Deriving the Poisson Limit of the Binomial

Why a Limit Connects Binomial and Poisson Models

A binomial model counts successes in a fixed number of independent trials, each with the same success probability. A Poisson model describes a count whose mean is fixed. The connection between them becomes precise when there are more and more trials, each with a smaller chance of success, while the expected number of successes stays constant. This tutorial derives that connection for the probability of every fixed count.

For example, imagine a hypothetical process with a large number of opportunities for an event, where each opportunity has a small chance of producing it. Counting events across those opportunities gives a binomial model. If the number of opportunities grows while the overall expected count remains unchanged, the binomial probabilities approach Poisson probabilities.

The success probability \(\lambda/n\) is valid when \(n\geq\lambda\), because it must lie between zero and one. In each binomial model, the expected count is \(np_n=\lambda\). The integer \(k\), by contrast, is held fixed as the number of trials grows: the derivation asks for the limiting probability of observing, for example, exactly two successes, not a count that also increases with \(n\).

Deriving the Probability of Exactly \(k\) Successes

For a binomial random variable with \(n\) trials and success probability \(p_n\), the probability of exactly \(k\) successes is \(\binom{n}{k}p_n^k(1-p_n)^{n-k}\), for \(k=0,1,\ldots,n\). In the sequence above, substitute \(p_n=\lambda/n\). For any fixed \(k\), this gives

Rewrite the binomial coefficient using the falling product \(n(n-1)\cdots(n-k+1)\), and separate the factors in the last power:

This expression has three factors whose limits can be evaluated separately. First, \(\frac{n(n-1)\cdots(n-k+1)}{n^k} =\prod_{j=0}^{k-1}(1-j/n)\). Since \(k\) is fixed, this is a product of a fixed number of terms, and each term approaches one. The product therefore approaches one. When \(k=0\), the product is empty and is defined to equal one, which gives the same conclusion.

For the second factor, use the standard logarithm limit \(\log(1-u)/u\to-1\) as \(u\to0\). Taking \(u=\lambda/n\), which approaches zero as \(n\) grows, gives

Exponentiating both sides shows that \((1-\lambda/n)^n\to e^{-\lambda}\). Finally, because \(k\) is fixed, \((1-\lambda/n)^{-k}\to1\). Combining these three limits with the constant factor \(\lambda^k/k!\) yields the result.

The right-hand side is the probability mass function of a Poisson random variable with parameter \(\lambda\). The fixed product \(np_n=\lambda\) thus becomes the Poisson parameter: it is the mean count of every binomial model in the sequence and the parameter in the limiting Poisson distribution.

Why the Limiting Probabilities Form a Distribution

Showing the limit for each fixed count gives the candidate probability at that count. We can also verify that these candidate probabilities add to one, as probabilities over all possible counts must. The exponential series states that \(e^\lambda=\sum_{k=0}^{\infty}\lambda^k/k!\). Therefore,

This also lets us describe the convergence of the count distributions, not only of one probability at a time. For any fixed nonnegative integer \(m\), the event \(X_n\leq m\) is the union of the finitely many events \(X_n=k\), for \(k=0,\ldots,m\). Adding their probabilities and taking limits gives

Thus the cumulative probabilities at each integer count approach those of a Poisson random variable with parameter \(\lambda\). The limiting probability mass function sums to one, so it describes a proper count distribution rather than losing probability as the trial count grows.

Worked Examples

Assumptions and the Scope of the Limit

The derivation is a statement about a particular sequence of probability models. In every model, the trials are independent, the number of trials is fixed at \(n\), and the success probability is the same for all trials, \(p_n=\lambda/n\). The parameter \(\lambda\) is held fixed as \(n\) increases. These conditions are what make the binomial probability formula apply and make the limiting calculation work.

The result is about probabilities for fixed counts. It does not say that a binomial model with any large trial count is automatically Poisson, or that the binomial and Poisson models have identical probabilities at finite \(n\). Rather, it establishes why the Poisson mass function arises from this specific limiting pattern.

Common Mistake

A common error is to let \(n\) grow while keeping the success probability \(p\) fixed, and then claim that the same Poisson limit follows with \(\lambda=np\). If \(p>0\) is fixed, however, \(np\) grows without bound; there is no fixed \(\lambda\) and the sequence does not meet the conditions used in the derivation. The limit instead requires \(p_n=\lambda/n\), so the success probability shrinks as the trial count increases.

Another tempting shortcut is to replace the binomial formula with the Poisson formula without showing how the terms behave. The derivation matters: it shows that the combinatorial factor tends to one after the powers of \(n\) are separated, the term \((1-\lambda/n)^n\) tends to \(e^{-\lambda}\), and the remaining fixed-power term tends to one. Those limits explain the form of the Poisson probability mass function.