Derivation of the Poisson Distribution as the Limiting Distribution of the Binomial Distribution
Theorem
Let $X_{n} \sim B(n,p)$.
If $\mu \approx np$, then $$ X_{n} \overset{D}{\to} \text{Poi} (\mu) $$
- $B(n,p)$ is the binomial distribution with $n$ trials and probability $p$.
- $\text{Poi} (\lambda)$ is the Poisson distribution with mean and variance $\lambda$.
- $\overset{D}{\to}$ denotes convergence in distribution.
Explanation
Note that the condition $\mu \approx np$ is required here. Since $ np \approx npq$, we have $q = (1-p) \approx 1$, that is, $p \approx 0$. This means that $p$ is very small.
Meanwhile, since $\displaystyle p \approx { {\mu} \over {n} }$, $n$ must be very large. The reason such conditions arise can be easily understood from the fact that the mean and variance of the Poisson distribution are equal.
Proof
Consider the moment generating function $M_{X} (t)$. $$ M_{X} (t) = \left\{ (1-p) + p e^{t} \right\} ^{n} = \left\{ 1 + p (e^{t} - 1 ) \right\} ^{n} $$ Since $\displaystyle p \approx { {\mu} \over {n} } $, $$ M_{X} (t) = \left\{ 1 + { {\mu (e^{t} - 1 )} \over {n} } \right\} ^{n} $$ Therefore $$ \lim_{n \to \infty} M_{X} (t) = e^{ \mu (e^{t} - 1 ) } $$ Since $ e^{ \mu (e^{t} - 1 ) }$ is the moment generating function of $\text{Poi}(\mu)$, $X_{n}$ converges in distribution to $ \text{Poi} (\mu)$.
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