Poisson Counts As A Rare-Event Limit

Why a Poisson Count Can Arise From Many Opportunities

A count of claims may be built from many individual opportunities: each covered item either has a claim during a period or does not. If the opportunities are independent and each has a small event probability, the total count is binomial. A useful limiting model appears when the number of opportunities increases while the expected number of events stays fixed. This tutorial derives that limit and shows exactly what is held constant.

The limit is not the assertion that every finite binomial count is Poisson. Rather, for a fixed expected count, the probability of each fixed count approaches a Poisson probability as the number of independent opportunities grows and each individual probability shrinks.

Definition: For a fixed $\lambda>0$, let $N_n$ be the number of events among $n$ independent opportunities, each with event probability $p_n=\lambda/n$, for integers $n\geq\lceil\lambda\rceil$. Then $N_n\sim\operatorname{Binomial}(n,\lambda/n)$, and $E[N_n]=np_n=\lambda$. The Poisson rare-event limit states that, for each fixed nonnegative integer $k$, $$ \lim_{n\to\infty}P(N_n=k)=e^{-\lambda}\frac{\lambda^k}{k!}. $$ These limiting probabilities are the probability function of a Poisson random variable with mean $\lambda$.

The restriction on $n$ ensures that $p_n\leq1$, as a probability must. Since $\lambda$ is fixed, all sufficiently large integers satisfy this restriction, which is all that is needed for the limit. At every finite $n$, the count is binomial; the Poisson model describes the limiting probabilities.

Deriving the Limiting Probability

For a binomial count, exactly $k$ events means choosing which $k$ of the $n$ opportunities have events, with events on those opportunities and no events on the rest. Thus, when $0\leq k\leq n$, $$ P(N_n=k)=\binom{n}{k}\left(\frac{\lambda}{n}\right)^k \left(1-\frac{\lambda}{n}\right)^{n-k}. $$ We now take the limit for a fixed $k$. The key is to separate factors with simple limits.

Rewrite the choice and event factors.
Using $\binom{n}{k}=n(n-1)\cdots(n-k+1)/k!$, combine it with $(\lambda/n)^k$. This gives $\frac{\lambda^k}{k!}\prod_{j=0}^{k-1}(1-j/n)$.

Separate the no-event factor.
Write $(1-\lambda/n)^{n-k}=(1-\lambda/n)^n(1-\lambda/n)^{-k}$.

Take the limits.
For fixed $k$, the finite product tends to $1$, $(1-\lambda/n)^{-k}$ tends to $1$, and $(1-\lambda/n)^n$ tends to $e^{-\lambda}$.

Together, these steps give the complete limiting calculation:

\begin{aligned} P(N_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 \left(1-\frac{\lambda}{n}\right)^{-k},\\ \lim_{n\to\infty}P(N_n=k) &=\frac{\lambda^k}{k!}\cdot1\cdot e^{-\lambda}\cdot1 =e^{-\lambda}\frac{\lambda^k}{k!}. \end{aligned}

The exponential limit follows, for example, by taking logarithms: $$ n\log\left(1-\frac{\lambda}{n}\right) \longrightarrow -\lambda, $$ because $\log(1-x)/x\to-1$ as $x\to0$. Exponentiating yields $(1-\lambda/n)^n\to e^{-\lambda}$. The derivation applies to each fixed $k$; it does not require $k$ to grow with $n$.

The limiting probabilities sum to one, since $$ \sum_{k=0}^{\infty}e^{-\lambda}\frac{\lambda^k}{k!} =e^{-\lambda}e^\lambda=1. $$ Consequently, they define the Poisson distribution with parameter $\lambda$, not merely a collection of separate pointwise limits. The finite binomial count also has mean $\lambda$ and variance $\lambda(1-\lambda/n)$, so its variance tends to $\lambda$, the variance of the limiting Poisson distribution.

Worked Applications of the Limit

Worked Example: A Fixed Mean Across a Large Portfolio

A hypothetical portfolio has 2,400 independent policy-period opportunities. For each opportunity, the probability of at least one claim in the period is $0.00125$. Use the rare-event limit to estimate the probability of exactly two claims, and check the mean used in the approximation.

Method and assumptions. Treat each opportunity as a Bernoulli trial with the same event probability, and assume independence. The count is binomial with $n=2{,}400$ and $p=0.00125$; the Poisson limit uses $\lambda=np$.

Step 1: Calculate the fixed mean. $$ \lambda=np=2{,}400(0.00125)=3. $$ Thus the corresponding limiting count is Poisson with mean $3$.

Step 2: Use the limiting probability for $k=2$. $$ P(N=2)\approx e^{-3}\frac{3^2}{2!} =e^{-3}\frac{9}{2} =4.5e^{-3} \approx 0.2240. $$ The probability is $0.2240$ to four decimal places.

Check. The input mean is $3$, so the probability uses $e^{-3}$ and $3^2$, not a parameter based on the individual probability alone. The calculation avoids the mistake of using $p=0.00125$ as the Poisson mean; the mean is the expected total count $np$.

Worked Example: Checking the Probability Sequence

For a hypothetical group of independent opportunities, suppose the expected event count is to remain $\lambda=2$ as the number of opportunities changes. Find the individual event probabilities for $n=10$ and $n=100$, then compare the exact and limiting probabilities of no events at $n=10$.

Method and assumptions. For each $n$, use $p_n=\lambda/n$. The trials are independent and have a common probability within each model.

Step 1: Calculate the probabilities and check admissibility. At $n=10$, $p_{10}=2/10=0.2$. At $n=100$, $p_{100}=2/100=0.02$. Both are valid probabilities, and both models have mean $np_n=2$.

Step 2: Calculate the exact probability of no events for $n=10$. $$ P(N_{10}=0)=(1-0.2)^{10}=0.8^{10}=0.1073741824. $$ To four decimal places, this is $0.1074$.

Step 3: Calculate the limiting probability. For a Poisson count with mean $2$, $$ P(N=0)=e^{-2}\frac{2^0}{0!}=e^{-2}\approx0.1353. $$

Check. The exact finite-$n$ result $0.1074$ differs from the limiting value $0.1353$. The theorem describes convergence as $n$ grows; it does not say that the value at $n=10$ must equal its limit. This avoids treating the limiting model as an exact finite binomial identity.

Worked Example: One Event in a Finite Model and in the Limit

A hypothetical system has 20 independent components, each with probability $0.05$ of an event during a specified period. Calculate the exact probability of exactly one event, then calculate the corresponding Poisson-limit probability using the same mean.

Method and assumptions. The exact count is binomial with $n=20$ and $p=0.05$. The rare-event limit keeps the mean $np$ fixed when considering larger $n$ and smaller individual probabilities.

Step 1: Calculate the mean. $$ \lambda=np=20(0.05)=1. $$

Step 2: Calculate the exact finite binomial probability. $$ \begin{aligned} P(N_{20}=1) &=\binom{20}{1}(0.05)(0.95)^{19}\\ &=20(0.05)(0.95)^{19}\\ &=(0.95)^{19}\\ &\approx0.3773536. \end{aligned} $$ Rounded to four decimal places, the exact probability is $0.3774$.

Step 3: Calculate the limiting probability. $$ P(N=1)=e^{-1}\frac{1^1}{1!}=e^{-1}\approx0.3679. $$ The two values differ by approximately $0.3773536-0.3678794=0.0094742$, or $0.0095$ to four decimal places.

Check. The exact calculation uses $n=20$ and $p=0.05$; the limiting calculation uses the mean $\lambda=1$. They are different models, so their probabilities need not match at finite $n$. The approximation is not obtained by replacing the binomial coefficient alone; all factors in the binomial probability contribute to the limit.

Assumptions and Common Exam Traps

The rare-event limit relies on a particular structure. The $n$ opportunities are independent, each has event probability $\lambda/n$, and $\lambda$ is fixed as $n$ increases. Equal probabilities and independence give the binomial model used in the derivation. If opportunities are dependent, or have unequal probabilities without a suitable model, the binomial expression above does not automatically apply. A Poisson count may still be useful in other settings, but that requires a separate justification.

One common trap is to say that the probability $p_n=\lambda/n$ is defined for every positive integer $n$. This can be invalid: if $n<\lambda$, then $\lambda/n>1$. Define the sequence for $n\geq\lceil\lambda\rceil$, or simply for all sufficiently large $n$, so the probabilities are valid. Another trap is to use the individual event probability as the Poisson mean. The parameter is the fixed expected total count $np_n=\lambda$.

A frequent algebra error occurs in the final probability. For exactly $k$ events, the limit is $e^{-\lambda}\lambda^k/k!$. For example, with $\lambda=4$ and $k=2$, it is $$ e^{-4}\frac{4^2}{2!}=8e^{-4}\approx0.1465, $$ not $0.0733$. The division by $2!$ is required, but it does not divide $8e^{-4}$ by another factor of two: $4^2/2!=8$.

The result matters because it explains the mathematical basis for using a Poisson count when many independent opportunities each have a small chance of an event. It also clarifies what such an approximation represents: a limit with a stable total mean, rather than a rule that every large-looking count or every small individual probability automatically justifies a Poisson model. Questions about practical accuracy for finite $n$ require attention to the actual model and are distinct from proving the limit.

Key takeaway: If $N_n\sim\operatorname{Binomial}(n,\lambda/n)$, with fixed $\lambda>0$ and valid probabilities, then for each fixed $k$, $P(N_n=k)\to e^{-\lambda}\lambda^k/k!$. Keep the total mean fixed, preserve the independence assumptions, and distinguish a finite binomial probability from its limit.

Check Your Understanding

For each problem, use the stated independent-trial model and the rare-event limit where requested. Probabilities in the answers are rounded to four decimal places.

A hypothetical set of 800 independent opportunities has event probability $0.0025$ per opportunity. Find the limiting probability of no events. (Answer: $\lambda=2$, $e^{-2}=0.1353$)
A model has 12,000 independent opportunities, each with event probability $0.00025$. Find the limiting probability of exactly one event. (Answer: $\lambda=3$, $3e^{-3}=0.1494$)
For a sequence with fixed mean $\lambda=2$, find $p_n$ when $n=50$, then find the limiting probability of exactly two events. (Answer: $p_{50}=0.04$, $e^{-2}2^2/2!=0.2707$)
Let $\lambda=4.5$. State the smallest integer $n$ for which $p_n=\lambda/n$ is valid, and find the limiting probability of exactly three events. (Answer: $n=5$, $e^{-4.5}4.5^3/3!=0.1687$)
For $n=2{,}000$ independent opportunities with probability $0.0015$, find the binomial mean and variance, then state the limiting Poisson mean and variance. (Answer: binomial mean $3$, binomial variance $2.9955$; limiting mean and variance are both $3$)

What You'll Learn