Uniform Integrability from Higher-Moment Bounds

Why Higher Moments Control Tails

Uniform integrability asks whether a family’s expected contribution from large values becomes small uniformly over the family. A bound on the first moments alone does not provide the same tail control. A uniform bound on a moment of order \(p>1\), however, makes large values costly enough that their contribution to the expectation must shrink.

The key is to compare a variable’s size on the event where it exceeds a threshold with its \(p\)-th power. This comparison gives explicit rates: probabilities of exceeding a threshold decay at order \(K^{-p}\), while tail expectations decay at order \(K^{-(p-1)}\).

Uniform Integrability by Tail Expectations

Let \((\Omega,\mathcal F,\mathbb P)\) be a probability space, and let \(\mathcal X\) be a family of real-valued random variables on it. Write \(\mathbb E\) for expectation with respect to \(\mathbb P\).

Definition: The family \(\mathcal X\) is uniformly integrable if every \(X\in\mathcal X\) is integrable 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 matters: the tail expectation must become small for the whole family at once, not merely for each variable separately.

A higher-moment bound gives quantitative control of both the probability of a large value and its contribution to the expected absolute value.

The Higher-Moment Tail Estimates

Proposition: Let \(p>0\), let \(X\) be a random variable with \(\mathbb E|X|^p<\infty\), and let \(K>0\). Then

\mathbb P(|X|>K)\le \frac{\mathbb E|X|^p}{K^p}.

If \(p>1\), then also

\mathbb E\!\left[|X|\mathbf 1_{\{|X|>K\}}\right] \le \frac{\mathbb E|X|^p}{K^{p-1}}.

Proof. On the event \(\{|X|>K\}\), we have \(|X|^p>K^p\). Thus \(\mathbf 1_{\{|X|>K\}}\le |X|^p/K^p\), and taking expectations gives the probability bound. For the tail expectation, on the same event, \(|X|=|X|^p/|X|^{p-1}\le |X|^p/K^{p-1}\). Multiplying by the event indicator and taking expectations gives the second bound. The assumption \(p>1\) makes the resulting power \(K^{-(p-1)}\) decrease to zero as \(K\) increases.

\(\square\)

The probability estimate is Markov’s inequality applied to \(|X|^p\). The expectation estimate uses the same idea but compares \(|X|\), on the tail event, directly with \(|X|^p\).

Worked Example: A Uniform Second-Moment Bound

Suppose \(\mathcal X\) is a family of random variables satisfying \(\sup_{X\in\mathcal X}\mathbb E|X|^2\le16\). Taking \(p=2\) and \(C=16\) in the proposition gives, for every \(X\in\mathcal X\) and every \(K>0\),

\mathbb P(|X|>K)\le\frac{16}{K^2}, \qquad \mathbb E\!\left[|X|\mathbf 1_{\{|X|>K\}}\right]\le\frac{16}{K}.

For example, at \(K=4\), the probability of exceeding the threshold is at most \(1\), and the tail expectation is at most \(4\). At \(K=8\), the corresponding bounds are \(1/4\) and \(2\). The probability bound becomes informative as \(K\) grows beyond \(4\), and both estimates decay to zero. These bounds hold simultaneously for every member of the family.

From a Uniform Moment Bound to Uniform Integrability

Theorem: Let \(\mathcal X\) be a family of random variables on a probability space. If, for some \(p>1\),

C:=\sup_{X\in\mathcal X}\mathbb E|X|^p<\infty,

then \(\mathcal X\) is uniformly integrable. More precisely, for every \(K>0\),

\sup_{X\in\mathcal X}\mathbb P(|X|>K)\le\frac{C}{K^p}, \qquad \sup_{X\in\mathcal X}\mathbb E\!\left[|X|\mathbf 1_{\{|X|>K\}}\right] \le\frac{C}{K^{p-1}}.

Proof. First, every \(X\in\mathcal X\) is integrable: for every real number \(x\), \(|x|\le 1+|x|^p\). Therefore \(\mathbb E|X|\le1+\mathbb E|X|^p\le1+C<\infty\). Next, apply the proposition to each \(X\in\mathcal X\). Since \(\mathbb E|X|^p\le C\), its two inequalities give the displayed bounds uniformly over \(X\). In particular,

0\le \sup_{X\in\mathcal X}\mathbb E\!\left[|X|\mathbf 1_{\{|X|>K\}}\right] \le \frac{C}{K^{p-1}}\longrightarrow 0 \quad\text{as }K\to\infty,

because \(p-1>0\). This is exactly the tail-expectation condition in the definition of uniform integrability.

\(\square\)

The theorem is useful because it converts a single uniform moment estimate into both integrability and uniform tail control. The explicit exponent is important: the probability bound has power \(p\), while the tail-expectation bound loses one power and has power \(p-1\).

Worked Example: Rare Events with Bounded \(p\)-th Moments

Let \(p>1\), and for each \(n\in\mathbb N\), let \(A_n\) be an event with \(\mathbb P(A_n)=1/n\). Define \(X_n=n^{1/p}\mathbf 1_{A_n}\). Then

\mathbb E|X_n|^p =n\,\mathbb P(A_n) =1.

Thus the family has a uniform \(p\)-th-moment bound with \(C=1\). For \(K>0\), the event \(\{|X_n|>K\}\) is \(A_n\) when \(n^{1/p}>K\), and is empty otherwise. Consequently,

\mathbb P(|X_n|>K)= \begin{cases} 1/n,&n>K^p,\\ 0,&n\le K^p. \end{cases}

If \(m=\lfloor K^p\rfloor+1\), the smallest integer strictly greater than \(K^p\), then the largest such probability is \(1/m\). In particular, \(1/m\le K^{-p}\). The tail expectation is

\mathbb E\!\left[|X_n|\mathbf 1_{\{|X_n|>K\}}\right] = \begin{cases} n^{-(p-1)/p},&n>K^p,\\ 0,&n\le K^p. \end{cases}

Since \(n^{-(p-1)/p}\) decreases with \(n\), its supremum is \(m^{-(p-1)/p}\le K^{-(p-1)}\). Both tail bounds tend to zero as \(K\to\infty\), confirming uniform integrability. Although the nonzero values grow with \(n\), their \(p\)-th moments remain bounded, which is precisely the control used by the theorem.

Worked Example: A Scaled Exponential Family

Let \(Y\) have the exponential distribution with density \(e^{-y}\) for \(y\ge0\), and set \(X_t=tY\) for \(t\in[0,1]\). Integration by parts gives

\mathbb E[Y^2]=\int_0^\infty y^2e^{-y}\,dy =2\int_0^\infty ye^{-y}\,dy =2.

The boundary terms vanish because \(y^2e^{-y}\) and \(ye^{-y}\) tend to zero at infinity, and the remaining integral \(\int_0^\infty e^{-y}\,dy\) equals \(1\). Therefore \(\mathbb E|X_t|^2=t^2\mathbb E[Y^2]=2t^2\le2\) for every \(t\in[0,1]\). With \(p=2\) and \(C=2\), the theorem gives

\sup_{0\le t\le1}\mathbb P(|X_t|>K)\le\frac{2}{K^2}, \qquad \sup_{0\le t\le1} \mathbb E\!\left[|X_t|\mathbf 1_{\{|X_t|>K\}}\right]\le\frac{2}{K}.

The parameter range includes \(t=0\), for which \(X_0=0\); the same bounds still hold. The estimate applies to the entire continuum of scale parameters, not just to a selected sequence.

How to Use the Bounds

In an application, the main task is usually to find a moment order \(p>1\) and a constant \(C\) that work for every variable in the family. Once those are known, the tail estimates follow directly. A larger available moment order can give a faster decay in the tail expectation, provided its uniform bound is finite.

Choose a moment order.
Identify \(p>1\) for which the \(p\)-th absolute moments are finite uniformly over the family.

Record the uniform constant.
Set \(C=\sup_{X\in\mathcal X}\mathbb E|X|^p\), and check that \(C<\infty\).

Apply the tail estimates.
For a threshold \(K>0\), use \(C/K^p\) for tail probabilities and \(C/K^{p-1}\) for tail expectations.

Take the threshold to infinity.
Since \(p-1>0\), the tail-expectation bound vanishes, proving uniform integrability.

A Common Pitfall: The Order Must Exceed One

For uniform integrability, it is not enough to obtain a tail-expectation estimate that stays bounded; it must tend to zero as \(K\) grows. The estimate here has the form \(C/K^{p-1}\), and its convergence to zero depends on \(p>1\). At \(p=1\), the displayed comparison does not produce a decreasing power of \(K\), so a uniform first-moment bound alone cannot be used in this argument to conclude uniform integrability.

It is also important not to confuse a small tail probability with a small tail expectation. A rare event can carry values large enough to contribute substantially to the expectation. The higher-moment estimate controls both quantities and provides separate bounds for each.

Check Your Understanding

Use the definition and estimates in this tutorial to answer the following questions.

If \(\sup_{X\in\mathcal X}\mathbb E|X|^3\le 27\), what bounds follow for \(\mathbb P(|X|>K)\) and \(\mathbb E[|X|\mathbf 1_{\{|X|>K\}}]\)?
Why does the tail-expectation estimate require \(p>1\) to imply uniform integrability?
For \(X_n=n^{1/p}\mathbf 1_{A_n}\) with \(\mathbb P(A_n)=1/n\), calculate \(\mathbb E|X_n|^p\).
In the scaled exponential example, what uniform constant works for the second moments of \(X_t=tY\), \(0\le t\le1\)?
Which part of the theorem’s conclusion follows from the probability tail bound, and which part establishes uniform integrability?