# If $Y$ = sum of $X$, what does the distribution look like when $X$ is poisson?

If $$X_i$$ is poisson. I know that the mgf of $$X_i$$ is $$e^{\lambda(e^t-1)}$$.

What would the distribution of $$Y$$ look like if $$Y$$ = the sum of all $$X_i$$?

is it Poisson itself?

• Without knowing their joint distribution, we cannot answer anything. On one extreme, where $X_i$'s are independent, then their sum again has Poisson distribution. On the other extreme, if $X_1 = \cdots = X_n$, then their sum is just $nX_1$, which is no longer Poisson. – Sangchul Lee Feb 19 at 1:29

It is true that if $$X_1, \dots, X_n$$ are independent Poisson variables with mean $$\lambda$$, then $$Y := X_1 + \dots + X_n$$ is Poisson with mean $$n\lambda$$.
To see this, observe that the moment generating function of $$Y$$ is $$\left( e^{\lambda (e^t - 1)} \right)^n = e^{n\lambda (e^t - 1)},$$ which is precisely the moment generating function of a Poisson variable with mean $$n\lambda$$.
This should feel intuitive. Imagine you have a Poisson process, where the expected number of events per unit time is $$\lambda$$. If for each $$i \in \{1, \dots, n\}$$, $$X_i$$ represents the number of events observed in time interval $$[i - 1 , i)$$, then $$Y = X_1 + \dots + X_n$$ represents the number of events observed in time interval $$[0, n)$$, and this is Poisson-distributed with mean $$n\lambda$$.
• how would it be different if it was Gamma assuming t he $$X_i$$ are iid? How would that look? – George Harrison Feb 19 at 1:33
• @GeorgeHarrison How about trying the same method? The mgf for the $\rm{Gamma}(\alpha, \beta)$ is $(1 - t / \beta)^{-\alpha}$. So if you have $n$ such variables, the mgf for the sum is $(1 - t /\beta)^{-n\alpha}$, which is the mgf for $\rm{Gamma}(n\alpha, \beta)$. – Kenny Wong Feb 19 at 1:37