# How to calculate the MGF for a poisson random variable

$\phi (t)=\sum_{n=0}^{\infty}e^{tn}\cdot \frac{\lambda^{n}}{n!}e^{-\lambda}$

$=e^{-\lambda}\sum_{n=0}^{\infty}\frac{(e^{t}\lambda)^{n}}{n!}$

$=e^{-\lambda+\lambda e^{t}}$ I'm confused here as how $\sum_{n=0}^{\infty}\frac{(e^{t}\lambda)^{n}}{n!}$ simplifies to $e^{\lambda e^{t}}$, can someone please explain?

• Taylor: $e^y=\sum_{n=0}^\infty y^n/n!$. Take $y=e^t\lambda$. – Zen Jun 10 '14 at 3:57

[Edit: Perhaps the easy way to see it is to try it backwards. Expand $\exp(\lambda e^t)$ as a series in $(\lambda e^t)$.]

Its straight from the series for $\exp(x)$

$$\exp(x) = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!} + ... = \sum_{n=0}^\infty \frac{x^n}{n!}$$

Note that in that series, $x$ is a constant; the dummy variable is $n$. So as long as the sum converges, any other constant will work just as well. Since the series converges for any finite argument, that works without any problem.

So for example $\sum_{n=0}^\infty \frac{f(w)^n}{n!} = \exp(f(w))$, as long as $f(w)$ is finite ... so it works for any finite value $f(w)$.

Now if $x= \lambda e^t$, you get your expression.

That is, that's the series for $\exp(\lambda e^t)$.