Let \(X \sim Poisson(\lambda)\). Its probability mass function is:
\[\mathbb{P}[X = k] = \frac{e^{-\lambda} \lambda^k}{k!}\]Suppose have a Binomial random variable \(X \sim Bin(N, p)\). We know that the expected value of \(X\) is \(Np\); for reasons that will soon become apparent, let’s call this variable \(\lambda := Np\). If we reparameterize the Binomial distribution as \(Bin(N, \frac{\lambda}{N})\) and take the limit as \(N\rightarrow \infty\), this Binomial variable will become \(Poisson(\lambda)\). Intuitively, this says that if we have an infinite number of trials, where each trial is infinitely unlikely to succeed, the number of success trials will be Poisson.
Proposition: Let \(X \sim Bin(N, p)\). Define \(\lambda := Np\) and reparameterize as \(X \sim Bin(N, \frac{\lambda}{N})\). As \(N \rightarrow \infty, X \rightarrow Poisson(\lambda)\)
Proof:
\[\begin{align*} P(X=k) &:= \begin{pmatrix} N \\k \end{pmatrix} p^{k} (1-p)^{N-k}\\ &= \frac{N!}{(N-k)! k!} \Big(\frac{\lambda}{N} \Big)^{k} \Big(1 - \frac{\lambda}{N} \Big)^{N-k}\\ &= \frac{O(N^k)}{k!} \frac{\lambda^k}{O(N^k)} \Big(1 - \frac{\lambda}{N} \Big)^{N-k} \end{align*}\]We need to recall two facts:
Taking the limit as \(N\rightarrow \infity\), we have
\[\begin{align*} P(X=k) &=^{N \rightarrow \infty} \frac{1}{k!} \lambda^k e^{-\lambda}\\ &=_D Poisson(\lambda) \end{align*}\]An alternative way to define the Poisson distribution is as:
\[\lim_{n \rightarrow \infty} Bin(\frac{\lambda}{n}, n)\]Why is \({n \choose k} (\lambda/n)^{k} ( 1 - \lambda/n)^{n-k}\) the same as \(e^{-\lambda}\lambda^k/k!\)?
Reorganizing the first term, per Stirling, \(\{n \choose k} \approx (en/k)^k\), and \((1-\lambda_n)^{n-k} a\approx e^{-\lambda / n}\). Then \((e/k)^k \lambda^k e^{-\lambda(1-k/n)}\) can be rewritten as \((1/k!)\lambda^k e^{-\lambda}\), giving us the PMF for the Poisson.
LeCam’s Theorem states that the sum of independent but not-necessarily-identically distributed Bernoulli random variables