📂Probability Distribution

The Poisson Distribution as a Limiting Distribution of the Binomial Distribution

Theorem

Let’s say $X_{n} \sim B(n,p)$.

If $\mu \approx np$ then $$X_{n} \overset{D}{\to} \text{Poi} (\mu)$$

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• $B(n,p)$ is a binomial distribution with trials $n$ and probability $p$.
• $\text{Poi} (\lambda)$ is a Poisson distribution with mean and variance $\lambda$.
• $\overset{D}{\to}$ means distribution convergence.

Description

Note that the condition $\mu \approx np$ is necessary here. Since $np \approx npq$, it implies $q = (1-p) \approx 1$, i.e., $p \approx 0$. This means that $p$ is very small.

On the other hand, because of $\displaystyle p \approx { {\mu} \over {n} }$, $n$ must be very large. The reason for this condition can be easily understood from the fact that the mean and variance are the same in a Poisson distribution.

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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