Uniform Integrability and Weak Laws for Triangular Arrays

Why Uniform Integrability Enters a Weak Law

For a row of independent observations, a variance bound can often show that the row average is close to its expectation. But finite first moments do not guarantee finite variances, and even when every entry is integrable, increasingly rare and increasingly large values can disrupt averages as the row length grows. The central issue is whether the family of entries has enough control over its tails.

Uniform integrability supplies that control. It lets us discard the large values with a uniformly small expected error, then use a variance argument on the remaining bounded observations. The result applies to a triangular array: the distribution of an entry may change with both the row and the position within the row.

Triangular Arrays and Uniform Integrability

Definition: A triangular array of random variables is a collection $\{X_{n,k}: n\geq 1,\ 1\leq k\leq n\}$, arranged in rows of lengths $n$. The row average is $n^{-1}\sum_{k=1}^n X_{n,k}$. The array is uniformly integrable if every $X_{n,k}$ is integrable and $$ \lim_{M\to\infty}\ \sup_{n\geq 1,\ 1\leq k\leq n} E\!\left[|X_{n,k}|\mathbf{1}_{\{|X_{n,k}|>M\}}\right]=0. $$

Here $\mathbf{1}_A$ is 1 on the event $A$ and 0 otherwise. The condition concerns expected tail size, not merely the probability of a large value. It requires one cutoff $M$ to make the tail expectation small for every entry in every row.

Uniform integrability also implies a uniform bound on first absolute moments. Indeed, choose $M_0$ so that the supremum of the tail expectations above is at most 1. For any entry, $$ E|X_{n,k}| =E[|X_{n,k}|\mathbf{1}_{\{|X_{n,k}|\leq M_0\}}] +E[|X_{n,k}|\mathbf{1}_{\{|X_{n,k}|>M_0\}}] \leq M_0+1. $$ This bound is useful, but the weak law below relies on the stronger fact that the tail expectations can be made arbitrarily small.

Worked Example: A Bounded, Nonidentically Distributed Array

Suppose the entries in each row are independent and $X_{n,k}$ is Bernoulli with success probability $p_{n,k}$, where the probabilities may vary freely across the array. Show that the row average minus its expectation converges in probability to zero.

Every entry lies between 0 and 1, so for $M\geq 1$ its tail expectation is zero. The array is therefore uniformly integrable. Also, $E[X_{n,k}]=p_{n,k}$ and $\operatorname{Var}(X_{n,k})=p_{n,k}(1-p_{n,k})\leq 1/4$. Independence within row $n$ gives $$ \operatorname{Var}\!\left(\frac{1}{n}\sum_{k=1}^n(X_{n,k}-p_{n,k})\right) =\frac{1}{n^2}\sum_{k=1}^n\operatorname{Var}(X_{n,k}) \leq \frac{1}{4n}. $$ For every $\varepsilon>0$, Chebyshev's inequality thus yields $$ P\!\left(\left|\frac{1}{n}\sum_{k=1}^n(X_{n,k}-p_{n,k})\right|>\varepsilon\right) \leq \frac{1}{4n\varepsilon^2}\longrightarrow 0. $$ The expectations need not be identical: the result centers each entry by its own expectation.

The Uniform-Integrability Weak Law

Theorem: Suppose that for each $n$, the variables $X_{n,1},\ldots,X_{n,n}$ are independent and integrable. If the full triangular array is uniformly integrable, then $$ \frac{1}{n}\sum_{k=1}^n\bigl(X_{n,k}-E[X_{n,k}]\bigr)\xrightarrow{p}0. $$ No identical-distribution assumption is required, and no independence between different rows is required.

The proof separates each entry into a bounded part and a tail. For a fixed cutoff $M>0$, define $$ Y_{n,k}^{(M)}=X_{n,k}\mathbf{1}_{\{|X_{n,k}|\leq M\}}, \qquad T_{n,k}^{(M)}=X_{n,k}\mathbf{1}_{\{|X_{n,k}|>M\}}. $$ Then $X_{n,k}=Y_{n,k}^{(M)}+T_{n,k}^{(M)}$. The truncated variables are bounded in absolute value by $M$, while uniform integrability controls the tail terms.

Fix the desired error.
Let $\varepsilon>0$. For any fixed $M$, split the centered row average into the centered truncated average and the centered tail average using $X_{n,k}=Y_{n,k}^{(M)}+T_{n,k}^{(M)}$.

Control the bounded part.
Within row $n$, the truncated variables remain independent. Since $|Y_{n,k}^{(M)}|\leq M$, their variances are at most $M^2$, so the variance of their centered average is at most $M^2/n$.

Control the tail part.
The expected absolute value of the centered tail average is at most twice the largest tail expectation in the array. Markov's inequality turns this into a probability bound.

Take the limits in order.
First let $n$ grow with $M$ fixed, making the bounded-part probability vanish. Then let $M$ grow, making the tail bound vanish by uniform integrability.

Here is the complete argument. Write $\delta(M)=\sup_{n,k}E[|X_{n,k}|\mathbf{1}_{\{|X_{n,k}|>M\}}]$, which tends to zero as $M\to\infty$. For fixed $M$, independence within row $n$, followed by Chebyshev's inequality, gives $$ P\!\left(\left|\frac{1}{n}\sum_{k=1}^n \bigl(Y_{n,k}^{(M)}-E[Y_{n,k}^{(M)}]\bigr)\right|>\frac{\varepsilon}{2}\right) \leq \frac{4M^2}{n\varepsilon^2}. $$ For the tail part, the triangle inequality and $|E[T]|\leq E|T|$ imply $$ E\!\left|\frac{1}{n}\sum_{k=1}^n \bigl(T_{n,k}^{(M)}-E[T_{n,k}^{(M)}]\bigr)\right| \leq \frac{2}{n}\sum_{k=1}^n E|T_{n,k}^{(M)}| \leq 2\delta(M). $$ Markov's inequality bounds the probability that this absolute value exceeds $\varepsilon/2$ by $4\delta(M)/\varepsilon$. If the absolute value of a sum exceeds $\varepsilon$, at least one of its two parts has absolute value greater than $\varepsilon/2$. Therefore $$ P\!\left(\left|\frac{1}{n}\sum_{k=1}^n (X_{n,k}-E[X_{n,k}])\right|>\varepsilon\right) \leq \frac{4M^2}{n\varepsilon^2}+\frac{4\delta(M)}{\varepsilon}. $$ For fixed $M$, take the limit superior as $n\to\infty$. The first term vanishes, leaving an upper bound $4\delta(M)/\varepsilon$. Letting $M\to\infty$ makes this bound zero. Since the probabilities are nonnegative, they converge to zero, which proves the theorem. $\square$

The order of limits matters: $M$ is held fixed while $n$ grows, so the bounded-part variance goes to zero. Only after that is $M$ increased to control the tails. Choosing a cutoff that grows with $n$ would require a separate argument balancing the two bounds.

Worked Example: Infinite-Variance Pareto Entries

For each $n$, let $Z_{n,1},\ldots,Z_{n,n}$ be independent with the Pareto density $f(z)=\tfrac{3}{2}z^{-5/2}$ for $z\geq 1$, and zero otherwise. Let $a_{n,k}=1+k/(n+1)$ and $X_{n,k}=a_{n,k}Z_{n,k}$. Show that the theorem applies, although the entries have infinite variance.

For $t\geq1$, direct integration gives $$ E[Z_{n,k}\mathbf{1}_{\{Z_{n,k}>t\}}] =\int_t^\infty z\frac{3}{2}z^{-5/2}\,dz =3t^{-1/2}, \qquad E[Z_{n,k}]=3. $$ The scale factors satisfy $1<a_{n,k}<2$. For $M\geq2$, $M/a_{n,k}\geq1$, and hence $$ E[|X_{n,k}|\mathbf{1}_{\{|X_{n,k}|>M\}}] =a_{n,k}E[Z_{n,k}\mathbf{1}_{\{Z_{n,k}>M/a_{n,k}\}}] =\frac{3a_{n,k}^{3/2}}{\sqrt{M}} \leq\frac{3\cdot2^{3/2}}{\sqrt{M}}. $$ This bound is uniform over all rows and positions and tends to zero. The array is uniformly integrable. Its entries are integrable with $E[X_{n,k}]=3a_{n,k}$, and they are independent within each row. Their second moments are infinite, since $\int_1^\infty z^2(3/2)z^{-5/2}\,dz=\infty$. Nevertheless, the theorem yields $$ \frac{1}{n}\sum_{k=1}^n(X_{n,k}-3a_{n,k})\xrightarrow{p}0. $$ Thus finite variances are not necessary for this weak law.

When Uniform Integrability Fails

Worked Example: Rare Spikes That Do Not Average Out

For each row $n$, let $B_{n,1},\ldots,B_{n,n}$ be independent Bernoulli variables with success probability $1/n$, and set $X_{n,k}=nB_{n,k}$. Each entry has expectation 1, so the centered row average is $n^{-1}\sum_{k=1}^nX_{n,k}-1$. Show that it does not converge in probability to zero.

If all Bernoulli variables in row $n$ are zero, then the row average is zero and its difference from its expectation is $-1$. Consequently, $$ P\!\left(\left|\frac{1}{n}\sum_{k=1}^n(X_{n,k}-1)\right|>\frac{1}{2}\right) \geq P(B_{n,1}=\cdots=B_{n,n}=0) =\left(1-\frac{1}{n}\right)^n \longrightarrow e^{-1}>0. $$ The probability of a deviation greater than $1/2$ therefore does not tend to zero. The array is not uniformly integrable: for any fixed $M$, take $n>M$. Then $$ E[|X_{n,k}|\mathbf{1}_{\{|X_{n,k}|>M\}}] = nP(B_{n,k}=1)=1. $$ Thus the supremum of the tail expectations cannot tend to zero as $M$ grows. Each row has rare values of size $n$; their expected contribution is not uniformly negligible.

Common Mistakes and the Role of the Assumptions

Controlling tail probabilities instead of tail expectations. A small probability of a large value does not by itself make its contribution to an average small. The rare-spike example has success probability $1/n$, but the spike has size $n$ and expected contribution 1.
Checking tails separately for each entry. Uniform integrability is a single condition over the whole array. Tail control that works for each fixed entry but deteriorates with $n$ is not enough.
Dropping independence within rows. Uniform integrability alone does not guarantee a weak law. For example, let $X_{n,k}=Z$ for every $k$ in a row, where $Z$ is a Bernoulli variable with success probability $1/2$. The array is bounded and uniformly integrable, but every centered row average equals $Z-1/2$, which does not converge in probability to zero. The theorem uses independence within each row to bound the variance of the truncated average.
Centering by one common mean. The theorem allows different expectations $E[X_{n,k}]$. In general, the correct centering is the average of the individual expectations, not a presumed shared mean.

Key takeaway: Within-row independence makes bounded truncations concentrate as the row length grows. Uniform integrability makes the discarded tails uniformly negligible. Together, these conditions give convergence in probability of the row average minus its own expectation without requiring identical distributions or finite variances.

Check Your Understanding

Use the truncation argument and the assumptions carefully.

State the uniform-integrability condition for a triangular array using expected tail size.
Why does the variance of the centered average of the truncated entries tend to zero when the cutoff $M$ is fixed?
In the Pareto example, why can the weak law hold even though the entries have infinite variance?
For the rare-spike array, compute the tail expectation for a fixed $M$ and $n>M$. What does it show?
Does the theorem require independence between different rows? Which independence condition does its proof use?