Moment Generating Function of Poisson

Solution 1:

To derive the MGF of poisson:

Poisson: $ \frac{\lambda^x e^{- \lambda}}{x!}$

$MGF = E[e^{t x}] = \sum_\limits{x=0}^{\infty} e^{tx} \frac{\lambda^x e^{- \lambda}}{x!}$

We don't care about anything not related to X so factor out $e^{-\lambda}$, we'll also group the two values with common powers i.e $e^{tx}$ and $\lambda^x$ are both to the power of x.

$E[e^{\lambda t}] = e^{-\lambda} \sum_\limits{x=0}^{\infty} \frac{(\lambda e^t)^x }{x!}$

Now the summation looks very similar to the exponential function from:

https://en.wikipedia.org/wiki/List_of_mathematical_series

i.e

$\sum_\limits{x=0}^{\infty} \frac{a^x}{x!} = e^a$

Thus sub the result of the exponential function in.

$E[e^{\lambda t}] = e^{-\lambda}{e^{\lambda e^t}}$

And simplify:

$E[e^{\lambda t}] = e^{\lambda(e^t -1)}$

Solution 2:

$$\sum\limits_{k=0}^\infty\mathrm e^{\theta k} \frac{\mathrm e^{-\lambda}\lambda^k }{k!} = \mathrm e^{-\lambda}\sum\limits_{k=0}^\infty \frac{(\mathrm e^{\theta}\lambda)^k }{k!} = \mathrm e^ {-\lambda}\,\mathrm e^{\mathrm e^\theta \lambda}=\mathrm e^{(\mathrm e^\theta-1) \lambda}$$