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.
This only holds if $X$ and $Y$ are independent, so we suppose this from now on. We have for $k \ge 0$:
\begin{align*}
  P(X+ Y =k) &amp;= \sum_{i = 0}^k P(X+ Y = k, X = i)\\
    &amp;= \sum_{i=0}^k P(Y = k-i , X =i)\\
    &amp;= \sum_{i=0}^k P(Y = k-i)P(X=i)\\
    &amp;= \sum_{i=0}^k e^{-\mu}\frac{\mu^{k-i}}{(k-i)!}e^{-\lambda}\frac{\lambda^i}{i!}\\
   &amp;= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \frac{k!}{i!(k-i)!}\mu^{k-i}\lambda^i\\
   &amp;= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \binom ki\mu^{k-i}\lambda^i\\
   &amp;= \frac{(\mu + \lambda)^k}{k!} \cdot e^{-(\mu + \lambda)}
\end{align*}
Hence,  $X+ Y \sim \mathcal P(\mu + \lambda)$.
Here's a much cleaner solution:

Consider a two Poisson processes occuring with rates $\lambda$ and $\mu$, where a Poisson process of rate $r$ is viewed as the limit of $n$ consecutive Bernoulli trials each with probability $\frac{r}{n}$, as $n\to\infty$.

Then $X$ counts the number of successes in the trials of rate $\lambda$ and $Y$ counts the number of successes in the trials of rate $\mu$, so the total number of successes is the same as if we had each trial succeed with probability $\frac{\lambda + \mu}{n}$, where we take $n$ to be large enough so that the event where the $i$th Bernoulli trial in both processes are successful has a negligible probability.
Then we are done.