Poisson Distribution of sum of two random independent variables $X$, $Y$ $X \sim \mathcal{P}( \lambda) $ and $Y \sim \mathcal{P}( \mu)$ meaning that $X$ and $Y$ are Poisson distributions. What is the probability distribution law of $X + Y$. I know it is $X+Y \sim \mathcal{P}( \lambda + \mu)$ but I don't understand how to derive it.

Using Moment Generating Function.
If $X \sim \mathcal{P}(\lambda)$, $Y \sim \mathcal{P}(\mu)$ and S=X+Y.
We know that MGF(Moment Generating Function) of $\mathcal{P}(\lambda)=e^{\lambda(e^t-1)}$(See the end if you need proof)
MGF of S would be
$$\begin{align}
M_S(t)&=E[e^{tS}]\\&=E[e^{t(X+Y)}]\\&=E[e^{tX}e^{tY}]\\&=E[e^{tX}]E[e^{tY}]\quad \text{given }X,Y\text{ are independent}\\&=e^{\lambda(e^t-1)}e^{\mu(e^t-1)}\\&=e^{(\lambda+\mu)(e^t-1)}
\end{align}$$
Thus S is a Poisson Distribution with parameter $\lambda+\mu$.  

MGF of Poisson Distribution
If $X \sim \mathcal{P}(\lambda)$, then by definition Probability Mass Function is
$$\begin{align}
f_X(k)=\frac{\lambda^k}{k!}e^{-\lambda},\quad k \in 0,1,2....
\end{align}$$
It's MGF is
$$\begin{align}
M_X(t)&=E[e^{tX}]\\&=\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}e^{-\lambda}e^{tk}\\&=e^{-\lambda}\sum_{k=0}^{\infty}\frac{\lambda^ke^{tk}}{k!}\\&=e^{-\lambda}\sum_{k=0}^{\infty}\frac{(\lambda e^t)^k}{k!}\\&=e^{-\lambda}e^{\lambda e^t}\\&=e^{\lambda e^t-\lambda}\\&=e^{\lambda(e^t-1)}
\end{align}$$
