Suppose $X$ has the $\mathrm{Poisson}(5)$ distribution considered earlier.
Then $P(X \in A) = \sum_{j\in A} \frac{e^{-5}5^j}{j!}$, which implies that $L(X) = \sum^\infty_{j=0} \left(\frac{e^{-5}5^j}{j!}\right)\delta_j$, a convex combination of point masses. The following propostion shows that we have $E(f(X)) = \sum_{j=0}^\infty\frac{f(j)e^{-5}5^j}{j!}$ for any function $f : \mathbb{R} \rightarrow \mathbb{R}$.
Prop. Suppose $\mu = \sum_i \beta \mu_I$ where $\{\mu_i\}$ are probability distributions, and $\{\beta_i\}$ are non-negative constants (summing to 1, if we want $\mu$ to also be a probability distribution). Then for Borel-measurable functions $f : \mathbb{R} \rightarrow \mathbb{R}$,
$$\int fd\mu = \sum_i \beta_i \int f \, d\mu_i,$$
provided either side is well-defined.
Using this proposition:
Let $X \sim \mathrm{Poisson}(5)$.
(a) compute *E*$(X)$ and *Var*$(X)$.
(b) compute *E*$(3^X)$.
I know that the answers from a previous question are: $E[X]=\lambda$, $E[X^2]=\lambda+\lambda^2$ (from the previous two, you could compute the variance) and $E[e^{3X}]= e^{2 \lambda}$.
However, I'm not sure as to how to get to this other than I'm supposed to use a Taylor series.