From Moment Bounds to Tail Control
Uniform integrability is a way to control the large values in a whole family of random variables at once. A bound on their first absolute moments does not provide that control: the expectation can remain fixed while more and more of it is carried by increasingly rare, increasingly large values. A uniform bound on a moment of order \(p>1\) rules out this behavior strongly enough to make the expected tails uniformly small.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). For an integrable real-valued random variable \(X\), its tail expectation at threshold \(K>0\) is \(\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\). The event is defined using the strict inequality \(|X|>K\); values exactly equal to \(K\) are not counted.
For a random variable \(X\), the \(p\)th absolute moment is \(\mathbb{E}|X|^p\), and its \(L^p\) norm is \(\|X\|_p=(\mathbb{E}|X|^p)^{1/p}\), when this moment is finite. The key condition is a common finite bound on these moments for some fixed real \(p>1\). No martingale assumption is needed for the implication. In a martingale application, one may take \((X_n)_{n\ge0}=(M_n)_{n\ge0}\), where \((M_n)\) is adapted to a filtration \((\mathcal{F}_n)_{n\ge0}\); the argument below uses only the moment bound, not the martingale property.
Proof. Fix \(K>0\) and \(X\in\mathcal{X}\). On the event \(\{|X|>K\}\), the exponent \(1-p\) is negative, so \(|X|^{1-p}\le K^{1-p}\). Therefore, on that event, \(|X|=|X|^p|X|^{1-p}\le K^{1-p}|X|^p\). Multiplying by the indicator and taking expectations gives $$ \mathbb{E}\big[|X|\mathbf{1}_{\{|X|>K\}}\big]\le K^{1-p}\mathbb{E}|X|^p\le \frac{C}{K^{p-1}}. $$ The right side does not depend on \(X\). Taking the supremum over the family preserves the bound. Since \(p-1>0\), \(C/K^{p-1}\) tends to zero as \(K\) tends to infinity. This is exactly the definition of uniform integrability. If \(C=0\), each \(X\) is zero almost surely because \(\mathbb{E}|X|^p=0\); the displayed bound and the conclusion still hold. \(\square\)
The same result can be stated using norms: if \(\sup_{X\in\mathcal{X}}\|X\|_p\le M<\infty\), take \(C=M^p\) to obtain the tail bound \(M^p/K^{p-1}\). The bound is quantitative: at a specified threshold it controls the largest expected tail among all members of the family. The power \(p-1\) also records how rapidly this guarantee improves as \(K\) increases.
Worked Tail Calculations
Worked Example: A Uniform Second-Moment Bound
Take \(\Omega=[0,1]\) with Lebesgue probability measure and \(\mathcal{F}\) the Lebesgue measurable sets. For each integer \(n\ge1\), define \(X_n=n\mathbf{1}_{[0,1/n^2]}\). Find the second moments and the exact uniform tail expectation at \(K=2\).
The interval \([0,1/n^2]\) has probability \(1/n^2\). Thus $$ \mathbb{E}|X_n|^2=n^2\frac{1}{n^2}=1 $$ for every \(n\), so the theorem applies with \(p=2\) and \(C=1\). At \(K=2\), the strict event \(\{|X_n|>2\}\) occurs on the interval precisely when \(n>2\). If \(n\le2\), its tail expectation is zero. If \(n>2\), it is \(n\) on an event of probability \(1/n^2\), and the tail expectation is \(n/n^2=1/n\). Its supremum over integer \(n>2\) is \(1/3\), attained at \(n=3\).
Check: the theorem gives the bound \(1/2^{2-1}=1/2\), and the exact supremum satisfies \(1/3\le1/2\). For any real \(K>0\), the same calculation gives a supremum of \(1/(\lfloor K\rfloor+1)\), since the smallest integer satisfying the strict condition \(n>K\) is \(\lfloor K\rfloor+1\). This tends to zero with \(K\), as uniform integrability requires.
Worked Example: How the Moment Order Changes the Rate
On the same probability space, define \(Y_n=n\mathbf{1}_{[0,n^{-3/2}]}\) for integers \(n\ge1\). Compute a uniform \(3/2\)-moment bound and the exact tail supremum at \(K=4\).
The interval has probability \(n^{-3/2}\), so $$ \mathbb{E}|Y_n|^{3/2}=n^{3/2}n^{-3/2}=1. $$ The theorem applies with \(p=3/2\) and \(C=1\). The strict tail event at \(K=4\) occurs exactly when \(n>4\). For such an \(n\), the tail expectation is \(n\cdot n^{-3/2}=n^{-1/2}\); for \(n\le4\), it is zero. Since \(n^{-1/2}\) decreases with \(n\), the supremum over integers \(n>4\) is \(1/\sqrt{5}\), attained at \(n=5\).
Check: the theorem's bound at \(K=4\) is \(1/4^{(3/2)-1}=1/2\). Indeed, \(1/\sqrt{5}\le1/2\), because \(5\ge4\). This family has a slower guaranteed rate, \(K^{-1/2}\), than a family with a uniform second-moment bound, for which the theorem gives \(K^{-1}\). A higher available moment order yields a faster power in this estimate.
Proof. On \([0,1]\) with Lebesgue probability measure, set \(X_n=n^{1/p}\mathbf{1}_{[0,1/n]}\) for integers \(n\ge1\). Direct calculation gives \(\mathbb{E}|X_n|^p=n(1/n)=1\). Fix \(K\ge1\) and choose \(n=\lfloor K^p\rfloor+1\). Then \(n>K^p\), so \(n^{1/p}>K\), and the strict tail event includes the whole interval \([0,1/n]\). Consequently, $$ \mathbb{E}\big[|X_n|\mathbf{1}_{\{|X_n|>K\}}\big]=n^{1/p}\frac1n=n^{-(p-1)/p}. $$ Also, \(n\le K^p+1\le2K^p\), where the last inequality uses \(K\ge1\). Because \((p-1)/p>0\), this implies $$ n^{-(p-1)/p}\ge 2^{-(p-1)/p}K^{-(p-1)}. $$ The supremum over the family is at least the tail expectation of this chosen member, proving the claimed lower bound. Thus no universally faster power of \(K\) can replace \(K^{-(p-1)}\) in the theorem, up to constant factors. \(\square\)
Why a First-Moment Bound Is Not Enough
Worked Example: Bounded First Moments Without Uniform Integrability
On \([0,1]\) with Lebesgue probability measure, let \(Z_n=n\mathbf{1}_{[0,1/n]}\) for integers \(n\ge1\). Show that the first moments are uniformly bounded but the family is not uniformly integrable.
The interval has probability \(1/n\), so \(\mathbb{E}|Z_n|=n(1/n)=1\) for every \(n\). Fix any finite \(K>0\). For every integer \(n>K\), the strict event \(\{|Z_n|>K\}\) is the interval \([0,1/n]\), and $$ \mathbb{E}\big[|Z_n|\mathbf{1}_{\{|Z_n|>K\}}\big]=n\frac1n=1. $$ There is always an integer \(n>K\), so the supremum over \(n\) is at least \(1\). It is also at most \(1\), since the tail expectation cannot exceed \(\mathbb{E}|Z_n|=1\). The supremum is therefore exactly \(1\) for every finite \(K>0\), not a quantity tending to zero.
Check: the family has \(\sup_n\mathbb{E}|Z_n|=1\) but is not uniformly integrable. The theorem does not apply: for any fixed \(p>1\), \(\mathbb{E}|Z_n|^p=n^p/n=n^{p-1}\), which is unbounded as \(n\) increases.
Common mistake. A tempting but incorrect calculation is to infer from \(\mathbb{E}|Z_n|=1\) that \(\mathbb{E}[|Z_n|\mathbf{1}_{\{|Z_n|>K\}}]\le1/K\). At \(K=2\) and \(n=3\), the actual tail expectation is \(3\cdot(1/3)=1\), which is greater than \(1/2\). The first-moment bound can control the probability of a large-value event, but it does not by itself control the amount of \(|Z_n|\) carried on that event. The proof of the theorem needs the stronger pointwise comparison \(|X|\le K^{1-p}|X|^p\) on \(\{|X|>K\}\); this comparison uses \(p>1\).
For a martingale sequence \((M_n)_{n\ge0}\) on \((\Omega,\mathcal{F},\mathbb{P})\), adapted to \((\mathcal{F}_n)_{n\ge0}\), the practical consequence is immediate: if \(\sup_{n\ge0}\mathbb{E}|M_n|^p<\infty\) for one fixed \(p>1\), then the family \(\{M_n:n\ge0\}\) is uniformly integrable, with the theorem's explicit tail bound. This implication requires the uniform moment bound, not just that each individual \(M_n\) has a finite \(p\)th moment. Uniform integrability is a property of the family as a whole, so the common bound is essential.
Check Your Understanding
Use the strict tail event \(\{|X|>K\}\) in each calculation.
- If \(\sup_{X\in\mathcal{X}}\mathbb{E}|X|^4\le16\), what tail bound does the theorem give at threshold \(K\)?
- For \(X_n=n\mathbf{1}_{[0,1/n^2]}\), compute the second moment and the tail expectation at a general threshold \(K>0\).
- Why does the proof require \(p>1\) when comparing \(|X|^{1-p}\) with \(K^{1-p}\) on the tail event?
- For \(Z_n=n\mathbf{1}_{[0,1/n]}\), compute the tail supremum at any fixed finite \(K>0\) and explain why bounded first moments do not imply uniform integrability here.
- What uniform \(L^p\) condition on a martingale sequence is sufficient to guarantee uniform integrability of its members?