Distribution of the sum of two Poisson random variables

Question

Let $X_1$ be a random variable with poisson distribution $\text{Poisson}(\lambda_1)$ (i.e. $f(x)=\frac{\lambda^x}{x!}e^{-\lambda}$ if $x \in \{ 0,1,2,3,\ldots\}$ and $0$ otherwise) and let $X_2$ be a random variable with poisson distribution $\text{Poisson}(\lambda_2)$. What is the distribution of $Y = X_1 + X_2$?

I tried using the convolution formula:

$$f_y(Y) = \Sigma_{x=1}^{\infty} f_2(y-x)f_1(x) = \lambda_2^ye^{-(\lambda_1+\lambda_2)}\Sigma_{x=1}^{\infty}\frac{\left(\frac{\lambda_1}{\lambda_2}\right)^x}{x!(y-x)!}$$

But I am kind of stuck here..

Any ideas?

Thanks!

Answer 1

\begin{align*}P(Y=y)&=P(X_1+X_2=y)=\sum_{k=0}^{y}P(X_1+X_2=y \mid X_2=k)P(X_2=k)\\&=\sum_{k=0}^{y}P(X_1=y-k)P(X_2=k)=\sum_{k=0}^{y}\frac{λ_1^{y-k}}{(y-k)!}e^{-λ_1}\frac{λ_2^k}{k!}e^{-λ_2}\\&=\frac1{y!}\sum_{k=0}^{y}\dbinom{y}{k}λ_1^{y-k}λ_2^ke^{-(λ_1+λ_2)}=\frac{(λ_1+λ_2)^y}{y!}e^{-λ_1+λ_2}\end{align*} so $Y \sim \text{Poisson}(λ_1+λ_2)$.