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.

In short, you can show this by using the fact that $$Pr(X+Y=k)=\sum_{i=0}^kPr(X+Y=k, X=i).$$
If $X$ and $Y$ are independent, this is equal to
$$
Pr(X+Y=k)=\sum_{i=0}^kPr(Y=k-i)Pr(X=i)
$$
which is
$$
\begin{align}
Pr(X+Y=k)&=\sum_{i=0}^k\frac{e^{-\lambda_y}\lambda_y^{k-i}}{(k-i)!}\frac{e^{-\lambda_x}\lambda_x^i}{i!}\\
&=e^{-\lambda_y}e^{-\lambda_x}\sum_{i=0}^k\frac{\lambda_y^{k-i}}{(k-i)!}\frac{\lambda_x^i}{i!}\\
&=\frac{e^{-(\lambda_y+\lambda_x)}}{k!}\sum_{i=0}^k\frac{k!}{i!(k-i)!}\lambda_y^{k-i}\lambda_x^i\\
&=\frac{e^{-(\lambda_y+\lambda_x)}}{k!}\sum_{i=0}^k{k\choose i}\lambda_y^{k-i}\lambda_x^i
\end{align}
$$
The sum part is just 
$$
\sum_{i=0}^k{k\choose i}\lambda_y^{k-i}\lambda_x^i=(\lambda_y+\lambda_x)^k
$$
by the binomial theorem.
So the end result is 
$$
\begin{align}
Pr(X+Y=k)&=\frac{e^{-(\lambda_y+\lambda_x)}}{k!}(\lambda_y+\lambda_x)^k
\end{align}
$$
which is the pmf of $Po(\lambda_y+\lambda_x)$.
