Uniform Integrability Versus Tightness of Probability Laws

Two Different Kinds of Tails

Tightness and uniform integrability both describe how a family behaves at large values, but they measure different things. Tightness asks whether the probability of leaving a sufficiently large bounded interval is uniformly small. Uniform integrability asks whether the expected size contributed by those large values is uniformly small. The distinction matters because a rare event can have little probability and still contribute substantially to an expectation.

We allow a family of random variables to live on different probability spaces. For each index $i\in I$, let $(\Omega_i,\mathcal{F}_i,\mathbb{P}_i)$ be a probability space, and let $X_i:\Omega_i\to\mathbb{R}$ be a measurable, real-valued random variable. Its law is the probability measure $\mu_i$ on $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$ given by $\mu_i(B)=\mathbb{P}_i(X_i\in B)$ for Borel sets $B$. Tightness concerns these laws; it does not require the variables to share a probability space.

Definition: The family of laws $(\mu_i)_{i\in I}$ is tight if, for every $\varepsilon>0$, there is a compact set $C\subseteq\mathbb{R}$ such that $\sup_{i\in I}\mu_i(C^c)<\varepsilon$. Equivalently, for every $\varepsilon>0$, there is an $R>0$ such that $\sup_{i\in I}\mathbb{P}_i(|X_i|>R)<\varepsilon$. The equivalence holds because every compact subset of $\mathbb{R}$ is contained in some interval $[-R,R]$, and each such interval is compact.

The strict inequality in the tail event is part of this formulation: the complement of $[-R,R]$ is $\{x:|x|>R\}$, not $\{x:|x|\ge R\}$. Using $\ge R$ instead gives the same notion of tightness, since one can enlarge the chosen radius. Uniform integrability is a stronger kind of tail control.

Definition: A nonempty family $(X_i)_{i\in I}$ of integrable real-valued random variables is uniformly integrable if $$ \lim_{K\to\infty}\ \sup_{i\in I}\mathbb{E}_i\!\left[|X_i|\,\chi_{\{|X_i|>K\}}\right]=0. $$ Here $\mathbb{E}_i$ denotes integration with respect to $\mathbb{P}_i$. The indicator uses the strict tail event $|X_i|>K$. Uniform integrability controls the expected magnitude on the tail, whereas tightness controls only the probability of the tail.

Uniform Integrability Implies Tightness

Theorem: Let $(X_i)_{i\in I}$ be a nonempty family of integrable, real-valued random variables, possibly defined on different probability spaces. If the family is uniformly integrable, then its family of laws on $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$ is tight.

Proof. Fix $\varepsilon>0$. By uniform integrability, there is a finite $K_0\ge 0$ such that $\sup_{i\in I}\mathbb{E}_i[|X_i|\chi_{\{|X_i|>K_0\}}]<1$. Choose $R\ge K_0$ with $R>1/\varepsilon$. On the event $\{|X_i|>R\}$, we have $|X_i|>R$ and $|X_i|>K_0$. Therefore $$ R\,\mathbb{P}_i(|X_i|>R) \le \mathbb{E}_i[|X_i|\chi_{\{|X_i|>R\}}] \le \mathbb{E}_i[|X_i|\chi_{\{|X_i|>K_0\}}] <1. $$ The first inequality follows by integrating the pointwise inequality $R\chi_{\{|X_i|>R\}}\le |X_i|\chi_{\{|X_i|>R\}}$; the second follows from inclusion of the two tail events. Thus $\mathbb{P}_i(|X_i|>R)<1/R<\varepsilon$ for every $i$. Taking the supremum over $i$ gives $\sup_i\mathbb{P}_i(|X_i|>R)\le 1/R<\varepsilon$. Hence the interval $[-R,R]$ has probability greater than $1-\varepsilon$ under every law, and the family is tight. $\square$

The estimate makes the difference between the two notions explicit. On a tail beyond $R$, each value of $|X_i|$ is at least $R$, so a bound on the tail expectation yields a probability bound after division by $R$. Tightness has no corresponding way to bound the tail expectation: a very large value can offset a very small probability.

Worked Example: A Uniformly Bounded Family

For each $p\in[0,1]$, take $\Omega=\{-2,2\}$, $\mathcal{F}$ its power set, and define $\mathbb{P}_p(\{2\})=p$, $\mathbb{P}_p(\{-2\})=1-p$. Let $X_p(\omega)=\omega$. This family illustrates tightness and uniform integrability directly, including the endpoint cases $p=0$ and $p=1$.

For every $p$, the law is supported on $\{-2,2\}$, so $\mathbb{P}_p(|X_p|>2)=0$. Its law assigns mass $p+(1-p)=1$ to the compact interval $[-2,2]$, which proves tightness. For any $K\ge2$, the strict event $\{|X_p|>K\}$ is empty. Consequently, $$ \sup_{p\in[0,1]}\mathbb{E}_p[|X_p|\chi_{\{|X_p|>K\}}]=0. $$ Thus the family is uniformly integrable. As a check on the distinction between signed and absolute expectations, $\mathbb{E}_p[X_p]=2p-2(1-p)=4p-2$, while $\mathbb{E}_p[|X_p|]=2p+2(1-p)=2$. At $p=0$, these values are $-2$ and $2$; at $p=1$, they are $2$ and $2$. In both endpoint cases the same tail calculation applies.

Tight Laws Can Have Large Expectations

Worked Example: Rare Values with Diverging Expectations

Let $\Omega=[0,1]$ with its Lebesgue sigma-algebra and Lebesgue probability measure $m$. For each integer $n\ge1$, set $X_n=n^2\chi_{(0,1/n)}$. These are measurable, nonnegative, integrable random variables on one probability space. The interval $(0,1/n)$ has measure $1/n$, so direct integration gives $$ \mathbb{E}[X_n]=\int_0^1 n^2\chi_{(0,1/n)}\,dm=n^2\frac1n=n. $$ For example, the expectations at $n=1,2,3$ are $1,2,3$, respectively; in general they diverge as $n\to\infty$.

For a fixed $R\ge0$, the event $\{|X_n|>R\}$ is $(0,1/n)$ if $n^2>R$, and is empty if $n^2\le R$. The smallest integer satisfying $n^2>R$ is $q=\lfloor\sqrt R\rfloor+1$. Among all integers $n\ge q$, the largest probability $1/n$ is $1/q$. Hence $$ \sup_{n\ge1}\mathbb{P}(|X_n|>R)=\frac{1}{\lfloor\sqrt R\rfloor+1}\longrightarrow 0 \quad\text{as }R\to\infty. $$ The laws are tight. This calculation also checks the strict tail condition: an index with $n^2=R$ contributes no tail probability.

Nevertheless, for every finite $K\ge0$, choose an integer $n>\sqrt K$. Then $\{|X_n|>K\}=(0,1/n)$, and $$ \mathbb{E}[|X_n|\chi_{\{|X_n|>K\}}]=n^2\frac1n=n. $$ As such integers $n$ can be arbitrarily large, the supremum of these tail expectations is infinite for each finite $K$; the family is not uniformly integrable. As an independent check, the law of $X_n$ puts mass $1/n$ at $n^2$ and mass $1-1/n$ at $0$, giving the same expectation $n^2(1/n)+0(1-1/n)=n$.

This example pinpoints a common mistake: “the tail probability tends to zero, so the tail expectation tends to zero.” Those are different quantities. In the calculation above, the tail probability at the value $n^2$ is $1/n$, but its contribution to the expectation is $n^2(1/n)=n$, not a quantity tending to zero. Tightness permits rare values to grow rapidly enough that their expected contribution stays large or even diverges.

Tightness Does Not Preserve Expectations

Worked Example: Convergence to Zero with Expectations Equal to One

On $([0,1],\mathcal{L},m)$, define $Y_n=n\chi_{(0,1/n)}$, for integers $n\ge1$. For each fixed $x\in[0,1]$, $Y_n(x)\to0$: if $x>0$, then $1/n\le x$ for all sufficiently large $n$, and at $x=0$ every $Y_n(0)=0$. Thus convergence holds everywhere. But direct integration gives $$ \mathbb{E}[Y_n]=\int_0^1 n\chi_{(0,1/n)}\,dm=n\frac1n=1 $$ for every $n$, whereas the expectation of the pointwise limit $0$ is $0$.

The laws are also tight. For $R\ge0$, $\{|Y_n|>R\}$ has probability $1/n$ when $n>R$, and probability $0$ when $n\le R$. The smallest integer greater than $R$ is $\lfloor R\rfloor+1$, so $$ \sup_{n\ge1}\mathbb{P}(|Y_n|>R)=\frac{1}{\lfloor R\rfloor+1}\longrightarrow0. $$ The expectation calculation can be checked from the two possible values: $Y_n=n$ with probability $1/n$ and $Y_n=0$ with probability $1-1/n$, giving $n(1/n)+0(1-1/n)=1$.

In fact, the laws converge weakly to the point mass at $0$. Here weak convergence means convergence of integrals against every bounded continuous function $g:\mathbb{R}\to\mathbb{R}$. In this example, $$ \mathbb{E}[g(Y_n)]-\mathbb{E}[g(0)] =\frac1n\bigl(g(n)-g(0)\bigr), $$ whose absolute value is at most $2\|g\|_\infty/n$, tending to zero. Yet the expectations of $Y_n$ remain $1$, rather than converging to the expectation $0$ of the limiting law. The test function $g(x)=x$ is not bounded, so it is not among the functions used to define weak convergence. The calculation shows why tightness, even together with convergence of laws, does not by itself control expectations of unbounded functions.

What Tightness Does—and Does Not—Say

Tightness is a statement about where probability mass is located, uniformly across a family of laws. It is useful when the aim is to prevent most of the mass from escaping every bounded interval. Uniform integrability adds a size-sensitive condition: not only must the tail probability be small, but the integral of $|X_i|$ over that tail must also be small uniformly in $i$.

The two examples with rare values illustrate separate failures. For $X_n=n^2\chi_{(0,1/n)}$, tightness coexists with expectations that diverge. For $Y_n=n\chi_{(0,1/n)}$, tightness and pointwise convergence to zero coexist with expectations that do not converge to the expectation of the limit. In both cases, the exceptional intervals have small measure, but their heights are large enough to matter after integration. Uniform integrability rules out precisely this kind of persistent tail contribution; tightness alone does not.

Check Your Understanding

Use the strict tail event $\{|X|>R\}$ when checking tightness, and distinguish its probability from a tail expectation.

State the definition of tightness for a family of probability laws on $\mathbb{R}$, and explain why intervals $[-R,R]$ suffice.
In the theorem, identify the pointwise inequality that turns a tail-expectation bound into a probability bound.
For $X_n=n^2\chi_{(0,1/n)}$, compute the tail probability supremum at a fixed $R\ge0$, including the case when $R$ is a perfect square.
For the same $X_n$, compute the tail expectation at a fixed finite threshold $K$ and explain why its supremum over $n$ is infinite.
For $Y_n=n\chi_{(0,1/n)}$, verify both the expectation of each variable and the expectation of its pointwise limit.