Why the Rare-Event Limit Holds np Fixed

Why Keep the Expected Count Fixed?

A Poisson model describes a count that can arise from many opportunities, each with a small chance of producing an event. The key limiting idea is not simply to make the number of opportunities large. As the number grows, the chance of an event on each opportunity must shrink in a coordinated way. The product of those two quantities—the expected total count—stays near a finite value.

For example, a hypothetical portfolio might contain many independent policies, each with a small probability of a specified claim during a year. If the number of policies increases while the individual claim probability decreases, the expected number of claims can remain constant. The Poisson limit formalizes this balance. It is a statement about a sequence of distributions, not a claim that any particular observed sequence of counts must settle down.

Definition: Suppose $X_n$ counts successes in $n$ independent trials, each with success probability $p_n$. Then $X_n\sim\operatorname{Bin}(n,p_n)$, and its expected count is $E[X_n]=np_n$. The rare-event limit holds when $n$ grows without bound, $p_n$ tends to zero, and $np_n$ tends to a finite constant $\lambda$. For $\lambda>0$, the limiting distribution is $\operatorname{Poisson}(\lambda)$, with probability mass function $P(Y=k)=e^{-\lambda}\lambda^k/k!$ for $k=0,1,2,\ldots$.

The finite constant $\lambda$ is the limiting expected count. In the especially clear case $p_n=\lambda/n$, the expected count is exactly $\lambda$ for every eligible $n$, and the success probability decreases as the number of trials increases. The formula $p_n=\lambda/n$ is a valid probability when $n\geq\lambda$. More generally, exact equality is not needed: it is sufficient that $np_n$ approaches $\lambda$.

The word rare refers to an individual trial: $p_n$ tends to zero. It does not mean that the total number of successes must tend to zero. Since the number of trials grows at the same time, a finite positive expected count can remain. Holding $np_n$ fixed is what prevents the total expected count from either vanishing or growing without bound in this limiting argument.

Deriving the Limit from the Binomial Model

Assume that for each $n$, the $n$ trials are independent and have the same success probability $p_n$. These assumptions give the binomial probability formula. First take $p_n=\lambda/n$, where $\lambda>0$ is fixed, and consider any fixed nonnegative integer $k$. For $n\geq k$,

P(X_n=k) =\binom{n}{k}\left(\frac{\lambda}{n}\right)^k \left(1-\frac{\lambda}{n}\right)^{n-k}.

Separate the factors that depend on $k$ from the factor that captures the large-$n$ behavior:

P(X_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}.

For fixed $k$, each factor in the finite product tends to 1, and $(1-\lambda/n)^{-k}$ tends to 1. The standard exponential limit gives $(1-\lambda/n)^n\to e^{-\lambda}$. Consequently, the probability of each fixed count $k$ tends to $e^{-\lambda}\lambda^k/k!$, the Poisson probability at $k$.

This calculation also explains why the product $np_n$, rather than $n$ or $p_n$ by itself, determines the limiting rate. With $p_n=\lambda/n$, the term contributed by the $k$ required successes approaches $\lambda^k/k!$, while the probability of no success across the large collection of remaining trials contributes the factor $e^{-\lambda}$.

Justification for the general form: If $np_n\to\lambda$ with $0<\lambda<\infty$, then $p_n\to0$. For fixed $k$, write the binomial probability as $$ P(X_n=k)=\frac{(np_n)^k}{k!} \left[\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)\right] (1-p_n)^n(1-p_n)^{-k}. $$ The finite product and $(1-p_n)^{-k}$ tend to 1. Also, $n\log(1-p_n)\to-\lambda$, so $(1-p_n)^n\to e^{-\lambda}$. Thus $P(X_n=k)\to e^{-\lambda}\lambda^k/k!$. These limiting probabilities sum to 1, since $e^{-\lambda}\sum_{k=0}^{\infty}\lambda^k/k!=1$; hence they define the Poisson limiting distribution.

This is convergence in distribution: probabilities for fixed count values, and therefore the resulting discrete distributions, approach those of the Poisson model. It does not assert almost-sure convergence of the counts along sample paths. A sequence of marginal binomial models does not by itself specify how the random variables are coupled on one probability space, so a sample-path claim would require additional assumptions.

Worked Examples: Maintaining a Finite Rate

Worked Example: A Growing Book of Hypothetical Policies

A hypothetical insurer models whether each policy has at least one specified type of claim during a year. Assume policy outcomes are independent and have the same claim probability within each portfolio. A portfolio has 800 policies with claim probability $0.0025$. Find the expected claim count, then identify probabilities for two larger portfolios that preserve that expected count. Finally, give the limiting Poisson probability of exactly one claim.

Method. Calculate the binomial mean $np$, preserve it as the trial count changes, and use the Poisson probability mass function for the limiting distribution.

For the original portfolio, $n=800$ and $p=0.0025$, so

$$ np=800(0.0025)=2. $$

The expected count is 2. For 1,600 policies, the success probability needed to keep the product equal to 2 is $p=2/1600=0.00125$. For 4,000 policies, it is $p=2/4000=0.0005$. The products verify the preservation:

1600(0.00125)=2,\qquad 4000(0.0005)=2.

In the limiting regime, $\lambda=2$. Thus the probability of exactly one claim in the Poisson limiting distribution is

P(Y=1)=e^{-2}\frac{2^1}{1!}=2e^{-2} =2(0.1353353)=0.2706706.

Rounded to four decimal places, the limiting probability is 0.2707. The expected count is 2 in each of the three binomial setups, and the per-policy probability decreases as the portfolio grows. The common mistake is to increase the number of policies while leaving the individual probability unchanged; that would change the expected count rather than preserve it.

Worked Example: Matching a Target Count Across Exposure Sizes

A hypothetical monitoring program treats each inspected component as an independent trial, with the same probability of detecting a specified defect for all components in a given inspection plan. The target expected number of detections is 4.5. Find the per-component probability for plans with 1,500, 3,000, and 9,000 components. Then find the limiting Poisson probability of zero detections.

Method. Set $np=4.5$, solve for $p$ in each plan, and apply the zero-count probability $e^{-\lambda}$ in the limiting Poisson model.

For 1,500 components,

p=\frac{4.5}{1500}=0.003.

For 3,000 components,

p=\frac{4.5}{3000}=0.0015.

For 9,000 components,

p=\frac{4.5}{9000}=0.0005.

The expected-count checks are $1500(0.003)=4.5$, $3000(0.0015)=4.5$, and $9000(0.0005)=4.5$. As the number of independent opportunities increases, the individual detection probability decreases, while the product remains at the target.

The limiting rate is $\lambda=4.5$. Therefore,

P(Y=0)=e^{-4.5}=0.0111089965\ldots

Rounded to four decimal places, the limiting zero-detection probability is 0.0111. The common mistake is to use the probability for one component as though it were the expected count. The expected count is the probability multiplied by the number of independent components.

Worked Example: Why Increasing the Trial Count Alone Is Not Enough

A hypothetical screening design has 400 independent opportunities, each with success probability $0.005$. A revised design has 1,600 opportunities but initially keeps the same success probability. Calculate both expected counts. Then adjust the revised probability to preserve the original expected count and find the limiting Poisson probability of exactly three successes under that preserved-rate regime.

Method. Compare the products $np$, then use $p=\lambda/n$ for the revised design and the Poisson mass function for the limiting probability.

In the original design,

$$ 400(0.005)=2. $$

If the revised design keeps $p=0.005$, its expected count is

$$ 1600(0.005)=8. $$

Thus increasing $n$ alone does not preserve the expected count. To keep the original rate $\lambda=2$, the revised success probability must instead be

p=\frac{2}{1600}=0.00125, \qquad 1600(0.00125)=2.

For the limiting Poisson distribution with $\lambda=2$, the probability of exactly three successes is

P(Y=3)=e^{-2}\frac{2^3}{3!} =e^{-2}\frac{8}{6} =0.1353353(1.3333333\ldots) =0.1804471\ldots

Rounded to four decimal places, this probability is 0.1804. The check is that the adjusted design again has expected count 2. The common mistake is to call any large-$n$ binomial model a finite-rate rare-event limit; if $p$ remains fixed and positive, $np$ grows without bound instead.

Reading the Limiting Regime Correctly

The three conditions in the limit work together. The number of trials grows, the probability of success on a single trial shrinks, and their product approaches a finite rate. Each condition has a role:

If $n$ grows while $p$ stays fixed and positive, then $np$ grows without bound. That is not a finite-$\lambda$ Poisson limit.
If $p$ shrinks so quickly that $np\to0$, the limiting count is zero with probability 1; this is the degenerate $\operatorname{Poisson}(0)$ case.
If $np\to\lambda$ for a finite positive $\lambda$, the individual successes become rare while the total expected count remains at a meaningful scale.

The expected count is central, but it is not a claim that the count will equal its expectation. It is a parameter that sets the scale of the limiting distribution. For the Poisson model, both the mean and variance equal $\lambda$; the rare-event limit preserves the finite mean scale as the binomial structure changes. The earlier material on binomial counts supplies the independent-trial model, while this limiting argument explains the particular way its parameters must vary.

Key takeaway: A nondegenerate Poisson rare-event limit requires $n\to\infty$, $p_n\to0$, and $np_n\to\lambda$ for some finite $\lambda>0$. Holding $np_n$ fixed, or making it approach a finite value, preserves the expected-count scale while individual opportunities become increasingly rare.

Check Your Understanding

For each problem, identify how the expected count is preserved or changed, and use the limiting Poisson model only where requested.

A hypothetical inspection plan has 2,400 independent items, each with probability $0.00125$ of a specified finding. Calculate the expected count. If a sequence of such plans increases the item count while preserving this rate, what is the limiting Poisson parameter? (Answer: Expected count and limiting parameter $\lambda=3$.)
A sequence has $n=2,000$ trials and $p=0.0025$, with $np$ held fixed as $n$ grows. Find $\lambda$ and the limiting probability of exactly two successes, rounded to four decimals. (Answer: $\lambda=5$, $P(Y=2)=e^{-5}5^2/2=0.0842$.)
A hypothetical event process has 500 independent opportunities with probability $0.004$ per opportunity. If the number of opportunities is tripled, what probability preserves the original expected count? Verify the new expected count. (Answer: Original mean $500(0.004)=2$; use $p=2/1500=1/750\approx0.0013333$, giving mean 2.)
A model has 1,000 independent trials with success probability $0.003$. Find the probability for 4,000 trials that preserves the mean, and state the finite Poisson rate for a sequence with this preserved mean. (Answer: Mean $3$; use $p=3/4000=0.00075$; limiting rate $\lambda=3$.)
For a sequence with $p_n=6/n$, calculate the probabilities for $n=3,000$, $6,000$, and $12,000$, then find the limiting probability of zero successes to four decimals. (Answer: The probabilities are $0.002$, $0.001$, and $0.0005$; $np_n=6$ each time, so $\lambda=6$ and $P(Y=0)=e^{-6}=0.0025$.)

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