When The Poisson Approximates The Binomial

Why Approximate a Binomial Count?

A portfolio with a fixed number of independent risks naturally leads to a binomial claim count: each risk either produces a claim or does not. When the portfolio is large and each individual claim probability is small, evaluating binomial probabilities can be less convenient than evaluating Poisson probabilities. The Poisson model can then provide a simpler approximation, but its quality depends on the parameters and on which probability is being estimated.

The key is to match the expected number of claims, not the probability for one risk. This tutorial develops that approximation, compares exact and approximate calculations, and gives ways to judge the size of the error. The discussion is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: If \(N\sim\operatorname{Binomial}(n,p)\), then \(N\) counts claims among \(n\) independent risks, each with the same claim probability \(p\). Its mean is \(np\), and its variance is \(np(1-p)\). The Poisson approximation uses \(Z\sim\operatorname{Poisson}(\lambda)\) with \(\lambda=np\), so that \(E[Z]=np=E[N]\).

This approximation is most plausible when \(n\) is large and \(p\) is small while \(np\) remains of a useful size. It does not require the expected count to be small: the important contrast is between a large number of low-probability opportunities and the resulting expected count. Matching means, however, does not make the two distributions identical. The binomial count cannot exceed \(n\), while the Poisson count has no finite upper bound.

Why the Poisson Formula Emerges

The approximation follows from the binomial probability function. Suppose \(n\) increases while the expected count \(\lambda=np\) stays fixed, so \(p=\lambda/n\). For a fixed nonnegative integer \(k\), the binomial probability can be written as follows:

P(N=k) =\binom{n}{k}\left(\frac{\lambda}{n}\right)^k \left(1-\frac{\lambda}{n}\right)^{n-k} =\frac{\lambda^k}{k!} \left\{\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)\right\} \left(1-\frac{\lambda}{n}\right)^{n-k}.

As \(n\) tends to infinity with \(k\) and \(\lambda\) fixed, each factor in the product tends to 1. Also, \((1-\lambda/n)^n\) tends to \(e^{-\lambda}\), and the remaining factor \((1-\lambda/n)^{-k}\) tends to 1. Therefore:

P(N=k)\longrightarrow e^{-\lambda}\frac{\lambda^k}{k!}.

The expression on the right is the Poisson probability with parameter \(\lambda\). This derivation explains the limiting approximation; in practice \(n\) is finite, so the binomial and Poisson probabilities can differ. A useful first check is to compare the actual \(p\) with the size of the portfolio and then calculate the particular probabilities needed.

Approximation rule: For \(N\sim\operatorname{Binomial}(n,p)\), approximate \(N\) by \(Z\sim\operatorname{Poisson}(\lambda)\) using \(\lambda=np\). A small \(p\) and a reasonably large \(n\) support the approximation, but no single cutoff guarantees accuracy for every probability or tail.

Worked Comparisons

Worked Example: Probabilities at Several Claim Counts

A hypothetical portfolio contains \(n=100\) independent policies. Each policy has probability \(p=0.02\) of producing a claim during the period. Compare the exact binomial probabilities and their Poisson approximations for zero, one, and two claims.

Method and parameter. The exact count is \(N\sim\operatorname{Binomial}(100,0.02)\). Its mean is \(np=100(0.02)=2\), so use \(Z\sim\operatorname{Poisson}(2)\).

For zero claims, the exact probability is \(P(N=0)=(0.98)^{100}=0.1326\) to four decimals. The approximation is \(P(Z=0)=e^{-2}=0.1353\). The absolute difference, using unrounded values, is approximately \(0.002716\).

For one claim, the exact calculation is \(P(N=1)=\binom{100}{1}(0.02)(0.98)^{99}=0.2707\). The Poisson calculation is \(P(Z=1)=e^{-2}(2)=0.2707\). The absolute difference is approximately \(0.000018\).

For two claims, the exact calculation is \(P(N=2)=\binom{100}{2}(0.02)^2(0.98)^{98}=0.2734\). The Poisson calculation is \(P(Z=2)=e^{-2}(2^2/2!)=0.2707\). The absolute difference is approximately \(0.002743\).

Check and trap avoided. The exact and approximate probabilities are close, but they are not interchangeable: in particular, the zero- and two-claim probabilities differ by about \(0.0027\). The Poisson parameter is the portfolio mean, \(2\), not the individual risk probability, \(0.02\).

Worked Example: Comparing a Tail Probability

For a hypothetical group of \(20\) independent insured devices, the probability of a claim on each device is \(0.10\). Find the exact probability of at least three claims and compare it with the Poisson approximation.

Method and parameter. Here \(N\sim\operatorname{Binomial}(20,0.10)\), with mean \(20(0.10)=2\). Approximate with \(Z\sim\operatorname{Poisson}(2)\). For either model, calculate the requested upper tail as one minus the probabilities of zero, one, and two claims.

The exact binomial probabilities are:

\begin{aligned} P(N=0)&=(0.9)^{20}=0.121577,\\ P(N=1)&=\binom{20}{1}(0.1)(0.9)^{19}=0.270170,\\ P(N=2)&=\binom{20}{2}(0.1)^2(0.9)^{18}=0.285180. \end{aligned}

Thus \(P(N\leq 2)=0.121577+0.270170+0.285180=0.676927\), and \(P(N\geq 3)=1-0.676927=0.323073\).

For the Poisson model:

P(Z\geq 3) =1-\left(e^{-2}+2e^{-2}+\frac{2^2}{2!}e^{-2}\right) =1-5e^{-2} =0.323324.

The absolute difference is approximately \(0.000250\). The approximation is close for this tail, although \(p=0.10\) is not especially small. That is a numerical result for this particular event, not a general guarantee that a probability of \(0.10\) is always acceptable.

Check and trap avoided. The tail is at least three, so the complement is the sum through two, not through three. Including \(P(N=3)\) in the complement would calculate a different event.

Worked Example: A Smaller Individual Probability

A hypothetical portfolio has \(400\) independent risks, each with claim probability \(0.005\). Compare the exact and approximate probabilities of at least one claim, and use an error bound as a conservative check.

Method and parameter. The exact count is \(N\sim\operatorname{Binomial}(400,0.005)\). Its mean is \(\lambda=400(0.005)=2\), so the approximation is \(Z\sim\operatorname{Poisson}(2)\).

The exact probability of no claims is \(P(N=0)=(1-0.005)^{400}=0.134658\). Therefore \(P(N\geq1)=1-0.134658=0.865342\). Under the Poisson model, \(P(Z=0)=e^{-2}=0.135335\), so \(P(Z\geq1)=1-e^{-2}=0.864665\). The absolute difference between the two probabilities of at least one claim is approximately \(0.000677\).

For independent risks with individual probabilities \(p_i\), one standard Le Cam bound, using total variation distance defined as the largest absolute difference in probability over events, is \(d_{\mathrm{TV}}\leq\sum_i p_i^2\). In this equal-probability example, the bound is:

d_{\mathrm{TV}}\leq 400(0.005)^2=0.010.

Consequently, the absolute error for any event, including at least one claim, is no more than \(0.010\) under this bound. The actual error, \(0.000677\), is much smaller. The bound is a conservative guarantee, not a prediction of the error for a particular event.

Check and trap avoided. The bound uses \(np^2\), not \(np\). Here the mean \(np=2\) is not an error bound; substituting it for \(np^2=0.010\) would give a bound that is both incorrect and uninformative.

Interpreting Approximation Quality

The probability \(p\) and the mean \(\lambda=np\) play different roles. The mean sets the Poisson distribution and determines its overall count scale. The individual probability \(p\) indicates how close the fixed-number-of-risks setup is to a rare-event setting. For a fixed mean, a larger \(n\) means a smaller \(p\), and generally supports a closer approximation.

The binomial and Poisson variances also help explain why matching the mean does not make the models identical:

\operatorname{Var}(N)=np(1-p)=\lambda(1-p), \qquad \operatorname{Var}(Z)=\lambda.

For \(p>0\), the Poisson variance is larger than the binomial variance with the same mean. The discrepancy in variance is \(\lambda p\), which becomes small when \(p\) is small. This comparison is useful context, but it does not by itself determine the error in a particular tail or point probability.

The finite support also matters. A binomial count is restricted to \(0,1,\ldots,n\), whereas a Poisson count can exceed \(n\). If \(n\) is very small or \(p\) is not small, the Poisson model can assign substantial probability to counts that are impossible under the binomial model. Even when the overall approximation is good, check the event the problem asks for rather than relying only on a rule of thumb.

Common Exam Traps

Using \(p\) as the Poisson parameter. The parameter is the expected count \(\lambda=np\), not the probability for a single risk.
Changing the event when taking a complement. For example, \(P(N\geq3)=1-P(N\leq2)\); subtracting through three instead would be wrong.
Treating a rule of thumb as a guarantee. Small \(p\) supports approximation, but the requested probability, tail, and finite portfolio size still matter.
Assuming equal means imply equal distributions. The binomial variance is \(\lambda(1-p)\), not \(\lambda\), and the two distributions have different supports.
Confusing a bound with an actual error. A total variation bound limits possible error over events; it need not be close to the error for the event being calculated.

Key takeaway: For \(N\sim\operatorname{Binomial}(n,p)\), use \(Z\sim\operatorname{Poisson}(np)\) when \(n\) is large and \(p\) is small. Match the mean, calculate the requested event under both models when accuracy matters, and treat any error bound as a conservative limit rather than an estimated difference.

Check Your Understanding

For each problem, identify the Poisson parameter from the binomial mean before calculating. Probabilities are rounded to four decimals.

A hypothetical set of \(80\) independent risks has claim probability \(0.0125\) per risk. Compare the exact binomial and Poisson probabilities of zero claims. (Answer: \(\lambda=1\); exact \(P(N=0)=0.3656\), Poisson \(P(Z=0)=0.3679\).)
A portfolio has \(50\) independent policies with claim probability \(0.04\) each. Compare the exact binomial and Poisson probabilities of exactly two claims. (Answer: \(\lambda=2\); exact \(P(N=2)\approx0.2762\), Poisson \(P(Z=2)=0.2707\).)
A hypothetical group contains \(12\) independent units, each with claim probability \(0.25\). Compare the probabilities of no claims under the binomial and Poisson models, and report their absolute difference. (Answer: \(\lambda=3\); exact \(0.75^{12}=0.0317\), Poisson \(e^{-3}=0.0498\), absolute difference \(0.0181\).)
A set of \(30\) independent risks has probability \(0.02\) of a claim per risk. Use a Poisson approximation to estimate the probability of exactly one claim. (Answer: \(\lambda=0.6\); \(P(Z=1)=0.6e^{-0.6}=0.3293\).)
A portfolio has \(1{,}000\) independent risks, each with claim probability \(0.001\). Compare the exact and Poisson probabilities of at least one claim. (Answer: \(\lambda=1\); exact \(1-0.999^{1000}\approx0.6323\), Poisson \(1-e^{-1}=0.6321\).)

What You'll Learn