Diagnosing Uniform Integrability in Rare-Event Models

Rare Events and Large Values

A random variable can be zero with high probability and still have a substantial expectation. The reason is that its nonzero values may be large enough to compensate for the small probability of the event on which they occur. This simple competition is central to diagnosing uniform integrability in rare-event models.

We study nonnegative random variables of the form $X_n=a_n\mathbf{1}_{A_n}$, where $a_n$ is a finite nonnegative number and $A_n$ is an event. Write $p_n=\mathbb{P}(A_n)$. The value of $X_n$ is $a_n$ on $A_n$ and zero outside it. The probability spaces may vary with $n$; all the criteria below use only $a_n$ and $p_n$.

Definition: A family $(X_n)$ of integrable random variables is uniformly integrable if its tail expectations vanish uniformly: for every positive tolerance, there is a finite cutoff such that every variable in the family has tail expectation below that tolerance. For the nonnegative variables considered here, this means $$ \lim_{K\to\infty}\sup_{n\in\mathbb{N}}\mathbb{E}\!\left[X_n\mathbf{1}_{\{X_n>K\}}\right]=0. $$

Three Quantities to Keep Separate

For these variables, the event probability, the amplitude, and their product play different roles. Convergence in probability to zero is determined by whether values above each fixed positive threshold occur with probabilities tending to zero. The first moment is determined by the product $a_np_n$. Uniform integrability asks whether the contribution from values above a threshold can be made small uniformly over the entire sequence.

The first moment has an immediate exact formula:

\mathbb{E}[X_n] =a_n\mathbb{P}(A_n) =a_np_n.

Indeed, $X_n$ is $a_n$ on an event of probability $p_n$ and zero otherwise. In contrast, for any fixed $K\geq 0$, its tail expectation is

\mathbb{E}\!\left[X_n\mathbf{1}_{\{X_n>K\}}\right] =a_np_n\mathbf{1}_{\{a_n>K\}}.

If $a_n\leq K$, the tail event is empty. If $a_n>K$, the tail event is exactly $A_n$, so its contribution is $a_np_n$. Thus uniform integrability depends not just on whether the means are small, but on whether mean contribution persists at amplitudes that escape to arbitrarily high levels.

A Criterion for Convergence in Probability

Theorem: For $X_n=a_n\mathbf{1}_{A_n}$ with $a_n\geq 0$, convergence in probability to zero holds if and only if, for every $\varepsilon>0$, $$ p_n\mathbf{1}_{\{a_n>\varepsilon\}}\longrightarrow 0. $$

Proof: Fix $\varepsilon>0$. If $a_n\leq\varepsilon$, then $X_n$ never exceeds $\varepsilon$, and $\mathbb{P}(X_n>\varepsilon)=0$. If $a_n>\varepsilon$, then $X_n>\varepsilon$ precisely on $A_n$, so $\mathbb{P}(X_n>\varepsilon)=p_n$. Therefore, for every $n$, $$ \mathbb{P}(|X_n|>\varepsilon) =p_n\mathbf{1}_{\{a_n>\varepsilon\}}. $$ By the definition of convergence in probability to zero, the left-hand side tends to zero for every $\varepsilon>0$. This is exactly the stated condition, in both directions. $\square$

A useful sufficient condition follows immediately: if $p_n\to 0$, then $X_n\to 0$ in probability, no matter how quickly $a_n$ grows. The converse need not hold. If the amplitudes eventually fall below every fixed positive threshold, convergence in probability can occur even when the event probabilities do not tend to zero.

A Criterion for Uniform Integrability

Theorem: The family $(X_n)$ is uniformly integrable if and only if $$ \lim_{K\to\infty}\sup_{n\in\mathbb{N}} a_np_n\mathbf{1}_{\{a_n>K\}}=0. $$

Proof: For every $n$ and $K$, the tail expectation is exactly $$ \mathbb{E}\!\left[X_n\mathbf{1}_{\{X_n>K\}}\right] =a_np_n\mathbf{1}_{\{a_n>K\}}. $$ Taking the supremum over $n$ gives the same expression that appears in the definition of uniform integrability. Its limit as $K$ tends to infinity is zero exactly when the family is uniformly integrable. $\square$

This criterion is especially simple when $a_n\to\infty$. In that case, uniform integrability is equivalent to $a_np_n\to 0$. To see why, if the family is uniformly integrable, then for every fixed $K$, all sufficiently large $n$ have $a_n>K$, and hence their means $a_np_n$ are bounded by the tail supremum at $K$. Taking $K$ arbitrarily large forces those means to tend to zero. Conversely, if $a_np_n\to 0$ and $a_n\to\infty$, then after any sufficiently large index the tail contributions are among the terms of a sequence tending to zero; the finitely many earlier amplitudes are eventually below the cutoff and contribute nothing.

Worked Examples: Same Probability, Different Amplitudes

Worked Example: A Persistent Mean from Rare Spikes

Let $p_n=1/n$, $a_n=n$, and $X_n=n\mathbf{1}_{A_n}$, where $\mathbb{P}(A_n)=1/n$. For any fixed $\varepsilon>0$, once $n>\varepsilon$, $$ \mathbb{P}(X_n>\varepsilon)=\frac{1}{n}\longrightarrow 0. $$ Thus $X_n\to 0$ in probability. But its first moment is $$ \mathbb{E}[X_n]=n\cdot\frac{1}{n}=1 $$ for every $n$, so the expectations do not tend to zero. For any $K\geq 0$, choose an integer $n>K$. Then $a_n>K$ and the tail expectation at that index is $$ a_np_n=n\cdot\frac{1}{n}=1. $$ Every tail supremum is therefore at least $1$, and the family is not uniformly integrable. Here a vanishing chance of a spike does not prevent its mean contribution from persisting.

Worked Example: A Smaller Spike with the Same Event Probabilities

Keep $p_n=1/n$, but take $a_n=\sqrt{n}$. For fixed $\varepsilon>0$, whenever $n>\varepsilon^2$, the amplitude exceeds $\varepsilon$, and $$ \mathbb{P}(X_n>\varepsilon)=\frac{1}{n}\longrightarrow 0. $$ The first moment is now $$ \mathbb{E}[X_n]=\sqrt{n}\cdot\frac{1}{n} =\frac{1}{\sqrt{n}}\longrightarrow 0. $$ For a cutoff $K$, only indices with $n>K^2$ can contribute to the tail. At each such index the contribution is $1/\sqrt{n}$, so $$ \sup_{n\in\mathbb{N}}\mathbb{E}\!\left[X_n\mathbf{1}_{\{X_n>K\}}\right] =\sup_{n>K^2}\frac{1}{\sqrt{n}} \leq\frac{1}{K}. $$ For $K>0$, this upper bound tends to zero as $K\to\infty$. The family is uniformly integrable. The same event probabilities as in the preceding example, combined with a slower-growing amplitude, produce a different conclusion for both the mean and uniform integrability.

Worked Examples: Changing the Event Probabilities

Worked Example: Rare Events with Vanishing Mean

Take $p_n=1/n^2$ and $a_n=n$. For fixed $\varepsilon>0$, when $n>\varepsilon$, $$ \mathbb{P}(X_n>\varepsilon)=\frac{1}{n^2}\longrightarrow 0, $$ so $X_n\to 0$ in probability. Its mean is $$ \mathbb{E}[X_n]=n\cdot\frac{1}{n^2}=\frac{1}{n}\longrightarrow 0. $$ For $K\geq 0$, the tail contribution is $1/n$ when $n>K$, and zero otherwise. Thus $$ \sup_{n\in\mathbb{N}}\mathbb{E}\!\left[X_n\mathbf{1}_{\{X_n>K\}}\right] =\sup_{n>K}\frac{1}{n} \leq\frac{1}{\lfloor K\rfloor+1}. $$ The right-hand side tends to zero, so the family is uniformly integrable. Reducing the event probabilities from $1/n$ to $1/n^2$, while retaining amplitude $n$, changes the means from a constant to a sequence tending to zero.

Worked Example: A Larger Amplitude Can Restore a Persistent Mean

Let $p_n=1/n^2$ and $a_n=n^2$. For every fixed $\varepsilon>0$, once $n^2>\varepsilon$, $$ \mathbb{P}(X_n>\varepsilon)=\frac{1}{n^2}\longrightarrow 0. $$ Yet $$ \mathbb{E}[X_n]=n^2\cdot\frac{1}{n^2}=1. $$ For any cutoff $K$, there is an integer $n$ with $n^2>K$, and at that index the tail expectation is $1$. Therefore the family is not uniformly integrable. The event probabilities still vanish, but the larger amplitude exactly offsets their decrease.

How to Read the Diagnostics

These calculations illustrate why the three questions should be answered separately. Convergence in probability tests the frequency of exceeding each fixed threshold. The first moment tests the product of amplitude and event probability. Uniform integrability tests whether that product remains significant among amplitudes above increasingly large cutoffs.

For example, a sequence can converge in probability to zero while its expectations stay equal to one, as in the first and fourth examples. That does not contradict the earlier material on Vitali’s Convergence Theorem: the relevant additional condition, uniform integrability, fails in those examples. When both convergence in probability and uniform integrability hold, Vitali’s theorem gives convergence of expectations, provided the limit is integrable.

A common pitfall is to infer too much from $p_n\to 0$. This condition ensures convergence in probability for the rare-event variables, but it says nothing by itself about the product $a_np_n$. Another pitfall is to inspect only the limiting means: even if the means are bounded, uniform integrability still requires that tail contributions vanish uniformly. The exact tail formula makes that final check explicit.

Check Your Understanding

Use the formulas in this tutorial to answer the following questions.

For $X_n=a_n\mathbf{1}_{A_n}$, what condition characterizes convergence in probability to zero?
If $p_n=1/n$ and $a_n=n$, why does convergence in probability not imply convergence of expectations?
Write the tail expectation at cutoff $K$ in terms of $a_n$, $p_n$, and $K$.
For $p_n=1/n^2$ and $a_n=n$, what are the first moment and the uniform-integrability conclusion?
When $a_n\to\infty$, what condition on $a_np_n$ characterizes uniform integrability?