The Poisson distribution Po(X=x|\mu) is
Po(X=x|\mu) = e^{-\mu} \frac{\mu^x}{x!}
where x \in \mathbb{N} , \mu > 0 is a constant.
The Poisson distribution can be derived from the limit of a binomial distribution.
\begin{align*}
\lim_{n \rightarrow \infty, np = \mu} Bin(x=X|n, p) &= \lim_{n \rightarrow \infty, np = \mu} \binom{n}{x} p^x (1-p)^{n-x} \\
&= \lim_{n \rightarrow \infty} \frac{n(n-1)\cdots(n-x+1)}{x!} (\frac{\mu}{n})^x (1-\frac{\mu}{n})^{n-x} \\
&= \lim_{n \rightarrow \infty} \frac{n^x(1-\frac{1}{n})\cdots(1-\frac{x-1}{n})}{x!} (\frac{\mu^x}{n^x}) (1-\frac{\mu}{n})^{n-x} \\
&= \lim_{n \rightarrow \infty} \frac{(1-\frac{1}{n})\cdots(1-\frac{x-1}{n})}{x!} \mu^x (1-\frac{\mu}{n})^{n-x} \\
&= \lim_{n \rightarrow \infty} \frac{(1-\frac{1}{n})\cdots(1-\frac{x-1}{n})}{x!} \mu^x (1+\frac{1}{-\frac{n}{\mu}})^{-\frac{n}{\mu}(-\mu)} (1-\frac{\mu}{n})^{-x}\\
&= \frac{\mu^x}{x!} e^{-\mu} \\
&= Po(X=x|\mu)
\end{align*}
The moment generating function of Po(X=x|\mu) is
\begin{align*}
M(\theta) &= E[e^{\theta X}] \\
&= \sum_{x} e^{\theta x} e^{-\mu} \frac{\mu^x}{x!} \\
&= e^{-\mu} \sum_{x} e^{\theta x} \frac{\mu^x}{x!} \\
&= e^{-\mu} \sum_{x} \frac{(e^\theta \mu)^x}{x!} \\
&= e^{-\mu} e^{e^\theta \mu}
\end{align*}
\begin{align*}
M'(\theta) &= e^{-\mu} e^{e^\theta \mu} (e^\theta \mu)' \\
&= \mu e^{-\mu} e^{e^\theta \mu + \theta} \\
M''(\theta) &= \mu e^{-\mu} e^{e^\theta \mu + \theta} (e^\theta \mu + \theta)' \\
&= \mu e^{-\mu} e^{e^\theta \mu + \theta} (e^\theta \mu + 1)
\end{align*}
\begin{align*}
M'(0) &= \mu \\
M''(0) &= \mu e^{-\mu} e^{\mu} (\mu + 1) \\
&= \mu (\mu + 1)
\end{align*}
therefore, the mean is
\begin{align*}
E[X] = \mu
\end{align*}
the variance is
\begin{align*}
V[X] &= E[X^2] - E[X]^2 \\
&= M''(0) - M(0)^2 \\
&= \mu
\end{align*}
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