Stability Under Sums and Bounded Multipliers

Why Stability Matters

Uniform integrability is a condition on the tails of an entire family of random variables. In applications, one often needs to combine several such families or rescale their members by random factors. The key question is whether these operations can create tails that are too large to control uniformly.

For a fixed finite number of summands, the answer is yes: uniform integrability is preserved. It is also preserved when each random variable is multiplied by a factor whose absolute value has one common almost-sure bound. The estimates behind these facts make clear why both the finiteness of the sum and the uniformity of the multiplier bound matter.

Uniform Integrability by Tail Expectations

Definition: A family $\mathcal{X}$ of integrable random variables is uniformly integrable if its tail expectations vanish uniformly: for every $\varepsilon>0$, there is a $K>0$ such that, for every $X\in\mathcal{X}$, $$ \mathbb{E}\bigl[|X|\mathbf{1}_{\{|X|>K\}}\bigr]<\varepsilon. $$ Equivalently, $\sup_{X\in\mathcal{X}}\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]$ tends to zero as $K$ tends to infinity.

Here $\mathbf{1}_A$ denotes the indicator of an event $A$. The threshold $K$ must work for every member of the family, which is why pointwise integrability of each member is not enough.

We will use a basic consequence. If $\mathcal{X}$ is uniformly integrable, then its first moments are uniformly bounded. Indeed, choose $R>0$ so that every tail expectation beyond $R$ is less than $1$. For each $X\in\mathcal{X}$, split according to whether $|X|\le R$:

\mathbb{E}|X| = \mathbb{E}\bigl[|X|\mathbf{1}_{\{|X|\le R\}}\bigr] + \mathbb{E}\bigl[|X|\mathbf{1}_{\{|X|>R\}}\bigr] \le R+1.

Thus $\sup_{X\in\mathcal{X}}\mathbb{E}|X|<\infty$. This bound will control the probability that a sum is large.

Finite Sums Preserve Uniform Integrability

Theorem: Let $m\in\mathbb{N}$, and suppose that for each $j\in\{1,\ldots,m\}$, $\mathcal{X}_j=\{X_{j,\alpha}:\alpha\in I\}$ is a uniformly integrable family of random variables. The index set $I$ may be arbitrary. Then the family of sums $$ \left\{\sum_{j=1}^{m}X_{j,\alpha}:\alpha\in I\right\} $$ is uniformly integrable.

Proof. Write $S_\alpha=\sum_{j=1}^{m}X_{j,\alpha}$, and let $E_\alpha=\{|S_\alpha|>K\}$. For each $j$, uniform integrability gives a finite bound $M_j=\sup_{\alpha\in I}\mathbb{E}|X_{j,\alpha}|$. If $|S_\alpha|>K$, the triangle inequality implies $\sum_{j=1}^{m}|X_{j,\alpha}|>K$. Consequently at least one summand has absolute value greater than $K/m$, so

\mathbb{P}(E_\alpha) \le \sum_{j=1}^{m}\mathbb{P}\left(|X_{j,\alpha}|>\frac{K}{m}\right) \le \frac{m}{K}\sum_{j=1}^{m}M_j.

The last inequality is Markov's inequality. Now fix $R>0$. On $E_\alpha$, split each $|X_{j,\alpha}|$ according to whether it exceeds $R$. Since $|S_\alpha|\le\sum_j|X_{j,\alpha}|$, this gives

\mathbb{E}\bigl[|S_\alpha|\mathbf{1}_{E_\alpha}\bigr] \le \sum_{j=1}^{m} \mathbb{E}\bigl[|X_{j,\alpha}|\mathbf{1}_{\{|X_{j,\alpha}|>R\}}\bigr] + mR\,\mathbb{P}(E_\alpha).

Given $\varepsilon>0$, choose $R$ large enough that each tail expectation in the sum is less than $\varepsilon/(2m)$, uniformly in $\alpha$. This makes their total less than $\varepsilon/2$. With this $R$ fixed, choose $K$ large enough that

mR\frac{m}{K}\sum_{j=1}^{m}M_j<\frac{\varepsilon}{2}.

The resulting bound on $\mathbb{E}[|S_\alpha|\mathbf{1}_{E_\alpha}]$ is less than $\varepsilon$, uniformly in $\alpha$. This is precisely uniform integrability of the sums. $\square$

The proof separates two sources of control. Large values of an individual summand are handled by that family’s tail condition. The remaining, bounded parts contribute at most $mR\mathbb{P}(E_\alpha)$, which becomes small because the probability of a large sum is uniformly small.

Worked Example: Adding Two Rare-Spike Families

Let $A_n$ be events with $\mathbb{P}(A_n)=1/n^2$, and define $X_n=n\mathbf{1}_{A_n}$ and $Y_n=2n\mathbf{1}_{A_n}$, for $n\in\mathbb{N}$. For $t>0$, the tail expectation of $X_n$ is zero unless $n>t$; when $n>t$, it is $n\mathbb{P}(A_n)=1/n<1/t$. Hence the family $\{X_n\}$ is uniformly integrable. For $Y_n$, the tail expectation is zero unless $2n>t$, and when $2n>t$ it is $2/n<4/t$. Thus $\{Y_n\}$ is also uniformly integrable.

The sum is $X_n+Y_n=3n\mathbf{1}_{A_n}$. Its tail expectation is zero unless $3n>t$, and in that case it equals $3n/n^2=3/n<9/t$. Therefore the sums are uniformly integrable as well. The estimate illustrates how the rare event keeps the tail expectation small even though the values on that event grow with $n$.

Uniformly Bounded Multipliers

Theorem: Let $\mathcal{X}=\{X_\alpha:\alpha\in I\}$ be uniformly integrable. Suppose $\{H_\alpha:\alpha\in I\}$ is a family of random variables for which there is a finite constant $C\ge 0$ satisfying $|H_\alpha|\le C$ almost surely for every $\alpha$. Then $\{H_\alpha X_\alpha:\alpha\in I\}$ is uniformly integrable.

\mathbb{E}\bigl[|H_\alpha X_\alpha|\mathbf{1}_{\{|H_\alpha X_\alpha|>K\}}\bigr] \le C\mathbb{E}\bigl[|X_\alpha|\mathbf{1}_{\{|X_\alpha|>K/C\}}\bigr].

For any $\varepsilon>0$, uniform integrability of $\mathcal{X}$ gives a threshold $L$ such that the expectation on the right, before multiplication by $C$, is less than $\varepsilon/C$ whenever its threshold is at least $L$. Taking $K=CL$ makes the displayed tail expectation less than $\varepsilon$, uniformly in $\alpha$. Hence the products are uniformly integrable. $\square$

No independence assumption is needed here. The multiplier can depend on $X_\alpha$; the estimate uses only the almost-sure bound on its absolute value.

Worked Example: A Bounded Random Sign and Scale

Let $\{X_n\}$ be any uniformly integrable family and let $H_n$ be random variables satisfying $|H_n|\le 2$ almost surely for every $n$. The multiplier theorem applies with $C=2$. In particular, for every $K>0$,

\mathbb{E}\bigl[|H_nX_n|\mathbf{1}_{\{|H_nX_n|>K\}}\bigr] \le 2\mathbb{E}\bigl[|X_n|\mathbf{1}_{\{|X_n|>K/2\}}\bigr].

As $K$ grows, $K/2$ grows, and the right side tends to zero uniformly in $n$. Thus $\{H_nX_n\}$ is uniformly integrable, whether $H_n$ is a deterministic scale, a random sign, or a scale depending on $X_n$.

Worked Example: Why the Common Bound Cannot Be Dropped

Let $X_n=1$ and $H_n=n$, regarded as constant random variables. The family $\{X_n\}$ is uniformly integrable: its tail expectation is zero for every threshold $K\ge1$. Each $H_n$ is bounded as a random variable, but there is no single finite bound that works for all $n$.

The products are $H_nX_n=n$. For any finite $K$, choose an integer $n>K$. Then

\mathbb{E}\bigl[|H_nX_n|\mathbf{1}_{\{|H_nX_n|>K\}}\bigr]=n.

As $n$ can be arbitrarily large, the supremum of these tail expectations is infinite for every finite $K$. The products are not uniformly integrable. It is the common bound, not merely boundedness of each multiplier separately, that the theorem requires.

What the Stability Results Do—and Do Not—Say

Both results depend on uniform estimates. In the sum theorem, the number of terms $m$ is fixed. Its proof uses thresholds and moment bounds for each of those finitely many families. It does not establish uniform integrability for sums with an unbounded or varying number of terms.

Similarly, in the multiplier theorem, the constant $C$ must be independent of the index. Individual multipliers may each be bounded while their bounds grow from one index to the next; the preceding example shows that this does not suffice.

Key takeaway: A fixed finite sum preserves uniform integrability, and a family of products preserves it when the multiplier magnitudes share one finite almost-sure bound. In each case, the uniformity of the hypothesis is essential.

Check Your Understanding

Use the tail-expectation definition and the estimates in the proofs to answer these questions.

Why does uniform integrability imply a uniform bound on the first moments?
In the finite-sum proof, why is the event $\{|S_\alpha|>K\}$ contained in the union of the events $\{|X_{j,\alpha}|>K/m\}$?
Where does the proof for bounded multipliers use the common constant $C$?
Can the multiplier theorem require independence between $H_\alpha$ and $X_\alpha$? Explain using its tail estimate.
Why does the family of products in the third worked example fail to be uniformly integrable?