Poisson Approximation for Rare Binomial Events

Why Rare Binomial Events Lead to a Poisson Law

Suppose a study follows many independent opportunities for an event, with the same small chance of success on each opportunity. A count of successes has a binomial distribution. When the number of opportunities is large and the success probability is small, exact binomial calculations can be cumbersome, while a Poisson distribution with a suitable rate can provide a close approximation.

The key quantity is the expected count, \(np\). The approximation is most useful when \(n\) is large, \(p\) is small, and \(np\) remains of moderate size. We will derive the limiting distribution under this regime, then use it to approximate probabilities and inspect the accuracy in concrete cases.

The rate is chosen to match the binomial mean: \(E[X]=np\). This is not an arbitrary substitution. The limiting argument below shows that, when \(np_n\) approaches a finite value, every fixed binomial count has a probability approaching the corresponding Poisson probability. The case \(\lambda=0\) is included, although it describes counts that are almost surely zero in the limit.

Deriving the Limiting Probabilities

For \(X_n\sim\operatorname{Bin}(n,p_n)\), the probability of exactly \(k\) successes is \(P(X_n=k)=\binom{n}{k}p_n^k(1-p_n)^{n-k}\), for \(k=0,\ldots,n\). The following result gives the precise limiting statement.

Proof. First suppose \(\lambda>0\). Since \(np_n\to\lambda\), for all sufficiently large \(n\), \(p_n>0\), and \(p_n\to0\). Fix an integer \(k\geq0\). For \(n\geq k\), rewrite the binomial probability as

For \(k=0\), the product is an empty product and equals one. For any fixed \(k\), the product tends to one, and \((1-p_n)^{-k}\to1\). To evaluate the remaining power, use the standard limit \(\log(1-u)/u\to-1\) as \(u\downarrow0\):

Thus \((1-p_n)^n\to e^{-\lambda}\). Substituting these limits in the factorization gives \(P(X_n=k)\to e^{-\lambda}\lambda^k/k!\).

Now suppose \(\lambda=0\). For \(k=0\), if \(p_n>0\), the same logarithm limit gives \(n\log(1-p_n)\to0\), because \(np_n\to0\); hence \((1-p_n)^n\to1\). If \(p_n=0\), then \(P(X_n=0)=1\) directly. These two cases together show \(P(X_n=0)\to1\). For each fixed \(k\geq1\), the union bound gives \(P(X_n\geq1)\leq np_n\to0\), so \(P(X_n=k)\to0\). This is the probability mass function of \(\operatorname{Pois}(0)\).

In either case, the probabilities at every fixed integer converge to the stated Poisson probabilities. At any continuity point \(x\) of the Poisson distribution function, the event \(X_n\leq x\) is either impossible if \(x<0\), or is a finite sum of probabilities \(P(X_n=k)\) if \(x\geq0\) is not an integer. Those finite sums converge to the corresponding Poisson probabilities. Therefore the distribution functions converge at every continuity point, which proves \(X_n\xrightarrow{d}X\). \(\square\)

The proof explains why the approximation works. The factor \((1-p_n)^n\) approaches the exponential term \(e^{-\lambda}\), while the combinatorial factor for any fixed count becomes the corresponding \(\lambda^k/k!\). The argument establishes convergence for fixed counts; it does not say that a particular finite \(n\) gives a specified error bound.

Using the Approximation in Calculations

For a finite binomial model, set \(\lambda=np\) and calculate the requested probability under \(X\sim\operatorname{Pois}(\lambda)\). For a point probability, the replacement is \(P(X=k)\approx e^{-np}(np)^k/k!\). For an event involving several fixed counts, add the corresponding Poisson probabilities. The examples show both the calculation and a direct comparison with the binomial answer.

What the Approximation Does—and Does Not—Say

For any fixed nonnegative integer \(m\), the theorem also gives a direct approximation for the probability of a bounded range of counts:

Indeed, the event \(X_n\leq m\) is the disjoint union of the \(m+1\) events \(X_n=k\), for \(k=0,\ldots,m\). Therefore

because a finite sum of convergent terms converges to the sum of their limits. This justifies approximating fixed lower-tail probabilities by adding Poisson masses. For a given finite model, however, the result alone does not guarantee a particular numerical accuracy. If accuracy matters, compare the approximation with an exact calculation or use an error bound appropriate to the situation.

The approximation is motivated by rare independent events, not merely by a large sample size. If \(p\) is not small, the shape of the binomial distribution need not be close to the Poisson shape even when \(n\) is large. Likewise, \(np\) determines the candidate Poisson rate, but matching the mean by itself is not proof that a finite-sample approximation is accurate.

Common Mistake

• Wrong reasoning: “The number of trials is large, so a Poisson approximation must be good.” A large \(n\) alone does not establish the rare-event regime.
• Correction: Check that \(p\) is small and use \(\lambda=np\) as the Poisson rate. The limiting theorem requires \(p_n\to0\) and \(np_n\to\lambda<\infty\). For a finite \(n\), these conditions motivate the approximation but do not supply a universal error guarantee.
• Another edge-case trap: Taking logarithms of \(1-p_n\) in a proof that allows \(p_n=0\) requires separate attention. At \(p_n=0\), the binomial count is exactly zero; the logarithm argument applies only when \(p_n>0\).