Two Different Ways to Control a Tail
A bound on \(\mathbb{E}|X|\) controls how much probability can lie far from zero. It does not, by itself, control how much expected magnitude lies there. That distinction is important in convergence arguments: a sequence can converge almost surely to zero while retaining a fixed amount of expectation in rarer and rarer, larger values. Uniform integrability is the condition that rules out this escape of expected magnitude.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(\mathcal{X}\) be a nonempty family of integrable real-valued random variables on that space. A sequence \((X_n)_{n\ge 1}\) is a particular case, with \(\mathcal{X}=\{X_n:n\ge1\}\). The absolute value is essential: a bound on \(\mathbb{E}X_n\) alone can conceal large positive and negative values that cancel.
For a single integrable random variable, its tail expectation tends to zero as \(K\) increases. Uniform integrability asks for this decay to hold uniformly over every member of the family. A uniform first-moment bound is the weaker statement \(\sup_{X\in\mathcal{X}}\mathbb{E}|X|<\infty\). The next result records one direction of the relationship.
Proof. By uniform integrability, there is a finite \(K>0\) such that \(\sup_{X\in\mathcal{X}}\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\le 1\). For each \(X\in\mathcal{X}\), split its absolute value according to whether it is at most \(K\) or greater than \(K\). The first part is bounded pointwise by \(K\), so \(\mathbb{E}|X|=\mathbb{E}[|X|\mathbf{1}_{\{|X|\le K\}}]+\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\le K+1.\) The same bound holds for every member of the family, proving the claim. \(\square\)
The converse is the central issue here. A first-moment bound does give a uniform estimate on tail probabilities, but it does not give the defining estimate on tail expectations.
Proof. On the event \(\{|X|>K\}\), one has \(|X|\ge K\). Therefore \(K\mathbf{1}_{\{|X|>K\}}\le |X|\) pointwise. Taking expectations gives \(K\mathbb{P}(|X|>K)\le\mathbb{E}|X|\le C\). Divide by \(K>0\), then take the supremum over \(X\in\mathcal{X}\). \(\square\)
A Sequence with Escaping Expected Magnitude
We now construct the basic counterexample on one fixed probability space. Its rare nonzero values grow with \(n\), while the probability of seeing them shrinks at exactly the rate needed to keep the first moment constant. This example shows why controlling tail probabilities is not enough.
Worked Example: Bounded First Moments but Not Uniform Integrability
Take \(\Omega=[0,1]\), with its Borel sets and Lebesgue probability measure. For each integer \(n\ge1\), define \(X_n=n\mathbf{1}_{[0,1/n]}\). The interval \([0,1/n]\) has probability \(1/n\), so \(X_n=n\) with probability \(1/n\) and \(X_n=0\) otherwise.
First calculate the absolute first moment: \(\mathbb{E}|X_n|=n\cdot\frac1n+0\cdot(1-\frac1n)=1.\) Thus \(\sup_{n\ge1}\mathbb{E}|X_n|=1\).
Now fix any finite threshold \(K\ge0\) and use that same threshold for every \(n\). If \(n\le K\), then \(X_n\) never exceeds \(K\), so \(\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]=0\). If \(n>K\), the event \(\{|X_n|>K\}\) is exactly \([0,1/n]\), and \(\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]=n\cdot\frac1n=1.\) There is an integer \(n>K\) for every finite \(K\). Consequently, \(\displaystyle \sup_{n\ge1}\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]=1\) for every finite \(K\). This supremum does not tend to zero, so the sequence is not uniformly integrable.
Check the probability tails separately. For \(n>K\), \(\mathbb{P}(|X_n|>K)=1/n\), and for \(n\le K\) it is zero. The largest probability among integers \(n>K\) occurs at \(n=\lfloor K\rfloor+1\), so \(\displaystyle \sup_{n\ge1}\mathbb{P}(|X_n|>K)=\frac{1}{\lfloor K\rfloor+1}.\) This tends to zero, even though the supremum of the tail expectations remains \(1\). The mass becomes less likely but its size increases enough to preserve its expected magnitude.
This sequence also illustrates why uniform integrability matters for convergence. For each \(\omega\in(0,1]\), eventually \(1/n<\omega\), so \(X_n(\omega)=0\) for all sufficiently large \(n\). At \(\omega=0\), the sequence equals \(n\), but this single point has probability zero. Hence \(X_n\to0\) almost surely, while \(\mathbb{E}|X_n|=1\) for every \(n\). The earlier material on uniform integrability and \(L^1\) convergence gives the broader convergence result: almost-sure convergence together with uniform integrability yields \(L^1\) convergence. This example fails the uniform-integrability hypothesis, and its expectations do not converge to \(\mathbb{E}|0|=0\).
Examples That Clarify the Boundary
Worked Example: A Uniformly Bounded Family
On the same probability space, let \(Y_n=2+\mathbf{1}_{[0,1/n]}\) for \(n\ge1\). Its values are \(3\) on an event of probability \(1/n\) and \(2\) on the complement. Therefore \(\mathbb{E}|Y_n|=3\cdot\frac1n+2\left(1-\frac1n\right)=2+\frac1n\le3.\) For every \(K\ge3\), neither possible value is strictly greater than \(K\). Thus \(\mathbb{E}[|Y_n|\mathbf{1}_{\{|Y_n|>K\}}]=0\) for every \(n\), and the supremum over \(n\) is zero. The family is uniformly integrable. The strict inequality in the definition matters at \(K=3\): the value \(3\) is not in the event \(\{|Y_n|>3\}\).
Check: the largest possible absolute value is \(3\), and \(\sup_n\mathbb{E}|Y_n|=3\), consistent with bounded first moments. Unlike a first-moment bound alone, the common pointwise bound directly makes all sufficiently distant tails vanish.
Worked Example: A Uniform Second-Moment Bound
Define \(Z_n=\sqrt{n}\mathbf{1}_{[0,1/n]}\) on \([0,1]\). Its second moment is \(\mathbb{E}|Z_n|^2=n\cdot\frac1n=1\) for every \(n\), and its first moment is \(\mathbb{E}|Z_n|=\sqrt n\cdot\frac1n=\frac1{\sqrt n}\le1.\)
For a fixed \(K\ge0\), the event \(\{|Z_n|>K\}\) is empty unless \(\sqrt n>K\), equivalently \(n>K^2\). When \(n>K^2\), \(\mathbb{E}[|Z_n|\mathbf{1}_{\{|Z_n|>K\}}]=\sqrt n\cdot\frac1n=\frac1{\sqrt n}.\) The smallest integer satisfying \(n>K^2\) is \(\lfloor K^2\rfloor+1\), and \(1/\sqrt n\) decreases as \(n\) increases. Hence \(\displaystyle \sup_{n\ge1}\mathbb{E}[|Z_n|\mathbf{1}_{\{|Z_n|>K\}}]=\frac1{\sqrt{\lfloor K^2\rfloor+1}},\) which tends to zero as \(K\to\infty\). Thus \((Z_n)\) is uniformly integrable.
Check: the computation used the strict tail event \(n>K^2\), not \(n\ge K^2\). The resulting supremum is over the same threshold \(K\) for all \(n\), and its displayed value tends to zero. This is one concrete example of how a stronger moment bound can prevent the escaping expected magnitude seen in the previous example.
Common Mistake: Replacing Tail Expectation with Tail Probability
A tempting but incorrect argument is: “The first moments are bounded, so Markov's inequality makes the tails small; therefore the family is uniformly integrable.” The conclusion does not follow. Markov's inequality bounds \(\mathbb{P}(|X|>K)\). Uniform integrability instead requires the expectation of \(|X|\) on that event to be small. The two quantities measure different things because the values on the tail event may themselves be large.
Worked Example: The Incorrect Estimate and Its Correction
Return to \(X_n=n\mathbf{1}_{[0,1/n]}\), for which \(\mathbb{E}|X_n|=1\). Markov's inequality gives, at each \(K>0\), \(\mathbb{P}(|X_n|>K)\le\frac1K.\) That estimate concerns probability only. Multiplying its right-hand side by \(K\) does not bound the tail expectation: the event can contain values much larger than \(K\).
For an exact check, take any integer \(n>K\). The tail event is \([0,1/n]\), so its probability is \(1/n\), while its tail expectation is \(\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]=n\cdot\frac1n=1.\) Thus the probability can be arbitrarily small as \(n\) grows, but the expected magnitude on the event stays equal to \(1\). The correction is to compute or bound the tail expectation itself; a bound on tail probability alone cannot establish uniform integrability.
There is a related issue with signed expectations. A bound on \(|\mathbb{E}X_n|\) is weaker even than a bound on \(\mathbb{E}|X_n|\), because positive and negative values can cancel. For example, let \(V_n\) equal \(n\) and \(-n\) with probability \(1/(2n)\) each, and equal zero otherwise. Then \(\mathbb{E}V_n=0\), but \(\mathbb{E}|V_n|=1\). In questions about uniform integrability, use the absolute-value definition and check the tail expectations, not only the signed means.
The practical distinction is concise: uniform integrability implies a uniform bound on first absolute moments, and that bound gives uniform control of tail probabilities. Neither fact reverses the implication from first moments to uniform integrability. To decide whether expected magnitude escapes, test \(\sup_{X\in\mathcal{X}}\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\) at a common threshold and then let \(K\) increase.
Check Your Understanding
For each question, distinguish the tail probability from the tail expectation and use the strict event \(|X|>K\).
- For \(X_n=n\mathbf{1}_{[0,1/n]}\), compute \(\mathbb{E}|X_n|\) and the tail expectation when \(n>K\).
- Why does the bound \(\sup_n\mathbb{P}(|X_n|>K)\to0\) not prove that \((X_n)\) is uniformly integrable?
- Prove that a uniformly integrable family has bounded expected absolute values by splitting each expectation at a fixed threshold.
- For \(Y_n=2+\mathbf{1}_{[0,1/n]}\), what is the tail expectation at \(K=3\), and why does strictness matter?
- For \(Z_n=\sqrt n\mathbf{1}_{[0,1/n]}\), which integers \(n\) contribute to the tail expectation at threshold \(K\)?