Rare Spikes and Failure of Uniform Integrability

When Rare Events Carry a Persistent Amount of Mass

A random variable can be large on an event of very small probability. It is tempting to conclude that such an event has little effect on expectation. But the size of the random variable and the probability of the event can compensate for one another: a spike of height \(n\) occurring with probability \(1/n\) contributes \(1\) to the first absolute moment, no matter how large \(n\) becomes.

Uniform integrability asks whether the expected contribution from values beyond a large cutoff becomes small uniformly over the family. Rare spikes show why this is a stronger requirement than merely bounding the first moments.

Definition: A family \(\mathcal{X}\) of integrable random variables is uniformly integrable if \[ \lim_{K\to\infty}\sup_{X\in\mathcal{X}}\mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right]=0. \] The expectation inside the supremum is the tail expectation at cutoff \(K\).

The absolute value matters: the tail expectation measures the size of the contribution, not its signed average. We will calculate it directly for several families and identify exactly when a family consisting of single spikes is uniformly integrable.

A Canonical Rare-Spike Family

Worked Example: A Fixed First Moment and a Persistent Tail

Take the probability space \([0,1]\) with Lebesgue probability measure. For each \(n\in\mathbb{N}\), let \(A_n=[0,1/n]\) and define \(X_n=n\mathbf{1}_{A_n}\). Since \(\mathbb{P}(A_n)=1/n\), the first absolute moment is

\mathbb{E}|X_n| =n\mathbb{P}(A_n) =n\cdot\frac{1}{n} =1.

Thus the family has uniformly bounded first absolute moments. For any \(K>0\), if \(n\le K\), then \(X_n\le K\) everywhere, so its tail expectation is zero. If \(n>K\), then \(X_n=n>K\) on \(A_n\) and is zero outside \(A_n\). Therefore

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

For every finite \(K\), there is an integer \(n>K\). Consequently the supremum of the tail expectations over \(n\) is \(1\), not a quantity tending to zero. The family is not uniformly integrable.

Its tail probabilities behave differently. When \(n>K\), \(\mathbb{P}(|X_n|>K)=1/n\), so

\sup_{n\in\mathbb{N}}\mathbb{P}(|X_n|>K) =\frac{1}{\lfloor K\rfloor+1},

which does tend to zero. Thus the events on which the variables exceed \(K\) become uniformly rare, but the expected size contributed by those events does not become small.

In this example, each spike has height \(n\), while its event has probability \(1/n\). Their product stays fixed. The definition of uniform integrability detects that persistent contribution even though the probability of seeing a spike becomes small.

Small Spike Contributions Can Give Uniform Integrability

Worked Example: Increasing Heights with Vanishing Tail Mass

On the same probability space, let \(B_n=[0,1/n^2]\) and set \(Z_n=n\mathbf{1}_{B_n}\). Now the spike still grows, but its probability is \(1/n^2\). The first absolute moment is

\mathbb{E}|Z_n| =n\cdot\frac{1}{n^2} =\frac{1}{n}.

For a cutoff \(K>0\), the tail expectation is zero when \(n\le K\), and equals \(1/n\) when \(n>K\). Hence

\sup_{n\in\mathbb{N}} \mathbb{E}\!\left[|Z_n|\mathbf{1}_{\{|Z_n|>K\}}\right] = \frac{1}{\lfloor K\rfloor+1},

which tends to zero as \(K\to\infty\). Therefore \(\{Z_n:n\in\mathbb{N}\}\) is uniformly integrable. The distinction from the previous example is not simply that the events are rare: both examples have shrinking event probabilities. Here the product of spike height and event probability also tends to zero.

An Exact Criterion for Single-Spike Families

The preceding calculations extend to any sequence of positive spike heights that tends to infinity. Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let \(A_n\in\mathcal{F}\), and let \(a_n>0\) with \(a_n\to\infty\). Define \(X_n=a_n\mathbf{1}_{A_n}\). The number \(a_n\mathbb{P}(A_n)\) is exactly the first absolute moment of \(X_n\).

Theorem: For \(X_n=a_n\mathbf{1}_{A_n}\) with \(a_n>0\) and \(a_n\to\infty\), the family \(\{X_n:n\in\mathbb{N}\}\) is uniformly integrable if and only if \[ a_n\mathbb{P}(A_n)\longrightarrow 0. \]

Proof. For every \(K>0\), the variable \(X_n\) equals \(a_n\) on \(A_n\) and zero on \(A_n^c\). Thus its tail expectation is exactly

\mathbb{E}\!\left[|X_n|\mathbf{1}_{\{|X_n|>K\}}\right] = a_n\mathbb{P}(A_n)\mathbf{1}_{\{a_n>K\}}.

First suppose \(a_n\mathbb{P}(A_n)\to0\). Given \(\varepsilon>0\), choose \(N\) so that \(a_n\mathbb{P}(A_n)<\varepsilon\) whenever \(n\ge N\). Since there are only finitely many indices \(n<N\), choose \(K\) at least as large as every \(a_n\) for those indices. Their tail expectations are then zero. For \(n\ge N\), the displayed identity shows that the tail expectation is either zero or less than \(\varepsilon\). It follows that the supremum over all \(n\) is at most \(\varepsilon\). As \(\varepsilon\) was arbitrary, the family is uniformly integrable.

Conversely, suppose the family is uniformly integrable. Given \(\varepsilon>0\), choose \(K\) so that every tail expectation at cutoff \(K\) is less than \(\varepsilon\). Because \(a_n\to\infty\), there exists \(N\) such that \(a_n>K\) for every \(n\ge N\). For these indices the identity gives \(a_n\mathbb{P}(A_n)<\varepsilon\). This proves \(a_n\mathbb{P}(A_n)\to0\), and establishes the equivalence. \(\square\)

Bounded First Moments Are Necessary but Not Sufficient

A uniformly integrable family must have bounded first absolute moments. To see this, choose \(K\) so large that the supremum of the tail expectations is at most \(1\). For every \(X\) in the family, split its absolute value according to whether it is at most \(K\):

\mathbb{E}|X| = \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|\le K\}}\right] + \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right] \le K+1.

The first term is at most \(K\) because \(|X|\le K\) on the event in its indicator; the second is at most \(1\) by the choice of cutoff. The bound \(K+1\) is independent of \(X\). The canonical example shows why the reverse implication fails: its first absolute moments all equal \(1\), but its tail expectations do not vanish uniformly.

Worked Example: Signed Spikes with Zero Mean

On \([0,1]\) with Lebesgue probability, for \(n\ge2\) let \(A_n=[0,1/(2n))\) and \(B_n=[1/(2n),1/n]\). These events are disjoint and each has probability \(1/(2n)\). Define \(W_n=n\mathbf{1}_{A_n}-n\mathbf{1}_{B_n}\). The signed expectation is zero:

\mathbb{E}W_n =n\cdot\frac{1}{2n}-n\cdot\frac{1}{2n} =0.

But cancellation does not reduce the first absolute moment:

\mathbb{E}|W_n| =n\mathbb{P}(A_n)+n\mathbb{P}(B_n) =\frac12+\frac12 =1.

Whenever \(n>K\), \(|W_n|=n\) on \(A_n\cup B_n\). Therefore the tail expectation is \(n(1/n)=1\); when \(n\le K\), it is zero. The family is not uniformly integrable. This example shows why a bound on signed expectations alone says nothing about absolute tails. Even a uniform bound on first absolute moments, as here, does not guarantee uniform integrability.

Reading the Tail Correctly

For a rare-spike family, the useful calculation is not just the probability of a large value. It is the product of the value’s size and the probability of the event on which it occurs. A probability can approach zero while that product stays bounded away from zero, as in the canonical example. Conversely, an increasing spike height does not by itself prevent uniform integrability: the second example has increasing heights and uniformly vanishing tail expectations.

A common pitfall is to argue that a family is harmless because its large values occur on events of small probability. That argument controls the chance of an exceedance, not the expected magnitude of the exceedance. Another pitfall is to inspect each fixed random variable separately: for any one integrable variable its tail expectation tends to zero as the cutoff grows, but uniform integrability requires one cutoff that works simultaneously for every member of the family.

Key takeaway: Bounded first absolute moments control the total expected size, but they do not prevent that size from remaining concentrated in increasingly high, increasingly rare spikes. Uniform integrability rules out precisely this persistent tail contribution.

Check Your Understanding

Use tail expectations, rather than tail probabilities alone, to answer these questions.

For \(X_n=n\mathbf{1}_{A_n}\) with \(\mathbb{P}(A_n)=1/n\), calculate the first absolute moment and the tail expectation at cutoff \(K\).
Why does a tail probability tending to zero uniformly not suffice to prove uniform integrability?
State the necessary and sufficient condition in the single-spike theorem.
Explain why a uniform bound on first absolute moments follows from uniform integrability.
In the signed-spike example, why does \(\mathbb{E}W_n=0\) fail to control the tail expectation?