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?


  • $\begingroup$ Assuming independence? $\endgroup$ – Did Nov 4 '15 at 12:11

\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 $Y \sim \text{Poisson}(λ_1+λ_2)$.

  • 3
    $\begingroup$ Assuming independence. $\endgroup$ – Did Nov 4 '15 at 12:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.