Divide a Chi-squared distribution by its degrees of freedom Say I have random variables $X_i,\;\; i=1...5$, where $X_i \sim \chi_{1}^2$.
I understand that $S\sim \chi_{5}^2$, because $S= \sum_{i=1}^5 X_i$. However what is the distribution of $\frac{S}{5}$. Do we know it? Because I am sure that $\frac{S}{5}\sim \chi_{1}^2$ is incorrect.
 A: I found the answer by myself, so I leave it here for others.
The sample mean of $n$ i.i.d. chi-squared variables of degree $k$ is distributed according to a gamma distribution with shape $\alpha$ and scale $\theta$ parameters: 
$$\bar{X} =\frac{1}{n} \sum_{i=1}^n X_i \sim Gamma(\alpha=\frac{nk}{2},\theta=\frac{2}{n}),\;\;\;\;\; where \;\; X_i\sim\chi_k^2$$
A: This is actually an extemely useful distributional form for use in sampling problems.  While I'm not sure if it has a commonly used name, I call it the scaled chi-squared distribution, and then I always make sure to define this clearly when I'm talking about it.  In general, if $Z_1,Z_2,Z_3,... \sim \text{N}(0,1)$ is a sequence of IID standard normal random variables then we have:
$$R_n \equiv \frac{1}{n} \sum_{i=1}^n Z_i^2 \sim \text{ScChiSq}(n) \equiv \text{Gamma}(\text{Shape} = \tfrac{n}{2},\text{Rate} = \tfrac{n}{2}),$$
where this distribution has density function:
$$\text{ScChiSq}(r|n) = \frac{(n/2)^{n/2}}{\Gamma(\tfrac{n}{2})} \cdot r^{n/2-1} e^{-rn/2}
\quad \quad \quad \text{for all } r \geqslant 0.$$
Using the scaled chi-squared distribution directly makes it easier to present certain results in sampling theory.  This distribution has a fixed unit mean, and if $R_n \sim \text{ScChiSq}(n)$ then $R_n \rightarrow 1$ in probability as $n \rightarrow \infty$.
