Independence of statistics from gamma distribution Let $X_1,...,X_n$ be a random sample from the gamma distribution $\mathrm{Gamma}\left(\alpha,\beta\right)$. 
Let $\bar{X}$ and $S^2$ be the sample mean and sample variance, respectively.
Then prove or disprove that $\bar{X}$ and $S^2/\bar{X}^2$ are independent.

My attempt: Since $S^2/\bar{X}^2 = \frac{1}{n-1} \sum_{i=1}^n \left(\frac{X_i}{\bar{X}}-1\right)^2 $, we need to check the independence of $\bar{X}$ and $\left(\frac{X_i}{\bar{X}} \right)_{i=1}^{n}$, but how should I establish the independence between them?
 A: There is a cute, simple, intuitively obvious demonstration for integral $\alpha.$  It relies only on well-known properties of the uniform distribution, Gamma distribution, Poisson processes, and random variables and goes like this:


*

*Each $X_i$ is the waiting time until $\alpha$ points of a Poisson process occur.

*The sum $Y = X_1+X_2+\cdots + X_n$ therefore is the waiting time until $n\alpha$ points of that process occur.  Let's call these points $Z_1, Z_2, \ldots, Z_{n\alpha}.$

*Conditional on $Y$, the first $n\alpha-1$ points are independently uniformly distributed between $0$ and $Y.$

*Therefore the ratios $Z_i/Y,\ i=1,2,\ldots, n\alpha-1$ are independently uniformly distributed between $0$ and $1.$  In particular, their distributions do not depend on $Y.$

*Consequently, any (measurable) function of the $Z_i/Y$ is independent of $Y.$

*Among such functions are $$\eqalign{X_1/Y &= Z_{[\alpha]}/Y\\ X_2/Y &= Z_{[2\alpha]}/Y - Z_{[\alpha]}/Y\\ \ldots\\ X_{n-1}/Y &= Z_{[(n-1)\alpha]}/Y - Z_{[(n-2)\alpha]}/Y\\ X_n/Y &= 1 - Z_{[(n-1)\alpha]}/Y}$$ (where the brackets $[]$ denote the order statistics of the $Z_i$). 
At this point, simply note that $S^2/\bar X^2$ can be written explicitly as a (measurable) function of the $X_i/Y$ and therefore is independent of $\bar X = Y/n.$
A: You want to prove that the mean $\bar{X}$ and the $n$ rv.s $X_i/\bar{X}$
are independent, or equivalently that the sum $U := \sum X_i$
and the $n$ ratios $W_i := X_i / U$ are independent. We can prove a slightly more
general result by assuming that the $X_i$ have possibly different shapes $\alpha_i$, but the
same scale $\beta>0$ which can be assumed to be $\beta = 1$.
Consider the joint Laplace transform of $U$ and $\mathbf{W}=[W_i]_{i=1}^n$
 i.e.,
$$\psi(t,\,\mathbf{z}) := \text{E}\{\exp[-tU - \mathbf{z}^\top \mathbf{W}\} =
 \text{E}\left\{ \exp\left[-t \sum_i X_i - \sum_i z_i \,\frac{X_i}{U} \right] \right\}
$$
This expresses as an $n$-dimensional integral over $(0, \infty)^n$
$$
 %% \psi(t,\,\mathbf{z} =
 \text{Cst} \, \int
 \exp \left[- (1 + t)(x_1 + \dots + x_n)  -
 \frac{z_1 x_1 + \dots + z_n x_n}{x_1 + \dots + x_n}
\right] \, x_1 ^{\alpha_1 - 1} \dots \, x_n^{\alpha_n - 1} 
 \text{d}\mathbf{x}
$$
where the constant is relative to $\mathbf{x}$.   If we introduce new variables under the integral sign by setting
$\mathbf{y} := (1 + t)\, \mathbf{x}$, we see easily that the integral can be written as a product of two
functions, one depending on $t$ the other depending on the vector $\mathbf{z}$. This
proves that $U$ and $\mathbf{W}$ are
independent.
Disclaimer. This question relates to Lukacs' theorem on
proportion-sum
independence,
hence to the article by Eugene Lukacs A Characterization of the Gamma
Distribution. I
just extracted here the relevant part of this article (namely p. 324), 
with some changes in the notations. I also replaced the use of the
characteristic function by that of the Laplace transform to avoid
changes of variables involving complex numbers.
A: Let $U=\sum_i X_i$. Note that $(X_i /U)_i$ is an ancillary statistic of $\beta$, i.e. its distribution does not depend on $\beta$.
Since $U$ is a complete sufficient statistic of $\beta$, it is independent to $(X_i /U)_i$ by Basu's theorem, so the conclusion follows.
I'm not sure of the construction of the ancillary statistic, since it is only independent of $\beta$, not $\alpha$.
