Tail Expectations Detect Escaping Mass
A sequence can have bounded expectations while placing a fixed amount of absolute expectation farther and farther out in its tails. The values responsible may occur with smaller and smaller probability, yet their size can grow enough to preserve that tail contribution. Uniform integrability is the condition that rules out this behavior: beyond a sufficiently large common threshold, the expected absolute contribution must be small for every member of the sequence.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((X_n)_{n\ge1}\) be a sequence of integrable real-valued random variables on it. We regard \(n\) as a discrete time index and the state space as \(\mathbb{R}\). No Markov or martingale assumption is needed here. For a threshold \(K>0\), define the tail expectation of \(X_n\) by \(\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]\). The event uses the strict inequality \(|X_n|>K\): values exactly equal to \(K\) are not counted.
The order of the supremum and limit matters. It is not enough that, for each fixed \(n\), the tail expectation becomes small as \(K\) grows: that is already true for every integrable \(X_n\). Uniform integrability requires one threshold to work for all \(n\). It also implies bounded first absolute moments. Indeed, for any \(K>0\), \(\mathbb{E}|X_n|\le K+\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]\); choose one threshold for which the tail expectations are bounded, and the right side is then bounded uniformly in \(n\). The converse does not hold, as the first example shows.
An Equivalent Tail Test
Another way to measure the part of \(|X_n|\) above \(K\) is the truncated excess \((|X_n|-K)^+=\max\{|X_n|-K,0\}\). The next result gives an equivalent test. It is useful when the excess above a threshold is easier to calculate than the tail expectation itself.
Proof. Write \(T_n(K)=\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]\) and \(A_n(K)=\mathbb{E}[(|X_n|-K)^+]\). For every outcome, \((|X_n|-K)^+\le |X_n|\mathbf{1}_{\{|X_n|>K\}}\): if \(|X_n|\le K\), both sides are zero; if \(|X_n|>K\), then \(|X_n|-K\le |X_n|\). Thus \(A_n(K)\le T_n(K)\), so if the supremum of the tail expectations tends to zero, then the supremum of the truncated excesses does too. Conversely, on the event \(|X_n|>2K\), we have \(|X_n|-K>|X_n|/2\). Therefore, pointwise, \(|X_n|\mathbf{1}_{\{|X_n|>2K\}}\le2(|X_n|-K)^+\). Taking expectations and then the supremum over \(n\) gives \(\sup_n T_n(2K)\le2\sup_n A_n(K)\). If the supremum of the excesses tends to zero, the supremum of the tail expectations at threshold \(2K\) tends to zero as well. Since \(2K\) ranges to infinity with \(K\), this is uniform integrability. \(\square\)
Worked Examples
Worked Example: Fixed Expectation Escaping to Larger Values
Let \(\Omega=(0,1)\) with Lebesgue probability measure. Set \(A_n=(0,1/n)\) and \(X_n=n\mathbf{1}_{A_n}\), for \(n\ge1\). Find the expectations, the almost-sure limit, and the uniform tail expectations.
The event \(A_n\) has probability \(1/n\), so \(\mathbb{E}|X_n|=n\mathbb{P}(A_n)=n(1/n)=1\). For each \(\omega\in(0,1)\), choose an integer \(N>1/\omega\). For every \(n\ge N\), \(\omega>1/n\), so \(\omega\notin A_n\) and \(X_n(\omega)=0\). Hence \(X_n\to0\) at every point of this probability space.
Now fix \(K>0\). If \(n\le K\), then \(X_n\le K\) everywhere and the tail expectation is zero. If \(n>K\), then \(X_n=n>K\) on \(A_n\), and it is zero outside \(A_n\). Consequently, \(\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]=n(1/n)=1\) for \(n>K\), and is zero for \(n\le K\). There is always an integer \(n>K\), so the supremum over \(n\) equals \(1\) for every finite \(K>0\). The sequence is not uniformly integrable, despite its constant first absolute moments and pointwise convergence to zero.
Check: Substituting \(X_n=n\mathbf{1}_{A_n}\) gives \(\mathbb{E}|X_n|=n\mathbb{P}(A_n)=1\), while for any \(n>K\) the tail expectation is also \(n\mathbb{P}(A_n)=1\). The tail contribution does not shrink; it moves to larger values.
Worked Example: Rare Values with Vanishing Tail Contribution
On \(\Omega=(0,1)\) with Lebesgue probability measure, let \(A_n=(0,1/n^2)\) and \(Y_n=n\mathbf{1}_{A_n}\). Determine whether \((Y_n)\) is uniformly integrable.
The event \(A_n\) has probability \(1/n^2\), so \(\mathbb{E}|Y_n|=n(1/n^2)=1/n\). For \(K>0\), the event \(|Y_n|>K\) is \(A_n\) when \(n>K\), and is empty when \(n\le K\). Thus the tail expectation equals \(n(1/n^2)=1/n\) for \(n>K\), and zero otherwise. The first integer \(n>K\) is \(\lfloor K\rfloor+1\). Since \(1/n\) decreases with \(n\), $$ \sup_{n\ge1}\mathbb{E}\big[|Y_n|\mathbf{1}_{\{|Y_n|>K\}}\big] =\frac{1}{\lfloor K\rfloor+1}. $$ This tends to zero as \(K\to\infty\), so the sequence is uniformly integrable.
Check: If \(K=2\), the first contributing index is \(n=3\), and the supremum is \(1/3\). The tail expectation at \(n=3\) is \(3\mathbb{P}(A_3)=3(1/9)=1/3\); for every \(n>3\), it is \(1/n\le1/3\). This verifies both the strict threshold rule and the supremum calculation.
Proof. On the event \(|X_n|>K\), the inequality \(1\le |X_n|^{p-1}/K^{p-1}\) holds. Multiplying by \(|X_n|\) gives \(|X_n|\mathbf{1}_{\{|X_n|>K\}}\le |X_n|^p/K^{p-1}\) on that event, and both sides are nonnegative outside it. Taking expectations yields \(\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]\le \mathbb{E}|X_n|^p/K^{p-1}\le C/K^{p-1}\). The bound is the same for every \(n\), and \(C/K^{p-1}\to0\) because \(p>1\). Taking the supremum proves uniform integrability. \(\square\)
Worked Example: A Uniform Moment Bound and Its Limitation
Let \(B_n=(0,1/n^3)\) on \(\Omega=(0,1)\) with Lebesgue probability measure, and define \(Z_n=n^2\mathbf{1}_{B_n}\). Verify uniform integrability using a moment bound, then compare the second moments.
For every \(n\ge1\), \(\mathbb{E}|Z_n|^{3/2}=(n^2)^{3/2}\mathbb{P}(B_n)=n^3(1/n^3)=1\). The proposition with \(p=3/2\) and \(C=1\) therefore applies. Directly, for \(K>0\), the tail event is \(B_n\) exactly when \(n^2>K\), and otherwise is empty. Its tail expectation is \(n^2(1/n^3)=1/n\) when \(n^2>K\), and zero otherwise. Writing \(m=\lfloor\sqrt K\rfloor+1\), the contributing indices are \(n\ge m\), so the supremum is \(1/m\), which tends to zero.
The second moments are \(\mathbb{E}|Z_n|^2=n^4(1/n^3)=n\), which are unbounded. Thus the higher-moment test is sufficient, not necessary: this sequence is uniformly integrable even though its second moments are not uniformly bounded.
Check: At \(K=4\), the first contributing index is \(n=3\), since \(3^2>4\) but \(2^2=4\) is not strictly greater than \(4\). The tail expectation at \(n=3\) is \(1/3\), and for \(n>3\) it is \(1/n\le1/3\). The supremum is therefore \(1/3\).
Worked Example: Centering Does Not Control Absolute Tails
For each \(n\ge1\), let \(C_n\) be an event of probability \(1/(2n)\) and \(D_n\) a disjoint event of the same probability, on a probability space supporting these events. Define \(W_n=n\mathbf{1}_{C_n}-n\mathbf{1}_{D_n}\). Does the zero mean imply uniform integrability?
The expectation is \(\mathbb{E}W_n=n(1/(2n))-n(1/(2n))=1/2-1/2=0\). But \(|W_n|=n\) on \(C_n\cup D_n\), an event of probability \(1/(2n)+1/(2n)=1/n\), and it is zero elsewhere. For \(n>K\), the strict tail event \(|W_n|>K\) is \(C_n\cup D_n\), so its tail expectation is \(n(1/n)=1\). For \(n\le K\), that tail expectation is zero. Since an integer \(n>K\) always exists, the supremum is \(1\) at every finite threshold. The sequence is not uniformly integrable.
Check: The positive and negative contributions cancel in \(\mathbb{E}W_n\), but not in \(\mathbb{E}|W_n|\): the latter is \(n(1/n)=1\). Uniform integrability uses absolute values, so the zero means do not make the tails small.
How to Apply the Test
For a sequence with rare, large values, first list the possible magnitudes and their probabilities. At a fixed threshold \(K\), include only outcomes whose absolute value is strictly greater than \(K\); multiply each included magnitude by its probability and add. Then take the supremum over the sequence index at that same \(K\), and only after that let \(K\) increase. This order prevents a common error: checking that each individual random variable has a small tail for large thresholds does not establish a bound that is uniform in the index.
When direct tail calculations are cumbersome, the proved higher-moment test can settle the question: a uniform bound on any absolute moment of order \(p>1\) is enough. If the first moments are bounded but no such higher-moment bound is available, return to the tail expectations themselves. The first example shows why: each expected absolute value is \(1\), yet for every threshold a later sequence member puts that entire expectation beyond the threshold.
Common mistake. A tempting but incorrect calculation is to use only \(\sup_n\mathbb{E}|X_n|<\infty\) and conclude that the tail expectations must vanish uniformly. A bound on the total expectation does not specify where that expectation lies. For the first example, \(\mathbb{E}|X_n|=1\) for all \(n\), but at each \(K\), choosing \(n>K\) gives \(\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]=n(1/n)=1\). The correction is to calculate the tail at a common threshold and then take the supremum over \(n\); bounded first moments alone do not pass that test.
Check Your Understanding
For each question, use absolute values and apply the strict tail event \(|X_n|>K\).
- For \(U_n=n\mathbf{1}_{A_n}\) with \(\mathbb{P}(A_n)=1/n^3\), calculate the tail expectation at a fixed \(K>0\), then determine whether the sequence is uniformly integrable.
- Explain why the sequence \(X_n=n\mathbf{1}_{(0,1/n)}\) is not uniformly integrable even though \(X_n\to0\) almost surely.
- State the equivalent truncated-excess test and identify the pointwise inequality that gives the reverse direction.
- If \(\sup_n\mathbb{E}|V_n|^p\le C\) for some \(p>1\), what common bound controls every tail expectation at threshold \(K\)?
- For centered random variables, why can \(\mathbb{E}V_n=0\) fail to provide information about uniform integrability?