# Poisson distribution of a sum.

Suppose the number of robberies of a clothing store in a random day is a random variable with Poisson distribution with $\lambda=5$. $X_i$ is the number of robberies in day $i$. $S_{30}=X_1+...X_{30}$. Given the days are independent, how does the variable $S_{30}$ distribute?

What I said is: $X_i\sim P(5=\lambda_i)$ and therefore $S_{30}\sim P(\lambda_1+...+\lambda_{30}=30*5=150)$. The answer, however, says $S_{30}\sim P(300)$. How is that? I could really use your help.

• How do you know that $S_{30}$ has a Poisson distribution? It is correct that $E(S_{30}) = E(X_1+ ... + X_{30}) = 30*5 = 150$, since each $X_i$ is independent with mean 5. – James Jun 11 '15 at 14:44
• How can I know, then, how it distributes? – Meitar Jun 11 '15 at 14:46
• It distributes as a Poisson with mean $\lambda_{30}$ as 150, just as you stated. – mandata Jun 11 '15 at 16:12
• The convolution of two independent Poisson distributions is another Poisson distribution whose mean is the sum of the means of the individual distributions. On a side note, that's evidently a crummy location for a clothing store! – Brian Tung Jun 11 '15 at 17:24