I'm unable to understand the proof behind determining the Moment Generating Function of a Poisson which is given below: $N \sim \mathrm{Poiss}(\lambda)$
$$
E[e^{\theta N}] = \sum\limits_{k=0}^\infty e^{\theta k} \frac{e^{-\lambda}\lambda^k }{k!} = e^ {\lambda[e^{\theta}-1]}
$$
Edit: Q.1 I don't understand how we go from $E[e^{\theta N}] = \sum\limits_{k=0}^\infty e^{\theta k} \frac{e^{-\lambda}\lambda^k }{k!}$. Q.2 Also, I didn't know how it goes from $\sum\limits_{k=0}^\infty \frac{(\mathrm e^{\theta}\lambda)^k }{k!} = \mathrm e^ {-\lambda}\,\mathrm e^{\mathrm e^\theta \lambda}$
Thank You
Best Answer
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:
$E[e^{\lambda t}] = e^{\lambda(e^t -1)}$