb E\!\left[Y\mathbf 1_{\{Y>K\}}\right]=0. $$ Proof. For each outcome, $Y\mathbf 1_{\{Y>K\}}$ tends to zero as $K\to\infty$. Also, $$ 0\le Y\mathbf 1_{\{Y>K\}}\le Y, $$ and $Y$ is integrable. The dominated convergence theorem therefore gives convergence of the expectations to zero. $\square$

This lemma concerns the tail expectation, not just the probability of the tail event. A small probability alone would not be enough to control an expectation if the values on that event were sufficiently large. Integrability of $Y$ controls both the values and the probability through the tail expectation.

Proof of the Envelope Criterion

Proof. Fix $X\in\mathcal X$. Since $|X|\le Y$ almost surely and $Y$ is integrable, $$ \mathbb E|X|\le \mathbb E Y<\infty. $$ Thus every family member is integrable. Next, on the event $\{|X|>K\}$, the inequality $|X|\le Y$ implies $Y>K$. Consequently, $$ |X|\mathbf 1_{\{|X|>K\}} \le Y\mathbf 1_{\{Y>K\}} \quad\text{almost surely}. $$ Taking expectations gives the same bound for each $X$; taking the supremum over the family gives $$ \sup_{X\in\mathcal X} \mathbb E\!\left[|X|\mathbf 1_{\{|X|>K\}}\right] \le \mathbb E\!\left[Y\mathbf 1_{\{Y>K\}}\right]. $$ The right-hand side tends to zero by the lemma. Hence the left-hand side tends to zero as well, which is precisely uniform integrability. $\square$

The estimate is useful because its right-hand side does not depend on $X$. It replaces a potentially complicated supremum over a family by the tail expectation of one random variable. The theorem does not require independence, identical distributions, or any relationship between different members.

Worked Examples

Worked Example: Multiplying an Integrable Variable by Event Indicators

Let $Z$ be an integrable random variable, and let $\{A_\alpha:\alpha\in I\}$ be any collection of events. Define $X_\alpha=Z\mathbf 1_{A_\alpha}$. Since indicators take values between zero and one, $$ |X_\alpha|=|Z|\mathbf 1_{A_\alpha}\le |Z| \quad\text{almost surely}. $$ The random variable $Y=|Z|$ is nonnegative and integrable, so it is an envelope. For $K\ge0$, the envelope estimate gives $$ \sup_{\alpha\in I} \mathbb E\!\left[|X_\alpha|\mathbf 1_{\{|X_\alpha|>K\}}\right] \le \mathbb E\!\left[|Z|\mathbf 1_{\{|Z|>K\}}\right]. $$ By the lemma, the right-hand side tends to zero. Therefore $\{Z\mathbf 1_{A_\alpha}:\alpha\in I\}$ is uniformly integrable, regardless of how the events in the collection are related.

Worked Example: A Continuously Indexed Scaled Indicator Family

Take $\Omega=(0,1)$ with Lebesgue probability measure, and let $U(\omega)=\omega$, so $U$ is uniform on $(0,1)$. For $0<t\le1$, define $$ X_t=t^{-1/2}\mathbf 1_{\{U\le t\}}. $$ For each $\omega=u\in(0,1)$, if $t<u$, then $X_t(u)=0$. If $t\ge u$, then $t^{-1/2}\le u^{-1/2}$. Thus, for every such $t$, $$ |X_t|\le Y,\qquad Y=U^{-1/2}. $$ This envelope is integrable because $$ \mathbb E Y=\int_0^1 u^{-1/2}\,du=2. $$ For completeness, each member is integrable as well: $$ \mathbb E|X_t|=t^{-1/2}\mathbb P(U\le t) =t^{-1/2}t=\sqrt t<\infty. $$ The envelope theorem now proves that the whole continuously indexed family $\{X_t:0<t\le1\}$ is uniformly integrable. The parameter ranges over infinitely many values, but the same integrable random variable controls every member.

Worked Example: A Damped Parametric Transform

Let $Z$ be any integrable real-valued random variable and, for $s\ge0$, set $$ X_s=\frac{Z}{1+s|Z|}. $$ The denominator is at least one, so $$ |X_s|=\frac{|Z|}{1+s|Z|}\le |Z| \quad\text{almost surely}. $$ Also, $\mathbb E|X_s|\le\mathbb E|Z|<\infty$ for every $s$. Taking $Y=|Z|$ as the common envelope yields, for every $K\ge0$, $$ \sup_{s\ge0} \mathbb E\!\left[|X_s|\mathbf 1_{\{|X_s|>K\}}\right] \le \mathbb E\!\left[|Z|\mathbf 1_{\{|Z|>K\}}\right]. $$ The right-hand side tends to zero, so the family indexed by all $s\ge0$ is uniformly integrable. Notice that the result covers the endpoint $s=0$, where $X_0=Z$, as well as every positive parameter value.

How to Use the Criterion

In applications, the main task is often finding the envelope. The family may be presented through a formula involving an index or parameter. One seeks a bound that removes that parameter and leaves a single integrable random variable.

Identify the family.
Write the random variables being considered and the full range of their indices or parameters.

Bound their absolute values.
Find a nonnegative random variable $Y$ such that $|X|\le Y$ almost surely for each family member.

Check integrability once.
Verify $\mathbb E Y<\infty$. This also establishes integrability of every member.

Compare the tails.
Use the estimate by $\mathbb E[Y\mathbf 1_{\{Y>K\}}]$, which vanishes as $K$ increases.

A common pitfall is to check only that each member is integrable. Uniform integrability is stronger: the tail expectations must vanish uniformly over the family. The envelope criterion supplies exactly this uniform control. Conversely, the criterion is sufficient, not a requirement in the definition: a family can be uniformly integrable even when no useful common integrable envelope has been identified.

The estimate also clarifies why the envelope must be integrable. A pointwise bound by a random variable with infinite expectation does not give a vanishing integrable tail bound through this argument. The useful conclusion comes from having a single dominator whose own expected tail becomes negligible.

Check Your Understanding

Use the envelope estimate and its hypotheses to answer the following questions.

Why does $|X|\le Y$ imply that $\{|X|>K\}$ is contained in $\{Y>K\}$, up to a null set?
Which part of the proof establishes that every member of the family is integrable?
For $X_t=t^{-1/2}\mathbf 1_{\{U\le t\}}$, what is the common envelope, and why is it integrable?
Does the envelope argument require the family to be countable or its members to be independent?
Why is the condition $\mathbb E Y<\infty$ relevant to the conclusion about tail expectations?