Why Tail Expectations Matter
A sequence of random variables can converge almost surely while its expectations fail to converge to the expectation of its limit. The obstacle is often a small probability of very large values: these values may disappear pointwise but still carry a fixed amount of expectation. Uniform integrability rules out that loss by controlling the expected size of the tails, not merely the probability of a tail event.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). A family \(\mathcal{X}\) of random variables is assumed to consist of integrable random variables. For \(K>0\), its tail expectation at threshold \(K\) is \(\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\), where the inequality in the event is strict.
This is stronger than saying each individual variable is integrable. For each fixed integrable \(X\), its own tail expectation tends to zero. Uniform integrability requires one threshold to work for the whole family. It also implies a uniform bound on first moments: choose \(K\) so the supremum of the tail expectations is at most \(1\). Then every \(X\in\mathcal{X}\) satisfies \(\mathbb{E}|X|\le K+1\), by splitting the expectation into \(|X|\le K\) and \(|X|>K\).
Uniform Integrability and Limits
The central reason for tracking tail expectations is that they let us control the part of a variable that truncation removes. Define the clipping function \(\phi_K(x)=\max(-K,\min(x,K))\). It leaves values in \([-K,K]\) unchanged and truncates values outside that interval. The following result states precisely how uniform integrability controls limits.
Proof. Write \(\alpha(K)=\sup_n\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]\), so \(\alpha(K)\to0\) as \(K\to\infty\). Convergence in probability gives a subsequence \((X_{n_j})\) converging to \(X\) almost surely. For each \(K>0\), whenever \(|X|>2K\), the almost-sure convergence implies that eventually \(|X_{n_j}|>K\) and \(|X_{n_j}|\to |X|\). Thus $$ |X|\mathbf{1}_{\{|X|>2K\}} \le \liminf_{j\to\infty}|X_{n_j}|\mathbf{1}_{\{|X_{n_j}|>K\}} $$ almost surely. Fatou's lemma therefore gives \(\mathbb{E}[|X|\mathbf{1}_{\{|X|>2K\}}]\le\alpha(K)\). In particular, \(\mathbb{E}|X|\le 2K+\alpha(K)<\infty\).
For \(K>0\), the clipping error satisfies \(|x-\phi_K(x)|\le |x|\mathbf{1}_{\{|x|>K\}}\). Also, \(\phi_K\) is continuous and bounded by \(K\) in absolute value, so \(\phi_K(X_n)\to\phi_K(X)\) in probability. For any \(\varepsilon>0\), $$ \mathbb{E}|\phi_K(X_n)-\phi_K(X)| \le \varepsilon+2K\,\mathbb{P}\big(|\phi_K(X_n)-\phi_K(X)|>\varepsilon\big). $$ Taking \(n\to\infty\) and then \(\varepsilon\downarrow0\) shows that this expectation tends to zero for fixed \(K\). The triangle inequality now yields $$ \mathbb{E}|X_n-X| \le \alpha(K)+\mathbb{E}|\phi_K(X_n)-\phi_K(X)| +\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]. $$ The last term is at most \(\alpha(K/2)\), by the tail bound already proved with threshold \(K/2\). First take the limit superior as \(n\to\infty\), then let \(K\to\infty\). Both tail bounds vanish, proving \(\mathbb{E}|X_n-X|\to0\). Finally, \(|\mathbb{E}X_n-\mathbb{E}X|\le\mathbb{E}|X_n-X|\), so the expectations converge as well. \(\square\)
Convergence in mean, \(\mathbb{E}|X_n-X|\to0\), is stronger than convergence of expectations alone: it controls the average absolute error. The theorem shows that convergence in probability plus uniform tail control is enough to obtain this stronger conclusion. In particular, almost-sure convergence implies convergence in probability, so the theorem also applies when almost-sure convergence is known.
Bounded and Dominated Families
Proof. Under condition (i), for every \(K\ge C\) the strict event \(\{|X|>K\}\) is empty up to a null set. Every tail expectation is therefore zero. Under condition (ii), \(|X|>K\) implies \(Y>K\), and hence $$ \mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}] \le \mathbb{E}[Y\mathbf{1}_{\{Y>K\}}]. $$ The right-hand side does not depend on \(X\) and tends to zero as \(K\to\infty\), since \(Y\) is integrable. Taking the supremum over \(\mathcal{X}\) proves uniform integrability. \(\square\)
A bounded stopping horizon gives a useful instance of domination. Let \((M_n)_{n\ge0}\) be a martingale adapted to a filtration \((\mathcal{F}_n)_{n\ge0}\), with each \(M_k\) integrable. A random time \(\tau\) is a stopping time if \(\{\tau\le n\}\in\mathcal{F}_n\) for each \(n\). If \(\tau\le N\) almost surely for a fixed finite integer \(N\), then for every \(n\), $$ |M_{n\wedge\tau}|\le \sum_{k=0}^{N}|M_k|\quad\text{almost surely}. $$ The right-hand side is integrable, since its expectation is the finite sum \(\sum_{k=0}^{N}\mathbb{E}|M_k|\). The proposition shows that the family \(\{M_{n\wedge\tau}:n\ge0\}\) is uniformly integrable. This conclusion concerns the stopped variables; boundedness of the stopping time does not by itself assert that the martingale values are bounded by a deterministic constant.
Worked Example: A Uniformly Bounded Stopped Family
Let \(\Omega=\{-1,1\}\) with both outcomes having probability \(\tfrac12\), and let \(Z(\omega)=\omega\). Set \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=\sigma(Z)\) for \(n\ge1\). Define \(M_0=0\) and \(M_n=Z\) for \(n\ge1\). This is a martingale: \(\mathbb{E}[M_1\mid\mathcal{F}_0]=\mathbb{E}Z=0=M_0\), and after time \(1\) the process is constant. Take the stopping time \(\tau=1\).
For \(n=0\), \(M_{n\wedge\tau}=0\); for every \(n\ge1\), \(M_{n\wedge\tau}=Z\). Thus every variable in the stopped family has absolute value at most \(1\). If \(K\ge1\), the event \(\{|M_{n\wedge\tau}|>K\}\) is empty, including at \(K=1\) because the tail event is strict. Hence, for every such \(K\), $$ \sup_{n\ge0}\mathbb{E}\big[|M_{n\wedge\tau}|\mathbf{1}_{\{|M_{n\wedge\tau}|>K\}}\big]=0. $$ This verifies uniform integrability directly. It also agrees with the bounded-family criterion.
Worked Example: An Unbounded Family with an Integrable Dominator
Let \(Z\) take positive integer values with \(\mathbb{P}(Z=k)=2^{-k}\) for \(k\ge1\). These probabilities sum to \(1\), because \(\sum_{k=1}^{\infty}2^{-k}=1\), and $$ \mathbb{E}Z=\sum_{k=1}^{\infty}\frac{k}{2^k}=2. $$ For example, the last sum follows by writing \(k=\sum_{r=1}^{k}1\), summing the nonnegative terms in the other order, and using \(\sum_{k=r}^{\infty}2^{-k}=2^{1-r}\), giving \(\sum_{r=1}^{\infty}2^{1-r}=2\).
Take \(\mathcal{F}_0\) to be trivial, \(\mathcal{F}_n=\sigma(Z)\) for \(n\ge1\), \(M_0=2\), and \(M_n=Z\) for \(n\ge1\). Since \(\mathbb{E}[Z\mid\mathcal{F}_0]=2\) and the process is constant after time \(1\), this is a martingale. With the stopping time \(\tau=1\), the stopped family consists of \(M_0=2\) and \(M_1=Z\), repeated at later indices. It is not uniformly bounded, but every member is dominated in absolute value by the integrable variable \(Y=2+Z\), whose expectation is \(4\). The domination criterion proves uniform integrability.
For a direct tail check, let \(K\ge2\) be an integer. The value \(2\) contributes nothing to the strict tail event. For \(Z\), the tail expectation is $$ \mathbb{E}[Z\mathbf{1}_{\{Z>K\}}] =\sum_{k=K+1}^{\infty}\frac{k}{2^k} =\frac{K+2}{2^K}. $$ To verify the sum, write \(k=K+1+r\); then it equals \(2^{-(K+1)}\sum_{r\ge0}(K+1+r)2^{-r}=2^{-(K+1)}(2(K+1)+2)=(K+2)/2^K\). For instance, at \(K=3\), the tail is \((3+2)/2^3=5/8\). The expression tends to zero as \(K\to\infty\), confirming the tail condition for the whole stopped family.
Stopping Does Not Guarantee Uniform Integrability
The word “stopped” alone does not provide tail control. An unbounded stopping time can produce a family of stopped variables with rare, increasingly large values. The next example also shows exactly what can go wrong when one tries to pass expectations through a limit.
Worked Example: A Stopped Martingale That Is Not Uniformly Integrable
Let \(\Omega=\{H,T\}^{\mathbb{N}}\) carry independent fair coin flips, and let \(\mathcal{F}_n\) contain the information from the first \(n\) flips. Define \(M_0=1\) and $$ M_n=2^n\mathbf{1}_{\{\text{the first }n\text{ flips are all }H\}},\qquad n\ge1. $$ On the event that the first \(n\) flips are all heads, \(M_n=2^n\); the next value is \(2^{n+1}\) with probability \(\tfrac12\) and \(0\) with probability \(\tfrac12\). Its conditional expectation is therefore \(\tfrac12 2^{n+1}+\tfrac12 0=2^n=M_n\). If a tail has already occurred, both the current and next values are zero. Thus \((M_n)\) is a martingale. Take the stopping time \(\tau=\infty\), so \(M_{n\wedge\tau}=M_n\).
For every \(n\), the all-heads event has probability \(2^{-n}\), and \(M_n=2^n\) on that event and zero otherwise. Therefore \(\mathbb{E}M_n=2^n2^{-n}=1\). Now fix any \(K>0\), and choose \(n\) with \(2^n>K\). The strict tail event \(\{M_n>K\}\) is exactly the all-heads event, so $$ \mathbb{E}[|M_n|\mathbf{1}_{\{|M_n|>K\}}]=2^n2^{-n}=1. $$ For every \(K>0\), the supremum over \(n\) is at least \(1\). The family is not uniformly integrable.
Almost surely, some tail eventually occurs: the probability of all heads in the first \(n\) flips is \(2^{-n}\), which tends to zero. After the first tail, \(M_n\) remains zero. Consequently \(M_n\to0\) almost surely, but \(\mathbb{E}M_n=1\) for every \(n\), whereas the expectation of the limit is \(0\). The persistent tail expectation explains the failure: the rare value \(2^n\) carries expectation \(1\) at every time.
A Common Mistake and a Practical Check
Common mistake: exchanging a limit and an expectation without tail control. In the preceding example, the incorrect calculation is “\(M_n\to0\) almost surely, so \(\lim_n\mathbb{E}M_n=\mathbb{E}0=0\).” The computed expectations show the error: \(\mathbb{E}M_n=1\) for every \(n\), so their limit is \(1\), not \(0\). Almost-sure convergence does not in general justify exchanging limit and expectation. The correction is to check a condition that controls the tails, such as uniform integrability; the example fails that condition because its tail expectations stay at least \(1\).
For a family of stopped variables, a useful order of checks is:
- If the stopping time is bounded by a deterministic \(N\), check that \(M_0,\ldots,M_N\) are integrable. Their finite sum of absolute values dominates all \(M_{n\wedge\tau}\).
- If the stopping time is not known to be bounded, look for a common integrable bound \(Y\), or verify the tail-expectation definition directly.
- For a proposed limit argument, keep the supremum over the family at the same threshold \(K\). Pointwise integrability of every member does not establish uniform integrability.
Uniform integrability is a condition on a family of random variables, not a guarantee attached automatically to a martingale or to a stopped process. Its role here is specific: it prevents expectation from disappearing into increasingly rare and increasingly large values when a limit is taken.
Check Your Understanding
For each question, pay attention to the strict tail event \(\{|X|>K\}\) and to whether the bound is uniform across the family.
- If \(|X|\le 5\) almost surely for every member of a family, what is its tail expectation for \(K=5\)? Why does the strict inequality matter?
- Suppose \(|X|\le Y\) almost surely for every \(X\) in a family, where \(Y\ge0\) and \(\mathbb{E}Y<\infty\). Give the common upper bound for the tail expectations at threshold \(K\).
- If \(\tau\le N\) almost surely and each \(M_k\), \(0\le k\le N\), is integrable, give an integrable random variable that dominates every \(M_{n\wedge\tau}\).
- For the fair-coin martingale in the final worked example, compute \(\mathbb{E}[|M_n|\mathbf{1}_{\{|M_n|>K\}}]\) when \(K<2^n\).
- What conclusion about \(\mathbb{E}X_n\) follows if a uniformly integrable sequence \(X_n\) converges in probability to \(X\)?