Uniform Integrability of Truncations of an Integrable Variable

Why Truncations Behave Uniformly

An integrable random variable may be unbounded, so its large values can contribute substantially to its expectation. Truncating those values changes the variable, but it does not create a new source of large magnitudes: the usual clipped and truncated versions are all bounded in absolute value by the original variable.

That pointwise comparison is useful even when the truncation levels vary without any pattern. The levels need not increase, converge, or share a finite upper bound. What matters is that the same integrable random variable controls every member of the family.

Three Ways to Truncate

Let \(X\) be a real-valued random variable and let \(K\geq 0\) be a finite number. The clipped version of \(X\) at level \(K\) replaces values below \(-K\) by \(-K\), values above \(K\) by \(K\), and leaves intermediate values unchanged:

C_K(X)=(-K)\vee(X\wedge K),

where \(a\vee b=\max(a,b)\) and \(a\wedge b=\min(a,b)\). Another convention, called hard truncation here, keeps values whose absolute value is at most \(K\) and replaces the others by zero:

H_K(X)=X\mathbf{1}_{\{|X|\leq K\}}.

The complementary tail truncation keeps only the values beyond the cutoff:

R_K(X)=X\mathbf{1}_{\{|X|>K\}}.

These definitions differ in how they treat values near or beyond the cutoff. For the uniform-integrability argument, however, each has the same essential property: its magnitude is no greater than \(|X|\).

Definition: A family \(\mathcal{Y}\) of integrable random variables is uniformly integrable if

\lim_{M\to\infty}\sup_{Y\in\mathcal{Y}} \mathbb{E}\!\left[|Y|\mathbf{1}_{\{|Y|>M\}}\right]=0.

The supremum requires the tail expectations to become small uniformly across the entire family, not just separately for each random variable.

The Common-Envelope Argument

The result we need is a direct consequence of domination. The envelope must be a single integrable random variable that bounds every member of the family.

Theorem: Let \(G\) be a nonnegative integrable random variable, and let \(\mathcal{Y}\) be a family of random variables satisfying \(|Y|\leq G\) almost surely for every \(Y\in\mathcal{Y}\). Then \(\mathcal{Y}\) is uniformly integrable.

Proof: For any \(Y\in\mathcal{Y}\) and any \(M>0\), the inequality \(|Y|\leq G\) implies that \(\{|Y|>M\}\subseteq\{G>M\}\), up to a null set. Therefore

|Y|\mathbf{1}_{\{|Y|>M\}} \leq G\mathbf{1}_{\{G>M\}} \quad\text{almost surely}.

Taking expectations and then the supremum over \(Y\in\mathcal{Y}\) gives

\sup_{Y\in\mathcal{Y}} \mathbb{E}\!\left[|Y|\mathbf{1}_{\{|Y|>M\}}\right] \leq \mathbb{E}\!\left[G\mathbf{1}_{\{G>M\}}\right].

As \(M\to\infty\), the integrand on the right tends to zero almost surely and is bounded by \(G\), which is integrable. The dominated convergence theorem makes the right-hand side tend to zero. This is exactly the defining condition for uniform integrability.

\square

This argument gives a useful quantitative bound as well: every tail expectation in the family is at most the same tail expectation for the envelope. In the applications below, the envelope will simply be \(|X|\).

Uniform Integrability at Arbitrary Levels

Theorem: If \(X\) is integrable, then each of the families

\{C_K(X):K\geq 0\},\qquad \{H_K(X):K\geq 0\},\qquad \{R_K(X):K\geq 0\}

is uniformly integrable. More generally, the union of these three families is uniformly integrable.

Proof: For \(K\geq 0\), clipping leaves a value unchanged when \(|X|\leq K\), and otherwise replaces it by a value of magnitude \(K\). Thus \(|C_K(X)|\leq |X|\) in both cases. For hard truncation, either \(H_K(X)=X\) or \(H_K(X)=0\), so \(|H_K(X)|\leq |X|\). For tail truncation, either \(R_K(X)=X\) or \(R_K(X)=0\), so \(|R_K(X)|\leq |X|\). Since \(X\) is integrable, \(|X|\) is a nonnegative integrable envelope for every member of the union. The common-envelope theorem proves the claim.

\square

The same proof covers arbitrary collections of levels. In particular, if \((K_n)_{n\in\mathbb{N}}\) is any sequence of nonnegative finite numbers, then \(\{C_{K_n}(X):n\in\mathbb{N}\}\), and the corresponding hard- and tail-truncated families, are uniformly integrable. No monotonicity or convergence assumption on \((K_n)\) is needed.

The conclusion even permits measurable random levels. If \(L\) is a nonnegative finite random variable, the definitions \(C_L(X)=(-L)\vee(X\wedge L)\), \(H_L(X)=X\mathbf{1}_{\{|X|\leq L\}}\), and \(R_L(X)=X\mathbf{1}_{\{|X|>L\}}\) still satisfy the same pointwise bounds. Any family formed from such versions of the fixed integrable \(X\) is uniformly integrable.

Worked Examples

Worked Example: Exponential Truncations

Let \(X\) have the exponential distribution with rate \(1\), so \(X\geq 0\) and its density is \(e^{-x}\) for \(x\geq 0\). Its expectation is \(\int_0^\infty xe^{-x}\,dx=1\), so it is integrable. For \(K\geq 0\), clipping gives \(C_K(X)=\min(X,K)\), and hard truncation gives \(H_K(X)=X\mathbf{1}_{\{X\leq K\}}\). Both are bounded by \(X\).

For any \(M>0\), the tail expectation of either version is at most

\mathbb{E}\!\left[X\mathbf{1}_{\{X>M\}}\right] =\int_M^\infty xe^{-x}\,dx =(M+1)e^{-M}.

The last equality follows by integration by parts: the boundary term at infinity is zero, and \(\int_M^\infty e^{-x}\,dx=e^{-M}\). Since \((M+1)e^{-M}\to 0\), this is a uniform tail bound for every clipping level and every hard-truncation level, including levels that change arbitrarily with the index.

Worked Example: A Discrete Integrable Variable

Let \(X\) take the value \(n\) with probability \(1/(\zeta(3)n^3)\), for each \(n\in\mathbb{N}\), where \(\zeta(3)=\sum_{n=1}^{\infty}n^{-3}\). These probabilities sum to one. Moreover,

\mathbb{E}[X] =\frac{1}{\zeta(3)}\sum_{n=1}^{\infty}\frac{1}{n^2} =\frac{\zeta(2)}{\zeta(3)} <\infty.

For \(M\geq 1\), put \(m=\lfloor M\rfloor+1\), the smallest integer strictly greater than \(M\). The tail expectation of \(X\) is

\mathbb{E}\!\left[X\mathbf{1}_{\{X>M\}}\right] =\frac{1}{\zeta(3)}\sum_{n=m}^{\infty}\frac{1}{n^2} \leq \frac{1}{\zeta(3)}\int_{m-1}^{\infty}\frac{dx}{x^2} =\frac{1}{\zeta(3)\lfloor M\rfloor}.

The inequality follows because \(x^{-2}\) is decreasing and, for each integer \(n\geq m\), \(n^{-2}\leq\int_{n-1}^{n}x^{-2}\,dx\). The displayed bound tends to zero. Every clipped, hard-truncated, or tail-truncated version at any level is dominated by \(X\), so the same bound proves uniform integrability for all of them.

Worked Example: Nonmonotone Levels

Take any integrable real-valued random variable \(X\), and define \(K_n=2\) when \(n\) is even and \(K_n=n\) when \(n\) is odd. This sequence of levels is not monotone: it repeatedly returns to \(2\). Nevertheless, for every \(n\),

|C_{K_n}(X)|\leq |X|, \qquad |H_{K_n}(X)|\leq |X|, \qquad |R_{K_n}(X)|\leq |X|.

Consequently, the combined family of all three types of truncation across these levels has, for every \(M>0\), tail expectations bounded by \(\mathbb{E}[|X|\mathbf{1}_{\{|X|>M\}}]\). That bound tends to zero because \(X\) is integrable. The levels’ failure to increase does not affect the argument: the uniform bound comes from \(X\), not from the sequence \((K_n)\).

What the Result Does—and Does Not—Require

A common temptation is to argue that the family is uniformly integrable because each clipped variable is bounded. That observation alone is insufficient: the bound for \(C_K(X)\) is \(K\), and \(K\) can be arbitrarily large. Uniform integrability is a statement about the whole family, so separate finite bounds for individual members do not automatically give a uniform bound on their tails.

The stronger and more relevant fact is that every member is bounded by the same integrable random variable \(|X|\). The common-envelope theorem then controls all tail expectations at once. In particular, the truncation levels do not need a common finite upper bound.

The integrability assumption matters. For example, if \(X\) is not integrable, the domination \(|C_K(X)|\leq |X|\) does not provide an integrable envelope. It cannot by itself establish uniform integrability. The theorem concerns truncations of an integrable variable, not an automatic regularizing effect that works for every variable.

The result is useful whenever a proof introduces approximations by clipping or truncating one integrable variable. Even if the chosen levels vary from one approximation to another, the resulting family has uniformly controlled tails. This conclusion is about the truncations themselves; it does not require the levels to tend to infinity or the truncated variables to converge.

Check Your Understanding

Use the common-envelope argument to answer the following questions.

State the tail-expectation definition of uniform integrability.
Why does \(|C_K(X)|\leq |X|\) hold both when \(|X|\leq K\) and when \(|X|>K\)?
What common integrable envelope controls hard truncations at a sequence of nonmonotone levels?
Does the uniform-integrability conclusion require the truncation levels to converge? Explain.
Why does boundedness of each individual clipped variable not, by itself, establish uniform integrability of the full family?