# Why is sum of r.v with po(1) distribution po(n)?

Let $U_i \sim Po(1)$ be independent, for $i=1,2,\ldots$ Set

$X_n = U_1 + U_2 + \cdots + U_n$

Then $X_n \sim Po(n)$

Why is that so? Is this valid for all distributions? Is there any formula we can use to get a general rule for all probabilities? I.e can we say that if the sum of independent Exp($1$) variables have the distribution Exp($n$) ?

• The result is for the Poisson, and does not apply, for example, to the exponential. The sum of $n$ iid exponetials is not exponential if $n\gt 1$. Oct 27 '15 at 17:11
• ok thanks! How could i prove that it is valid for poisson distributions?
– Ptru
Oct 27 '15 at 17:19
• You are welcome. I would prefer not to write an answer, since it has been proved man times on MSE that the sum of any two independent Poisson, parameters $\lambda$ and $\mu$, is Poisson with parameter $\lambda+\mu$, and that is all that is needed. Please search the site for "sum of independent Poisson." Oct 27 '15 at 17:23

Mathematically, we would write these examples as follows: $$X_i \sim \operatorname{Gamma}(a_i, b); \quad f_{X_i}(x) = \frac{b^{a_i} x^{a_i - 1} e^{-b x}}{\Gamma(a_i)}$$ implies $$T_n = \sum_{i=1}^n X_i \sim \operatorname{Gamma}\left(\sum_{i=1}^n a_i, b\right).$$ For the Normal distribution, $$X_i \sim \operatorname{Normal}(\mu_i, \sigma_i^2) \quad \text{implies} \quad T_n \sim \operatorname{Normal}\left(\sum_{i=1}^n \mu_i, \sum_{i=1}^n \sigma_i^2\right).$$ And for the Binomial case, $$X_i \sim \operatorname{Binomial}(n_i, p) \quad \text{implies} \quad T_n = \operatorname{Binomial}\left(\sum_{i=1}^n n_i, p\right).$$ These are all provable using moment generating functions, or by an induction argument, as is also the case with the Poisson distribution. But it is not true for the exponential distribution example. Rather, because $$\operatorname{Exponential}(\lambda) = \operatorname{Gamma}(1,\lambda),$$ we can see from the above that the sum of $n$ IID exponential variables with common rate $\lambda$ will be Gamma distributed with shape $n$ and rate $\lambda$.
The proof for the Poisson case using MGFs is straightforward. We first calculate the MGF of a single Poisson variable $X$ with rate $\lambda$: $$M_X(t) = \operatorname{E}[e^{tX}] = \sum_{x=0}^\infty e^{tx} e^{-\lambda} \frac{\lambda^x}{x!} = \sum_{x=0}^\infty e^{\lambda(e^t-1)} e^{-\lambda e^t} \frac{(\lambda e^t)^x}{x!} = e^{\lambda(e^t-1)}.$$ Therefore, if $X_i \sim \operatorname{Poisson}(\lambda_i)$ are independent, the MGF of their sum is $$M_{T_n}(t) = \prod_{i=1}^n M_{X_i}(t) = \prod_{i=1}^n e^{\lambda_i (e^t - 1)} = \exp \left( (e^t-1) \sum_{i=1}^n \lambda_i \right),$$ which is clearly Poisson with rate $$\lambda_T = \sum_{i=1}^n \lambda_i.$$