Poisson Approximation for Rare Extreme Exceedances

Why Count Rare Exceedances?

In reliability and extreme-value problems, a useful question is often not “What is the largest observation?” but “How many observations exceed a demanding threshold?” For example, a system may be assessed by counting components whose loads exceed a design limit, or a monitoring program may count unusually large measurements in a fixed period. If each observation has only a small chance of exceeding the threshold, the count can often be approximated by a Poisson random variable.

The approximation is especially useful because it turns a collection of small probabilities into a simple summary: their sum. That sum determines the approximate distribution of the count, including the probability of at least one exceedance. Independence is essential to the result below; small marginal probabilities alone do not guarantee a Poisson approximation.

Exceedance Counts

Let \(X_1,\ldots,X_n\) be independent random variables, not necessarily identically distributed, and fix a threshold \(u\). Define the exceedance indicator \(I_i\) and total exceedance count \(N_u\) by

Each \(I_i\) is a Bernoulli random variable with success probability \(p_i\). The count \(N_u\) is therefore a sum of independent Bernoulli variables, also called a Poisson-binomial count when the probabilities differ. The quantity \(\lambda=E[N_u]\) is its expected count. A Poisson approximation uses a Poisson random variable with this same mean.

The defining probability at zero immediately gives \(P(Z\geq 1)=1-e^{-\lambda}\). Thus, if the exceedance count is well approximated by \(\operatorname{Poisson}(\lambda)\), the chance of at least one exceedance is approximately \(1-e^{-\lambda}\). The next result quantifies when that approximation is justified.

A Poisson Approximation and Its Error Bound

For two distributions on the nonnegative integers, their total variation distance is the largest difference they assign to any event. If \(W\) and \(Z\) have such distributions, write

In particular, a bound on this distance controls the error in approximating any probability about the count—not only the probability of zero or of at least one.

Here is a coupling proof of the bound. For each \(i\), let \(Z_i\sim\operatorname{Poisson}(p_i)\). The total variation distance between a Bernoulli\((p_i)\) variable and a Poisson\((p_i)\) variable is

To verify the equality, at zero the Bernoulli probability is \(1-p_i\), which is no greater than \(e^{-p_i}\); at one its probability exceeds the Poisson probability by \(p_i-p_i e^{-p_i}\); and at values at least two only the Poisson distribution has positive mass. The total positive difference is therefore \(p_i(1-e^{-p_i})\). The inequality follows from \(1-e^{-x}\leq x\) for \(x\geq0\).

For discrete distributions, there is a coupling—two variables defined together with the specified marginal distributions—whose probability of disagreement equals their total variation distance. To see the construction, match the common mass at each value, namely the smaller of the two probabilities there. The total matched mass is \(1-d_{\mathrm{TV}}\); the remaining mass can be paired using the unmatched parts of the distributions. Apply this construction separately to each pair \((I_i,Z_i)\), with the pairs independent across \(i\). Then

For any event \(A\), the difference between the probabilities that two coupled counts belong to \(A\) is at most the probability that the counts differ. Finally, independent Poisson variables add: for two such variables with parameters \(a,b\), their sum has probability at \(k\)

Repeated application gives \(\sum_i Z_i\sim\operatorname{Poisson}(\sum_i p_i)\). The coupling inequality now proves the stated bound. \(\square\)

The bound is small when no individual probability is large and the sum of their squares is small. For identically distributed observations with exceedance probability \(p\), it becomes \(np^2\), while the approximating parameter is \(np\). In the common rare-event regime \(p=p_n\to0\) and \(np_n\to\lambda\), the error bound \(np_n^2=(np_n)p_n\) tends to zero. Thus the count converges in distribution to \(\operatorname{Poisson}(\lambda)\).

At Least One Exceedance and the Sample Maximum

If \(X_1,\ldots,X_n\) are independent and \(M_n=\max_i X_i\), then the event \(M_n\leq u\) is exactly the event of zero exceedances. Independence gives an exact expression, even before approximation:

The Poisson approximation replaces this product by \(e^{-\lambda}\). Hence

For identically distributed observations, \(p_i=p\) and \(\lambda=np\). If thresholds \(u_n\) are chosen so that \(nP(X_1>u_n)\to\tau\), with \(0\leq\tau<\infty\), then the error bound tends to zero and

This connects rare exceedance counts to the limiting behavior of maxima: the probability of no exceedance is the probability that the maximum stays at or below the threshold. The count approximation supplies a direct route to that probability when the expected number of exceedances stays on a finite scale.

Common Pitfalls and Practical Interpretation

A common mistake is to use \(1-e^{-\lambda}\) merely because each exceedance probability is small. Small individual probabilities are not sufficient if dependence is strong. For example, if all indicators are identical, each can have a small marginal probability while the count is either zero or a large block of exceedances; that count need not resemble a Poisson variable. The theorem requires independence. With dependent observations, a separate approximation argument that accounts for their dependence is needed.

A second pitfall is to confuse the mean count with the probability of at least one exceedance. The mean is \(\lambda\), whereas the Poisson approximation to that probability is \(1-e^{-\lambda}\). These values are close only when \(\lambda\) is itself small. For instance, at \(\lambda=0.5\), the mean count is \(0.5\), but the chance of at least one exceedance is about \(0.3935\).

Finally, the total variation bound is an absolute error bound. If the probability being estimated is extremely close to zero, an error that is small in absolute terms may still be large relative to that probability. In that setting, check whether the bound is adequate for the intended use rather than interpreting “small error” as a guarantee of small relative error.