# Distribution family of the mean of iid random variables

Let $X_1,X_2,...,X_n$ be $n$ iid random variables following some distribution family $A$ (beta, gamma, normal, etc.).

Does the mean $\bar{X}=\frac{1}{n}\sum{X_i}$ also follow a $A$-like distribution (with different parameters)?

I know it is approximately normal due to the Central Limit Theorem, but does it follow the same family $A$? I'm particularly interested in Beta distributions.

Thanks

• One quick way to answer this question yourself is to look up the form of the CF or MGF of the beta (or whatever) family on the Wikipedia page and ascertain whether it is closed under multiplication. (It's not.) If it is, then sums in your family stay in your family, implying that means are (at a minimum) rescaled versions of members of your family. If it isn't, then sums aren't in the family and so (in almost all situations) means aren't in it, either. – whuber Mar 17 '13 at 21:29
• No, this doesn't hold in general or for the specific case of betas. There are families that are closed under iid sums (Binomial, Poisson, Gamma, Normal) and there are some that are closed under arbitrary linear combinations. These latter ones are called stable distributions and have been fully characterized. The Gamma distribution is an interesting case: It is closed under sums, but is not stable, yet it is also closed under means! – cardinal Mar 17 '13 at 21:32
• @whuber,@cardinal thank you for the responses, but if I understand correctly, you reach the "no" answer based on the fact that Beta is not closed under sums. This is clear in this case because a sum of Beta's goes out of the $[0,1]$ support, but the mean never does. So the question is: can we use the closure argument for the mean instead of the sum? In fact, whuber metions "(in almost all situations)". – Julián Urbano Mar 17 '13 at 21:52
• No, that wasn't the logic used. At any rate: As a simple case, take the uniform distribution on $[0,1]$. This is a beta distribution. But, $(X_1+X_2)/2$ has a triangular density and so is not beta. QED. :-) – cardinal Mar 17 '13 at 22:02
• quick and painful :-) Write the answer if you want so we can close this question. – Julián Urbano Mar 17 '13 at 22:05

Let $X_1,X_2$ follow a uniform distribution on $[0,1]$, which is a Beta distribution as well. Variable $(X_1+X_2)/2$ has a triangular density function, which is not Beta.