Defining Uniform Integrability by Tail Expectations

Why Tail Expectations Matter

An integrable random variable has finite first absolute moment: its expected absolute value is finite. For a family of random variables, however, knowing that each member is integrable does not by itself describe whether their large values can be controlled uniformly across the family.

Uniform integrability makes that uniform control precise. Instead of asking only whether each variable has a finite expected absolute value, we examine the expected contribution from values whose absolute size exceeds a threshold. The threshold is allowed to grow, but the resulting tail expectations must become small uniformly over the entire family.

Throughout, let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \(\mathcal{X}\) be a nonempty family of real-valued integrable random variables on that space. Write \(\mathbf{1}_A\) for the indicator of an event \(A\).

Definition by Tail Expectations

Definition: The family \(\mathcal{X}\) is uniformly integrable if \[ \lim_{K\to\infty}\sup_{X\in\mathcal{X}}\mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right]=0. \] Equivalently, for every \(\varepsilon>0\), there is a finite \(K>0\) such that, for every \(X\in\mathcal{X}\), \[ \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right]<\varepsilon. \]

The order of the quantifiers is essential. First choose the desired tolerance \(\varepsilon\); then choose one threshold \(K\) that works for every member \(X\) of the family. The threshold may depend on \(\varepsilon\) and on the family, but it cannot depend on which \(X\) is being checked.

The event \(\{|X|>K\}\) selects the part of the sample space where \(X\) has large magnitude. Multiplying by \(|X|\) and taking an expectation measures how much that tail contributes to the first absolute moment. Thus the condition is not merely that large values are rare: it controls their expected magnitude as well.

For a fixed \(X\), the tail expectation is nonincreasing in \(K\). Indeed, if \(K_2\ge K_1\), then \(\{|X|>K_2\}\subseteq\{|X|>K_1\}\), so \[ \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K_2\}}\right] \le \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K_1\}}\right]. \] Uniform integrability asks for this decay to be uniform in \(X\).

Worked Example: A Uniformly Bounded Family

Suppose there is a finite constant \(M\ge 0\) such that \(|X|\le M\) almost surely for every \(X\in\mathcal{X}\). Take any \(K\ge M\). Then \(\mathbb{P}(|X|>K)=0\) for each member of the family, and therefore \[ \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right]=0. \] The supremum of these tail expectations is also zero. Since this holds for every \(K\ge M\), the family is uniformly integrable.

The boundary case \(K=M\) is valid because the defining event uses the strict inequality \(|X|>K\). This example also shows that no moment calculation is needed once a common almost-sure bound is known.

Worked Tail Calculations

Worked Example: A Single Exponential Random Variable

Let \(X\) have the exponential distribution with rate \(1\), so its density on \([0,\infty)\) is \(e^{-x}\). For \(K\ge0\), its tail expectation is \[ \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right] =\int_K^\infty x e^{-x}\,dx. \] Integration by parts gives \[ \int_K^\infty x e^{-x}\,dx =\left[-xe^{-x}\right]_K^\infty+\int_K^\infty e^{-x}\,dx =Ke^{-K}+e^{-K} =(K+1)e^{-K}. \] Here \(xe^{-x}\) tends to zero as \(x\to\infty\). Also, \((K+1)e^{-K}\to0\), so the singleton family consisting of \(X\) is uniformly integrable.

This calculation measures the expected size of \(X\) on its tail, not just the probability of that tail.

Worked Example: Exponential Rates in a Bounded Interval

For each \(\lambda\in[1,2]\), let \(X_\lambda\) have density \(\lambda e^{-\lambda x}\) on \([0,\infty)\). The family is \(\mathcal{X}=\{X_\lambda:1\le\lambda\le2\}\). For \(K\ge0\), direct integration yields \[ \mathbb{E}\!\left[|X_\lambda|\mathbf{1}_{\{|X_\lambda|>K\}}\right] =\int_K^\infty x\lambda e^{-\lambda x}\,dx =\left(K+\frac{1}{\lambda}\right)e^{-\lambda K}. \] Since \(\lambda\ge1\), we have \(1/\lambda\le1\) and \(e^{-\lambda K}\le e^{-K}\). Hence, for every \(\lambda\in[1,2]\), \[ \left(K+\frac{1}{\lambda}\right)e^{-\lambda K} \le (K+1)e^{-K}. \] Taking the supremum over \(\lambda\) preserves this upper bound. As \((K+1)e^{-K}\to0\), the whole family is uniformly integrable.

The common bound is the key point: calculating the tail for each rate separately would not, by itself, establish the uniform condition. The displayed inequality gives one bound that applies to every rate in the specified interval.

Uniform Integrability Bounds First Moments

The tail condition immediately gives an important consequence: the expected absolute values of all members of a uniformly integrable family are bounded by one finite constant.

Theorem: If \(\mathcal{X}\) is uniformly integrable, then \[ \sup_{X\in\mathcal{X}}\mathbb{E}|X|<\infty. \]

Proof. By uniform integrability, there is a finite \(K>0\) such that, for every \(X\in\mathcal{X}\), \[ \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right]<1. \] For each \(X\), split its absolute value according to whether it is at most \(K\) or exceeds \(K\): \[ |X|=|X|\mathbf{1}_{\{|X|\le K\}}+|X|\mathbf{1}_{\{|X|>K\}}. \] The first term is bounded above by \(K\) pointwise, so its expectation is at most \(K\), since \(\mathbb{P}(\Omega)=1\). The expected second term is less than \(1\) by the choice of \(K\). Consequently, \[ \mathbb{E}|X| \le K+\mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right] <K+1 \] for every \(X\in\mathcal{X}\). The right-hand side does not depend on \(X\), and therefore \(\sup_{X\in\mathcal{X}}\mathbb{E}|X|\le K+1<\infty\). \(\square\)

The proof uses a simple but useful decomposition. Values up to the chosen threshold contribute at most \(K\) to the expectation on a probability space; values above it contribute less than \(1\), uniformly over the family. The exact tolerance \(1\) is not important. Any fixed positive tolerance would produce a finite uniform bound.

Every Single Integrable Variable Is Uniformly Integrable

For comparison, when there is only one random variable, integrability already ensures that its tail expectations vanish. The convergence need not be uniform over a family of different variables; for a singleton, there is no family-wide variation to control.

Proposition: If \(X\) is integrable, then the singleton family \(\{X\}\) is uniformly integrable.

Proof. For each \(n\in\mathbb{N}\), define \(Y_n=|X|\mathbf{1}_{\{|X|\le n\}}\). The sequence \((Y_n)\) increases pointwise to \(|X|\), since \(X\) is real-valued and thus \(|X|\) is finite almost surely. By the Monotone Convergence Theorem, \[ \mathbb{E}Y_n\longrightarrow\mathbb{E}|X|. \] Because \(X\) is integrable, \(\mathbb{E}|X|<\infty\). Splitting \(|X|\) at the threshold \(n\) therefore gives \[ \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>n\}}\right] =\mathbb{E}|X|-\mathbb{E}Y_n \longrightarrow 0. \] For any real \(K\ge n\), the event \(\{|X|>K\}\) is contained in \(\{|X|>n\}\), so its tail expectation is no larger. Thus the tail expectation tends to zero as \(K\to\infty\), which proves uniform integrability of the singleton family. \(\square\)

What the First-Moment Bound Does and Does Not Say

Uniform integrability always gives a uniform first-moment bound, as the theorem shows. The definition, however, asks for more than that bound: it asks that the contributions from values beyond a sufficiently large threshold be small uniformly over the family. A bound on the total first moments alone does not state this tail requirement.

This distinction is worth keeping explicit when checking a family. The total expectation combines contributions from all sizes of \(|X|\), whereas the defining condition isolates the portion above \(K\). A family may have finite first absolute moment for every member, and even have a common bound on those moments, without the definition having been verified. To establish uniform integrability directly from its definition, one must control the supremum of the tail expectations.

Fix a tolerance.
Choose the desired error \(\varepsilon>0\) for the tail expectation.

Choose one threshold.
Find a finite \(K\) that works for every member of the family, not a different threshold for each variable.

Check the tail contribution.
Verify that the supremum of \(\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\) over the family is below \(\varepsilon\).

Uniform integrability is useful precisely because it packages this uniform tail control into a single condition. The first-moment bound is an immediate consequence, but tail expectations retain more information about how the family behaves at large magnitudes.

Check Your Understanding

Use the definition and proofs above to answer the following questions.

In the definition of uniform integrability, which quantities may depend on \(\varepsilon\), and which must work for every member of the family?
Why does a common almost-sure bound \(M\) imply that the tail expectation is zero when \(K=M\)?
For the exponential-rate family, what common upper bound controls all tail expectations, and why does it tend to zero?
In the proof that uniform integrability bounds first moments, how is the expectation split at the threshold \(K\)?
Why does the Monotone Convergence Theorem establish that the tail expectation of one integrable random variable tends to zero?