Three Ways to Control a Family of Random Variables
Uniform integrability is a way to rule out a particular failure of integrability: a family may have bounded expectations while more and more of its expected magnitude is concentrated on events of smaller and smaller probability. Tail expectations detect that concentration directly. Truncation errors measure the same effect by clipping large values, while a small-event criterion describes it without referring to a particular threshold. This tutorial proves the equivalence of these tests and shows what each one reveals.
Fix a probability space \((\Omega,\mathcal{F},\mathbb{P})\), an arbitrary index set \(I\), and a family \((X_\alpha)_{\alpha\in I}\) of real-valued integrable random variables on that space. Expectations and event statements below are understood up to almost-sure equality. For \(K\ge0\), define the clipped value \(C_K(X)=\operatorname{sgn}(X)(|X|\wedge K)\), where \(a\wedge b=\min(a,b)\) and \(\operatorname{sgn}(0)=0\). Thus \(C_K(X)\) agrees with \(X\) when \(|X|\le K\) and clips its magnitude to \(K\) otherwise.
Tail Expectations and Truncation Errors
The tail expectation counts the full magnitude on the event that the magnitude exceeds \(K\). The truncation error instead counts only the amount clipped off. For every real \(x\), \(|x-C_K(x)|=(|x|-K)^+\), where \(y^+=\max(y,0)\). These quantities are different at the same threshold, but bounds at thresholds \(K\) and \(K/2\) show that their uniform limits agree.
- \(\lim_{K\to\infty}\sup_{\alpha\in I}\mathbb{E}[|X_\alpha|\mathbf{1}_{\{|X_\alpha|>K\}}]=0\).
- \(\lim_{K\to\infty}\sup_{\alpha\in I}\mathbb{E}[|X_\alpha-C_K(X_\alpha)|]=0\).
Proof. For any real \(x\) and \(K>0\), if \(|x|\le K\), both \((|x|-K)^+\) and \(|x|\mathbf{1}_{\{|x|>K\}}\) are zero. If \(|x|>K\), then \(0\le |x|-K\le |x|\). Therefore \[ (|x|-K)^+\le |x|\mathbf{1}_{\{|x|>K\}}. \] In the other direction, if \(|x|>K\), then \(|x|/2<|x|-K/2\), so \(|x|\le2(|x|-K/2)^+\); if \(|x|\le K\), the product \(|x|\mathbf{1}_{\{|x|>K\}}\) is zero. Consequently, \[ |x|\mathbf{1}_{\{|x|>K\}}\le2(|x|-K/2)^+. \] Apply these pointwise bounds to each \(X_\alpha\), take expectations, and then take the supremum over the same index set \(I\). The first bound shows that vanishing tail expectations imply vanishing truncation errors. The second shows that vanishing truncation errors imply vanishing tail expectations, since \(K/2\to\infty\) as \(K\to\infty\). The identity \(|X_\alpha-C_K(X_\alpha)|=(|X_\alpha|-K)^+\) completes both directions. \(\square\)
The second inequality deliberately uses a lower truncation threshold. It does not assert that the tail expectation and the clipping error are equal at \(K\). This distinction matters when calculating a criterion for a particular family.
Worked Example: A Constant Family with an Exponential Distribution
On \(([0,\infty),\mathcal{B}([0,\infty)),\mathbb{P})\), let \(X\) have density \(e^{-x}\) for \(x\ge0\), and take \(X_\alpha=X\) for every \(\alpha\in I\). Its expectation is \(\int_0^\infty xe^{-x}\,dx=1\). Compute the tail expectation and truncation error and verify that this family is uniformly integrable.
For \(K\ge0\), the strict tail event is \(X>K\). Integration by parts gives \[ \mathbb{E}[X\mathbf{1}_{\{X>K\}}] =\int_K^\infty xe^{-x}\,dx =\left[-(x+1)e^{-x}\right]_K^\infty =(K+1)e^{-K}. \] The clipping error is \[ \mathbb{E}[|X-C_K(X)|] =\int_K^\infty(x-K)e^{-x}\,dx =(K+1)e^{-K}-K e^{-K} =e^{-K}, \] where \(\int_K^\infty e^{-x}\,dx=e^{-K}\). Both expressions tend to zero, so both criteria hold. As a check, the tail expectation minus \(K\mathbb{P}(X>K)\) equals the clipping error: \((K+1)e^{-K}-Ke^{-K}=e^{-K}\). The probability used here is \(\mathbb{P}(X>K)=e^{-K}\), obtained by integrating the density.
Bounded First Moments and Small Events
Uniform integrability can also be expressed using all events of small probability. The relevant condition is stronger than merely bounding \(\mathbb{E}|X_\alpha|\): the integrals must become small uniformly over both the family and the event.
- \(\sup_{\alpha\in I}\mathbb{E}|X_\alpha|<\infty\).
- \(\lim_{\delta\downarrow0}\sup_{\alpha\in I}\sup_{A\in\mathcal{F}:\,\mathbb{P}(A)\le\delta}\mathbb{E}[|X_\alpha|\mathbf{1}_A]=0\).
Proof. First suppose the family is uniformly integrable. Choose a finite \(K>0\) for which \(\sup_\alpha\mathbb{E}[|X_\alpha|\mathbf{1}_{\{|X_\alpha|>K\}}]\le1\). Splitting according to \(|X_\alpha|\le K\) and \(|X_\alpha|>K\) gives \[ \mathbb{E}|X_\alpha| \le K+\mathbb{E}[|X_\alpha|\mathbf{1}_{\{|X_\alpha|>K\}}]\le K+1. \] Thus the first condition holds. To prove the second, let \(\varepsilon>0\). Choose \(K>0\) so that the supremum of the tail expectations at \(K\) is less than \(\varepsilon/2\). For any event \(A\), \[ \mathbb{E}[|X_\alpha|\mathbf{1}_A] \le K\mathbb{P}(A)+\mathbb{E}[|X_\alpha|\mathbf{1}_{\{|X_\alpha|>K\}}]. \] Indeed, on \(A\cap\{|X_\alpha|\le K\}\) the integrand is at most \(K\), and the remaining part is bounded by the full tail integral. If \(\mathbb{P}(A)\le\delta=\varepsilon/(2K)\), this bound is less than or equal to \(\varepsilon\), uniformly in \(\alpha\). Hence the small-event condition holds.
Conversely, suppose both conditions hold, and write \(C=\sup_\alpha\mathbb{E}|X_\alpha|<\infty\). If \(C=0\), then \(|X_\alpha|=0\) almost surely for every \(\alpha\), so the family is uniformly integrable. Suppose \(C>0\). Given \(\varepsilon>0\), the small-event condition supplies \(\delta>0\) such that every \(A\in\mathcal{F}\) with \(\mathbb{P}(A)\le\delta\) satisfies \(\sup_\alpha\mathbb{E}[|X_\alpha|\mathbf{1}_A]<\varepsilon\). For any \(K>C/\delta\), Markov's inequality gives \[ \mathbb{P}(|X_\alpha|>K)\le\frac{\mathbb{E}|X_\alpha|}{K} \le\frac{C}{K}<\delta. \] Apply the small-event condition to \(A=\{|X_\alpha|>K\}\), separately for each \(\alpha\). It follows that \(\sup_\alpha\mathbb{E}[|X_\alpha|\mathbf{1}_{\{|X_\alpha|>K\}}]\le\varepsilon\). Since \(\varepsilon\) was arbitrary, the tail expectations tend uniformly to zero. This proves the converse and the equivalence. \(\square\)
The bounded-first-moment assumption in this theorem cannot be dropped. Small-event control alone says nothing about a variable that is large on the whole probability space. Conversely, bounded first moments alone allow the expectation to concentrate on rarer events, as the next example shows.
Worked Example: Rare Spikes with Bounded First Moments
Let \(U\) be uniform on \((0,1)\) on the probability space \(((0,1),\mathcal{B}((0,1)),\mathbb{P})\), and define \(X_n=n\mathbf{1}_{\{U\le1/n\}}\) for integers \(n\ge1\). Every variable is integrable. Check the first moments, the tail criterion, the truncation error, and the small-event condition.
The event \(\{U\le1/n\}\) has probability \(1/n\), so \(\mathbb{E}|X_n|=n(1/n)=1\) for every \(n\). Thus the first moments are uniformly bounded. For a fixed \(K\ge0\), the strict tail event \(|X_n|>K\) occurs precisely on \(\{U\le1/n\}\) when \(n>K\); for \(n\le K\) it is empty. Therefore \[ \mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}] =\begin{cases}1,&n>K,\\0,&n\le K.\end{cases} \] There are integers \(n>K\) for every finite \(K\), so the supremum over \(n\) is exactly \(1\), not a quantity tending to zero. The family is not uniformly integrable.
For \(n>K\), the clipping error is \((n-K)\) on an event of probability \(1/n\), and is zero elsewhere. Hence \(\mathbb{E}|X_n-C_K(X_n)|=(n-K)/n=1-K/n\); for \(n\le K\), it is zero. As \(n\) increases beyond \(K\), \(1-K/n\) approaches \(1\), so the supremum over \(n\) is \(1\). This agrees with the truncation criterion. Finally, for any \(\delta>0\), choose an integer \(n\ge1\) large enough that \(1/n\le\delta\), and set \(A=\{U\le1/n\}\). Then \(\mathbb{P}(A)\le\delta\), but \(\mathbb{E}[|X_n|\mathbf{1}_A]=1\). Thus the small-event integrals do not tend uniformly to zero, despite the bounded first moments.
Check: At \(K=2\) and \(n=3\), the tail event has probability \(1/3\), its tail expectation is \(3(1/3)=1\), and its clipping error is \((3-2)(1/3)=1/3\). These are different at the same threshold, as the general inequalities allow.
Worked Example: Spikes That Become Small in Expectation
On the same uniform probability space, define \(Y_n=n\mathbf{1}_{\{U\le1/n^2\}}\) for integers \(n\ge1\). This family also has increasingly tall spikes, but their expected magnitudes decrease. Verify uniform integrability directly.
The spike event has probability \(1/n^2\), so \(\mathbb{E}|Y_n|=n/n^2=1/n\le1\). At a fixed \(K\ge0\), the strict tail event is nonempty only when \(n>K\), and in that case \(\mathbb{E}[|Y_n|\mathbf{1}_{\{|Y_n|>K\}}]=n(1/n^2)=1/n\). The smallest integer greater than \(K\) is \(\lfloor K\rfloor+1\). Since \(1/n\) decreases with \(n\), the supremum of the tail expectations is exactly \(1/(\lfloor K\rfloor+1)\), which tends to zero.
For \(n>K\), the clipping error is \((n-K)/n^2\), and it is zero for \(n\le K\). Since \(0\le(n-K)/n^2\le1/n\), the supremum of the clipping errors is at most \(1/(\lfloor K\rfloor+1)\), so it too tends to zero. For instance, at \(K=2\) and \(n=3\), the event has probability \(1/9\), the tail expectation is \(3/9=1/3\), and the clipping error is \((3-2)/9=1/9\). This verifies the two calculations directly while illustrating why not every sequence of tall spikes prevents uniform integrability.
What the Criteria Do—and Do Not—Say
The tail criterion is often the most direct test when the distribution of \(|X_\alpha|\) is available. The truncation formulation is useful when clipped variables or excesses above a threshold are easier to calculate. The small-event criterion is useful when a problem naturally controls integrals over events, for example through a bound that is uniform in the event's probability.
Common mistake. It is not enough to show that \(\mathbb{P}(|X_n|>K)\) becomes small uniformly as \(K\) grows. That controls how often large values occur, not how much expected magnitude they carry. In the rare-spike example, for fixed \(K\), the tail probability for \(n>K\) is \(1/n\), which tends to zero as \(n\to\infty\); nevertheless, the tail expectation is \(1\) for every such \(n\), and its supremum over the family is \(1\). The correction is to calculate \(\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]\), keeping the supremum over \(n\) at the same threshold \(K\). A small tail probability is not a substitute for a small tail expectation.
The equivalent criteria can be summarized as follows: uniform integrability means that large-value contributions vanish uniformly; equivalently, clipping at a common threshold produces vanishing expected error; equivalently, bounded first moments cannot concentrate on events whose probabilities are uniformly small. Each formulation controls magnitudes as well as probabilities.
Check Your Understanding
For each question, keep the threshold fixed while taking a supremum over the family.
- For a real number \(x\), compare \((|x|-K)^+\) with \(|x|\mathbf{1}_{\{|x|>K\}}\). Why does the reverse bound use threshold \(K/2\)?
- In the exponential example, calculate \(\mathbb{P}(X>K)\) and verify the relation between the tail expectation and clipping error.
- For \(X_n=n\mathbf{1}_{\{U\le1/n\}}\), compute the tail expectation at a fixed \(K\). Why does its supremum fail to vanish?
- Why are both bounded first moments and the small-event condition needed in the small-event criterion?
- For \(Y_n=n\mathbf{1}_{\{U\le1/n^2\}}\), compute the tail expectation at \(K=3\) and identify the first integer \(n\) that contributes a nonzero value.