Compute $E[e^{tX}]$ as a function of t ∈ R when
(1) X is Poisson λ,
(2) X is Binomial n, p. [Hint: you may find the Binomial Theorem to be useful].
The PMF of Poisson is $$ p_x(k)=e^{-λ}λ^k/k! $$
I know if X is Poisson, E[X] = λ. So how would i incorporate that into $E[e^{tX}]$?
Let $X$ be poisson $\text{Poi}(\lambda)$ distributed. Then for the moment generating function $$m_X(t)=\mathbb{E}e^{tX}=\sum_{k=0}^\infty e^{tk} \mathbb{P}(X=k)=\sum_{k=0}^\infty e^{tk} \frac{\lambda^k e^{-\lambda}}{k!}=e^{-\lambda}\sum_{k}\frac{(e^t \lambda)^k}{k!}$$ The last sum is the series representation of $e^{(e^t\lambda)}$, so $$m_X(t)=e^{-\lambda}e^{e^t\lambda}=e^{\lambda (e^t-1)}.$$
Similarly let $X$ be binomially distributed with parameters $(n,p)$- We have $$m_X(t)=\mathbb{E}e^{tX}=\sum_{k}e^{tx}\mathbb{P}(X=k)=\sum_{k}e^{kt}{n \choose k}p^k(1-p)^{n-k}=\sum_{k}{n \choose k}(pe^t)^{k}(1-p)^{n-k}$$ Using the binomial theorem: $$\mathbb{E}e^{tX}=(1-p+pe^t)^n$$