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Integrability and Uniform Integrability · Tutorial 144 of 946

Uniform Integrability and Truncation Errors

Learn how clipping errors characterize uniform integrability, and how to calculate those errors in concrete examples.

Intermediate 9 min read

What You'll Learn

  • Define clipping at a level and express its pointwise error exactly.
  • Prove the equivalence between uniform integrability and uniformly vanishing clipping errors in L1.
  • Compare clipping errors with tail expectations using explicit bounds.
  • Calculate clipping errors for bounded variables and a Pareto variable.
  • Diagnose failure of uniform integrability in a rare-spike family.

Why clipping gives a useful criterion

Uniform integrability is defined by controlling the expected contribution of values far out in the tails. Clipping a random variable at a fixed level gives another way to inspect those tails: values inside the level are unchanged, while values beyond it are brought back to the boundary.

The difference between a variable and its clipped version is therefore concentrated in the tail. The size of this difference, measured in \(L^1\), turns out to characterize uniform integrability exactly. This equivalence is useful because it turns a tail condition into an approximation statement.

Clipping and its error

Let \((X_\alpha)_{\alpha\in A}\) be a family of integrable real-valued random variables on a probability space. For \(K>0\), define the clipping function \(T_K:\mathbb R\to\mathbb R\) by

$$ T_K(x)= \begin{cases} -K, & x<-K,\\ x, & -K\le x\le K,\\ K, & x>K. \end{cases} $$

Thus \(T_K(x)\) is \(x\) when \(x\) lies between \(-K\) and \(K\), and otherwise is the nearer endpoint. The pointwise error has a simple exact form:

$$ |x-T_K(x)|=(|x|-K)_+, \qquad u_+:=\max(u,0). $$

In particular, the \(L^1\) clipping error of \(X_\alpha\) is

$$ \mathbb E\bigl[|X_\alpha-T_K(X_\alpha)|\bigr] = \mathbb E\bigl[(|X_\alpha|-K)_+\bigr]. $$

The clipping error is not the same as the whole tail contribution \(\mathbb E[|X_\alpha|\mathbf 1_{\{|X_\alpha|>K\}}]\). A value of size \(x>K\), for example, contributes \(x-K\) to the clipping error but \(x\) to the tail expectation. The two quantities are comparable, which is what makes the characterization work.

Definition: The family \((X_\alpha)_{\alpha\in A}\) is uniformly integrable if
$$ \lim_{K\to\infty}\sup_{\alpha\in A} \mathbb E\bigl[|X_\alpha|\mathbf 1_{\{|X_\alpha|>K\}}\bigr]=0. $$

The clipping characterization

Theorem: A family \((X_\alpha)_{\alpha\in A}\) of integrable random variables is uniformly integrable if and only if
$$ \lim_{K\to\infty}\sup_{\alpha\in A} \mathbb E\bigl[|X_\alpha-T_K(X_\alpha)|\bigr]=0. $$
Proof. For every real \(x\) and every \(K>0\),
$$ (|x|-K)_+\le |x|\mathbf 1_{\{|x|>K\}}. $$
Indeed, if \(|x|\le K\), both sides are zero, and if \(|x|>K\), then \(|x|-K\le |x|\). Substituting \(x=X_\alpha\), taking expectations, and then taking the supremum over \(\alpha\) gives
$$ \sup_{\alpha\in A}\mathbb E[|X_\alpha-T_K(X_\alpha)|] \le \sup_{\alpha\in A}\mathbb E[|X_\alpha|\mathbf 1_{\{|X_\alpha|>K\}}]. $$
Consequently, uniform integrability implies that the supremum of the clipping errors tends to zero as \(K\) tends to infinity. For the reverse implication, fix \(K>0\). If \(|x|>K\), then
$$ |x|\le 2(|x|-K/2), $$ because this inequality is equivalent to \(K\le |x|\). If \(|x|\le K\), then \(|x|\mathbf 1_{\{|x|>K\}}=0\). Combining these cases yields, for every real \(x\),
$$ |x|\mathbf 1_{\{|x|>K\}} \le 2(|x|-K/2)_+. $$
Using the pointwise clipping-error identity at level \(K/2\), we obtain
$$ \sup_{\alpha\in A}\mathbb E[|X_\alpha|\mathbf 1_{\{|X_\alpha|>K\}}] \le 2\sup_{\alpha\in A}\mathbb E[|X_\alpha-T_{K/2}(X_\alpha)|]. $$
If the supremum of the clipping errors tends to zero with the clipping level, the right-hand side tends to zero as \(K\) tends to infinity. This is precisely uniform integrability. \(\square\)

The proof gives more than an equivalence of limits: it supplies explicit inequalities in both directions. The first says that clipping error is no larger than the tail expectation at the same level. The second controls the tail expectation at level \(K\) by twice the clipping error at level \(K/2\).

Worked Example: A uniformly bounded family

Worked Example: A uniformly bounded family

Suppose there is a finite constant \(M\) such that \(|X_\alpha|\le M\) almost surely for every \(\alpha\in A\). Take any \(K\ge M\). Then \(|X_\alpha|\le K\) almost surely, so clipping changes none of the variables:

$$ T_K(X_\alpha)=X_\alpha \quad\text{almost surely}, \qquad \mathbb E[|X_\alpha-T_K(X_\alpha)|]=0. $$

The supremum of these errors is zero for every \(K\ge M\), and therefore tends to zero. The theorem implies that the family is uniformly integrable. Directly, its tail expectations are also zero for every \(K\ge M\), since the event \(\{|X_\alpha|>K\}\) has probability zero.

Worked Example: A Pareto variable

Worked Example: A Pareto variable

Let \(X\) be a positive random variable with Pareto density \(f(x)=\alpha x^{-\alpha-1}\) for \(x\ge1\), where \(\alpha>1\). The condition \(\alpha>1\) ensures that \(X\) is integrable. For \(K\ge1\), its clipping error is

$$ \begin{aligned} \mathbb E[|X-T_K(X)|] &=\int_K^\infty (x-K)\alpha x^{-\alpha-1}\,dx\\ &=\alpha\left( \int_K^\infty x^{-\alpha}\,dx -K\int_K^\infty x^{-\alpha-1}\,dx \right)\\ &=\alpha\left( \frac{K^{1-\alpha}}{\alpha-1} -\frac{K^{1-\alpha}}{\alpha} \right)\\ &=\frac{K^{1-\alpha}}{\alpha-1}. \end{aligned} $$

In the third line, the two integrals are finite because \(\alpha>1\), and their values follow by evaluating the corresponding antiderivatives at \(K\) and at infinity. The final expression tends to zero as \(K\to\infty\), since \(1-\alpha<0\). Thus the singleton family consisting of \(X\) has vanishing clipping error.

For comparison, the tail expectation is

$$ \begin{aligned} \mathbb E[X\mathbf 1_{\{X>K\}}] &=\int_K^\infty x\alpha x^{-\alpha-1}\,dx\\ &=\frac{\alpha}{\alpha-1}K^{1-\alpha}. \end{aligned} $$

It is larger than the clipping error, as expected: the tail expectation counts the full value \(x\), whereas the clipping error counts only the excess \(x-K\).

Worked Example: Rare spikes and nonuniform errors

Worked Example: Rare spikes and nonuniform errors

For each \(n\in\mathbb N\), let \(X_n=n\) with probability \(1/n\), and let \(X_n=0\) with probability \(1-1/n\). Each variable is integrable, and

$$ \mathbb E[|X_n|]=n\cdot\frac1n=1. $$

Fix \(K>0\). If \(n\le K\), clipping at level \(K\) makes no change. If \(n>K\), the nonzero value \(n\) is clipped to \(K\), so its error is \(n-K\), occurring with probability \(1/n\). Hence

$$ \mathbb E[|X_n-T_K(X_n)|] = \begin{cases} 0, & n\le K,\\[4pt] \dfrac{n-K}{n}=1-\dfrac{K}{n}, & n>K. \end{cases} $$

For fixed \(K\), the values \(1-K/n\) approach \(1\) as integers \(n>K\) increase. They are all less than \(1\), so their supremum is \(1\). Thus

$$ \sup_{n\in\mathbb N}\mathbb E[|X_n-T_K(X_n)|]=1 \quad\text{for every }K>0. $$

The supremum does not tend to zero. By the theorem, this family is not uniformly integrable. The tail calculation shows the same failure directly: for every \(n>K\), the event \(\{|X_n|>K\}\) occurs exactly when \(X_n=n\), and therefore

$$ \mathbb E[|X_n|\mathbf 1_{\{|X_n|>K\}}] =n\cdot\frac1n=1. $$

For each fixed \(n\), the clipping error does eventually become zero when \(K\ge n\). The failure is that the level required to make the error small cannot be chosen uniformly over all \(n\).

What the criterion does and does not say

Uniform integrability is a uniform approximation property: one clipping level must make the \(L^1\) error small for every member of the family. It is not enough that each variable can be approximated well at some level depending on that variable. The rare-spike example makes this distinction explicit.

Another useful distinction is between clipping and deleting the tail. Clipping replaces values beyond \(K\) by \(K\) or \(-K\); deleting the tail would instead replace them by zero. These operations have different pointwise errors. For clipping, the error is \((|X|-K)_+\). For deletion, the error is \(|X|\mathbf 1_{\{|X|>K\}}\), the tail quantity in the definition. The theorem works because the two errors bound one another after, at most, a change of level.

In applications, the clipping formulation can be convenient when one wants to approximate a family by bounded variables while controlling the approximation error in \(L^1\). The theorem identifies exactly when the errors can be made small uniformly. It also warns against checking only individual variables: uniformity is the essential part of the condition.

Check Your Understanding

Use the pointwise clipping-error formula and the theorem to answer the following questions.

  1. For a real number \(x\), what is the exact value of \(|x-T_K(x)|\)?
  2. Why does uniform integrability imply uniformly vanishing clipping errors at the same level?
  3. What inequality bounds the tail expectation at level \(K\) by the clipping error at level \(K/2\)?
  4. For the Pareto density in the worked example, why does the clipping error tend to zero when \(\alpha>1\)?
  5. In the rare-spike example, why does the clipping error of each fixed \(X_n\) tending to zero fail to establish uniform integrability?

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