Expected Mass Beyond a Threshold
A random variable can be integrable even when it sometimes takes very large values. For a family of random variables, the issue is whether those rare large values can carry a substantial part of the expectation as the family varies. Uniform integrability answers this by asking whether one common threshold makes the expected contribution from the tails small for every member of the family.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). A real-valued random variable \(X\) is integrable if \(\mathbb{E}|X|<\infty\). Let \(\mathcal{X}\) be a family of integrable random variables on this space; the family may be finite, countable, or uncountable. For a threshold \(K\ge 0\), the tail event for \(X\) is \(\{|X|>K\}\). The strict inequality matters: values exactly equal to \(K\) are not in this event.
For each fixed integrable \(X\), its expected tail mass tends to zero as \(K\) increases. Uniform integrability is stronger: the threshold cannot be chosen separately for each \(X\). The supremum compares all family members at the same \(K\). For fixed \(X\), the tail expectation is nonincreasing in \(K\), since increasing \(K\) can only shrink the tail event. The supremum over the family is also nonincreasing, so the limit in the definition is well-defined.
Why Tail Mass Is the Right Quantity
The probability \(\mathbb{P}(|X|>K)\) records how often \(X\) enters the tail, but not how large its values are there. The expectation \(\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\) accounts for both frequency and magnitude. A family can have uniformly small tail probabilities while retaining a fixed amount of expected magnitude in its tails; a worked example below shows this explicitly.
Tail expectations also control the error from clipping a random variable at a threshold. Define the clipping function \(c_K(x)=\max(-K,\min(x,K))\), which leaves values in \([-K,K]\) unchanged and replaces values outside that interval by the nearest endpoint.
Proof. Fix a real number \(x\). If \(|x|\le K\), then \(c_K(x)=x\), so \(|x-c_K(x)|=0\). If \(|x|>K\), clipping gives \(|x-c_K(x)|=|x|-K\), which is at most \(|x|\). In both cases, $$ |x-c_K(x)|\le |x|\mathbf{1}_{\{|x|>K\}}. $$ Substituting \(X\) for \(x\) and taking expectations gives $$ \mathbb{E}|X-c_K(X)|\le \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right]. $$ By uniform integrability, choose \(K\) so that the supremum of the right-hand side over \(X\in\mathcal{X}\) is less than \(\varepsilon\). Taking the supremum in the inequality proves the claimed bound. \(\square\)
This proposition explains one practical use of tail control: the whole family can be replaced by bounded, clipped variables with uniformly small error in expected absolute value. The conclusion does not say that a single clipped variable works for every threshold; it says that one threshold \(K\) works for the entire family once the desired error is fixed.
Worked Examples
Worked Example: A Uniformly Bounded Family
Let \(U\) be uniform on \([0,1]\), on the probability space \(([0,1],\mathcal{B}([0,1]),\mathbb{P})\) with Lebesgue probability. For each integer \(n\ge1\), set \(X_n=2(-1)^n\mathbf{1}_{[0,1/2]}(U)\). Determine whether the family is uniformly integrable.
For every \(n\), \(|X_n|=2\) when \(U\in[0,1/2]\), an event of probability \(1/2\), and \(|X_n|=0\) otherwise. Thus $$ \mathbb{E}|X_n|=2\cdot\tfrac12+0\cdot\tfrac12=1. $$ If \(0\le K<2\), the event \(\{|X_n|>K\}\) is exactly \(\{U\in[0,1/2]\}\), so $$ \mathbb{E}\!\left[|X_n|\mathbf{1}_{\{|X_n|>K\}}\right] =2\cdot\tfrac12=1. $$ If \(K\ge2\), the strict event \(\{|X_n|>K\}\) is empty up to null sets, including at \(K=2\), and the tail expectation is \(0\). These values do not depend on \(n\), so the supremum over \(n\) is \(1\) for \(0\le K<2\) and \(0\) for \(K\ge2\). It tends to zero; the family is uniformly integrable.
Check: Every member is bounded in absolute value by \(2\), and the strict tail event is empty at \(K=2\). The exact tail calculation agrees with that bound.
Worked Example: An Unbounded Uniformly Integrable Family
On the same uniform probability space, define \(X_n=n\mathbf{1}_{(0,1/n^2)}(U)\) for each integer \(n\ge1\). The possible nonzero values grow without bound. Determine whether this family is uniformly integrable.
The interval \((0,1/n^2)\) has probability \(1/n^2\). Therefore each variable is integrable and $$ \mathbb{E}|X_n|=n\cdot\tfrac{1}{n^2}=\tfrac1n. $$ For a fixed threshold \(K\ge0\), if \(n\le K\), the nonzero value \(n\) does not exceed \(K\), so the strict tail event is empty. If \(n>K\), the tail event is \(\{U\in(0,1/n^2)\}\), and $$ \mathbb{E}\!\left[|X_n|\mathbf{1}_{\{|X_n|>K\}}\right] =n\cdot\tfrac{1}{n^2}=\tfrac1n. $$ The smallest positive integer \(n\) satisfying \(n>K\) is \(\lfloor K\rfloor+1\). Since \(1/n\) decreases with \(n\), the supremum over all integers \(n>K\) is \(1/(\lfloor K\rfloor+1)\). Including \(n\le K\), whose tail expectations are zero, gives $$ \sup_{n\ge1}\mathbb{E}\!\left[|X_n|\mathbf{1}_{\{|X_n|>K\}}\right] =\frac{1}{\lfloor K\rfloor+1}. $$ This tends to zero as \(K\to\infty\), so the family is uniformly integrable despite having no common finite bound on its possible values.
Check: At an integer threshold \(K=m\ge0\), the first index whose value is strictly above the threshold is \(n=m+1\). Its tail expectation is \((m+1)/(m+1)^2=1/(m+1)\), as in the supremum formula.
Worked Example: Small Tail Probabilities but No Uniform Integrability
On \(([0,1],\mathcal{B}([0,1]),\mathbb{P})\) with \(U\) uniform, let \(Y_n=n\mathbf{1}_{(0,1/n)}(U)\) for \(n\ge1\). Compare the tail probability and expected tail mass at a common threshold \(K\ge0\).
The interval \((0,1/n)\) has probability \(1/n\). For each \(n\), $$ \mathbb{E}|Y_n|=n\cdot\tfrac1n=1. $$ If \(n\le K\), then \(Y_n\) never exceeds \(K\), so both its tail probability and tail expectation are zero. If \(n>K\), then $$ \mathbb{P}(|Y_n|>K)=\tfrac1n,\qquad \mathbb{E}\!\left[|Y_n|\mathbf{1}_{\{|Y_n|>K\}}\right] =n\cdot\tfrac1n=1. $$ The smallest integer \(n>K\) is \(\lfloor K\rfloor+1\), so $$ \sup_{n\ge1}\mathbb{P}(|Y_n|>K)=\frac{1}{\lfloor K\rfloor+1}, \qquad \sup_{n\ge1}\mathbb{E}\!\left[|Y_n|\mathbf{1}_{\{|Y_n|>K\}}\right]=1. $$ The supremum tail probability tends to zero, but the supremum expected tail mass does not. Hence \((Y_n)_{n\ge1}\) is not uniformly integrable.
Check: At the first index above an integer threshold \(K=m\), the tail probability is \(1/(m+1)\), while the tail magnitude is \(m+1\). Their product is exactly \(1\). The small probability does not remove the expected mass because the tail value grows to compensate.
Basic Consequences and a Common Mistake
Proof. Choose any \(K\ge C\). For every member \(X\), the event \(\{|X|>K\}\) has probability zero, because \(|X|\le C\le K\) almost surely. Therefore \(\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]=0\) for every \(X\) in the family, and the supremum of these tail expectations is zero. The same holds for every larger \(K\), so the defining limit is zero. \(\square\)
A finite family of integrable random variables is also uniformly integrable. For each fixed member \(X\), the standard dominated convergence theorem gives \(\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\to0\), since the integrand tends to zero pointwise and is bounded by the integrable variable \(|X|\). For a finite family, the maximum of finitely many quantities tending to zero also tends to zero. This argument does not extend directly to an infinite family: the threshold needed for one member may fail for another.
Common mistake. In the last example, one might calculate \(\mathbb{P}(|Y_n|>K)=1/n\) for \(n>K\) and conclude that the expected tail mass is also small. That substitutes a probability for an expectation. The actual calculation is $$ \mathbb{E}\!\left[|Y_n|\mathbf{1}_{\{|Y_n|>K\}}\right] =n\cdot\frac1n=1, $$ not \(1/n\). The correction is to multiply each tail value by its probability, or, more generally, to calculate the tail expectation itself. Uniform integrability concerns this expected magnitude, not only the chance of entering the tail.
In martingale arguments, uniform integrability is useful because it rules out expectation being carried by increasingly rare, increasingly large values. The tail calculation is therefore a diagnostic as well as a definition: it asks whether a family’s large-value contribution can be made negligible uniformly. Small tail probabilities, bounded expectations considered one variable at a time, or convergence of typical values do not by themselves answer that question.
Check Your Understanding
Use the strict tail event \(\{|X|>K\}\), and compare family members at the same threshold whenever a supremum appears.
- For a family satisfying \(|X|\le 5\) almost surely for every member, what is the tail expectation at \(K=5\), and why does the strict inequality matter?
- For \(X_n=n\mathbf{1}_{(0,1/n^2)}(U)\), compute the supremum tail expectation at \(K=3\).
- For \(Y_n=n\mathbf{1}_{(0,1/n)}(U)\), compute both tail quantities at \(K=4\). Do they give the same conclusion about uniform integrability?
- Explain why the threshold in the definition must be common to the whole family rather than chosen separately for each random variable.
- How does the tail expectation bound the expected absolute error after clipping at \(K\)?