Expectations Can Miss Rare, Large Values
An expectation averages both the size of a random variable and the probability of its values. A family can therefore have bounded expectations even when a small amount of probability carries a substantial amount of its expected value farther and farther into the tail. Uniform integrability is the condition that rules out this persistent tail contribution. The distinction matters: controlling the probability of large values is not the same as controlling their contribution to an expectation.
Throughout, random variables are measurable, real-valued functions on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with \(\mathbb{P}(\Omega)=1\). Unless stated otherwise, the families below consist of nonnegative random variables. Expectations are Lebesgue integrals with respect to \(\mathbb{P}\). We use the strict tail event \(\{X>K\}\) for nonnegative \(X\).
For a single integrable random variable, its tail integral tends to zero as \(K\) tends to infinity; this follows from the Dominated Convergence Theorem, since \(|X|\chi_{\{|X|>K\}}\to0\) pointwise and is bounded by the integrable function \(|X|\). Uniform integrability asks for this tail control uniformly over a family. The issue in this tutorial is whether a bound on \(\mathbb{E}[X]\) alone provides that uniform control.
A Family With Bounded Expectations but Persistent Tails
Worked Example: Rare Spikes With Expectation One
On \(\Omega=[0,1]\) with the Lebesgue sigma-algebra and probability measure \(m\), define \(A_n=(0,1/n)\) and \(X_n=n\chi_{A_n}\) for each positive integer \(n\). These are measurable, nonnegative random variables.
The interval \(A_n\) has probability \(1/n\). Thus \(X_n=n\) on \(A_n\) and \(X_n=0\) off \(A_n\), so its expectation is $$ \mathbb{E}[X_n]=n\,m(A_n)+0\cdot m(A_n^c)=n\cdot\frac1n=1. $$ In particular, the expectations are uniformly bounded.
Now fix any finite \(K\ge0\). If \(n\le K\), then \(X_n\le K\) everywhere, so the strict tail event \(\{X_n>K\}\) is empty and its tail integral is zero. If \(n>K\), then \(\{X_n>K\}=A_n\), and $$ \mathbb{E}\!\left[X_n\chi_{\{X_n>K\}}\right] =n\,m(A_n)=n\cdot\frac1n=1. $$ For every finite \(K\), there is a positive integer \(n>K\). Consequently, $$ \sup_{n\ge1}\mathbb{E}\!\left[X_n\chi_{\{X_n>K\}}\right]=1. $$ This supremum does not tend to zero as \(K\to\infty\); the family is not uniformly integrable. Check: each individual tail integral is either zero or one, and for every finite threshold at least one index gives one, so the displayed supremum is exactly one.
The same example separates tail probability from tail expectation. For fixed \(\varepsilon>0\), if \(n\le\varepsilon\), then \(X_n>\varepsilon\) never occurs; if \(n>\varepsilon\), then \(\{X_n>\varepsilon\}=A_n\). Therefore $$ \mathbb{P}(X_n>\varepsilon)= \begin{cases} 0,&n\le\varepsilon,\\[2pt] 1/n,&n>\varepsilon. \end{cases} $$ For fixed \(\varepsilon\), all sufficiently large \(n\) satisfy \(n>\varepsilon\), and then \(1/n\to0\). Thus \(X_n\to0\) in probability. In fact, for every \(x\in[0,1]\), \(X_n(x)\to0\): at \(x=0\) every value is zero, and at \(x>0\), once \(1/n\le x\), \(x\notin(0,1/n)\). Nevertheless, \(\mathbb{E}[X_n]=1\) for every \(n\). Rare values vanish in probability while retaining all the expected value in the tail.
Why Higher Moments Give Uniform Tail Control
Proof. First, \(X\le 1+X^p\) for every nonnegative real number \(X\): if \(X\le1\), then \(X\le1\); if \(X>1\), then \(X\le X^p\), since \(p>1\). Hence \(\mathbb{E}[X]\le1+\mathbb{E}[X^p]\le1+M<\infty\), so each \(X\) is integrable. On the event \(\{X>K\}\), the inequality \(X^{p-1}>K^{p-1}\) gives \(X\le X^p/K^{p-1}\). Multiplying by the indicator and integrating yields $$ \mathbb{E}\!\left[X\chi_{\{X>K\}}\right] \le\frac{\mathbb{E}[X^p]}{K^{p-1}} \le\frac{M}{K^{p-1}}. $$ Taking the supremum over \(X\in\mathcal{C}\) preserves the bound. Since \(p-1>0\), \(M/K^{p-1}\to0\) as \(K\to\infty\), which is precisely uniform integrability. If \(M=0\), then every \(X^p\) has integral zero and the same displayed bound is zero; the conclusion still holds. \(\square\)
Worked Example: Rare Spikes With Bounded Second Moments
On the same probability space \(([0,1],\mathcal{L},m)\), set \(Z_n=\sqrt n\,\chi_{(0,1/n)}\) for positive integers \(n\). Its expectation and second moment are $$ \mathbb{E}[Z_n]=\sqrt n\cdot\frac1n=\frac1{\sqrt n}\le1, \qquad \mathbb{E}[Z_n^2]=n\cdot\frac1n=1. $$ The second moments are uniformly bounded, so the theorem with \(p=2\) and \(M=1\) gives a tail bound of \(1/K\) for \(K>0\).
We can also compute the supremum exactly. If \(\sqrt n\le K\), then the strict tail event \(\{Z_n>K\}\) is empty. If \(\sqrt n>K\), then this event is \((0,1/n)\), and its tail integral is $$ \mathbb{E}\!\left[Z_n\chi_{\{Z_n>K\}}\right] =\sqrt n\cdot\frac1n=\frac1{\sqrt n}. $$ The least positive integer satisfying \(n>K^2\) is \(\lfloor K^2\rfloor+1\). Since \(1/\sqrt n\) decreases as \(n\) increases, it follows that $$ \sup_{n\ge1}\mathbb{E}\!\left[Z_n\chi_{\{Z_n>K\}}\right] =\frac1{\sqrt{\lfloor K^2\rfloor+1}}. $$ This tends to zero as \(K\to\infty\), in agreement with the theorem. For example, at \(K=2\), the first index with \(\sqrt n>2\) is \(n=5\), and the supremum is \(1/\sqrt5\); the theorem's bound at that threshold is \(1/2\), which is larger, as required. The family is uniformly integrable.
The crucial difference from the first family is not that the spikes disappear: both families have large values on small events. It is that the second moments of the new family remain bounded. For the first family, \(\mathbb{E}[X_n^2]=n^2(1/n)=n\), which is unbounded. The higher-moment theorem converts that stronger control into a uniform tail-integral estimate.
Worked Example: A Uniformly Bounded Family
On \(([0,1],\mathcal{L},m)\), let \(Y_n=3\chi_{(0,1/(n+1))}\) for \(n\ge1\). Each \(Y_n\) is nonnegative and measurable, with $$ \mathbb{E}[Y_n]=3\cdot\frac1{n+1}. $$ Since \(n\ge1\), \(n+1\ge2\), so \(\mathbb{E}[Y_n]\le3/2\). For a threshold \(K\ge3\), \(Y_n\le K\) everywhere, and the strict tail event \(\{Y_n>K\}\) is empty; every tail integral is zero. For \(0\le K<3\), the tail event is exactly \((0,1/(n+1))\), and its tail integral is \(3/(n+1)\). The largest of these values over \(n\ge1\) is \(3/2\), attained at \(n=1\). Thus the supremum tail integral is \(3/2\) for \(0\le K<3\) and zero for \(K\ge3\). It tends to zero, so this family is uniformly integrable. Check: the endpoint \(K=3\) belongs to the zero-tail case because the definition uses \(Y_n>K\), not \(Y_n\ge K\).
A Common Mistake: Confusing Probability Bounds With Tail Bounds
A tempting but invalid calculation is to infer from \(\mathbb{E}[X_n]=1\) that $$ \mathbb{E}\!\left[X_n\chi_{\{X_n>K\}}\right]\le\frac{\mathbb{E}[X_n]}{K}. $$ That inequality does not follow: the factor \(X_n\) remains inside the tail expectation. For the rare-spike example, whenever \(n>K\), the left side is \(1\), while the proposed right side is \(1/K\); for instance, with \(K=2\) and \(n=3\), the tail expectation is \(3\cdot(1/3)=1\), whereas \(1/K=1/2\). This is a concrete contradiction to the proposed bound.
What the first-moment bound does yield is a bound on tail probability: on \(\{X>K\}\), \(X>K\), so \(K\mathbb{P}(X>K)\le\mathbb{E}[X]\) for \(K>0\). For \(X_n\) with \(n>K\), this says \(\mathbb{P}(X_n>K)=1/n\le1/K\), which is correct. But a small probability can multiply a large value: in this example the probability is \(1/n\), the value is \(n\), and their product remains one. Uniform integrability concerns that product as captured by the tail integral, not probability alone.
Thus a uniform bound on expectations is useful, but insufficient: it controls the total average and does not prevent that average from being carried by rarer and larger values. A uniform bound on a strictly higher moment does prevent this, through the explicit power-of-\(K\) estimate proved above. Uniform boundedness of the random variables is another sufficient condition, since if \(|X|\le C\) for every member, then all strict tail integrals vanish for \(K\ge C\). When assessing a family, calculate or bound the tail integrals at a common threshold rather than relying only on its expectations or on the probabilities of its tail events.
Check Your Understanding
For each question, keep the strict tail event and the family-wide supremum at a common threshold explicit.
- For \(X_n=n\chi_{(0,1/n)}\), compute \(\mathbb{E}[X_n\chi_{\{X_n>K\}}]\) for the cases \(n\le K\) and \(n>K\). Why does the family fail to be uniformly integrable?
- For fixed \(\varepsilon>0\), write the exact value of \(\mathbb{P}(X_n>\varepsilon)\) in the cases \(n\le\varepsilon\) and \(n>\varepsilon\). What does it imply as \(n\to\infty\)?
- Suppose \(X\ge0\) and \(\mathbb{E}[X^p]\le M\) for some \(p>1\). Derive the bound on \(\mathbb{E}[X\chi_{\{X>K\}}]\) for \(K>0\).
- For \(Z_n=\sqrt n\,\chi_{(0,1/n)}\), compute its second moment and explain how the higher-moment theorem applies.
- If every member of a family satisfies \(|X|\le C\), for which thresholds \(K\) do all strict tail integrals vanish?