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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?

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    $\begingroup$ Consider the joint Laplace transform of the sum $U:= \sum_i X_i$ and the vector $\mathbf{W}$ of proportions $W_i := X_i / U$. This is $\text{E}\{\exp[-tU - \mathbf{z}^\top \mathbf{W}]\}$; you can show that this is the product of a function of $t$ and a function of $\mathbf{z}$. $\endgroup$ – Yves Jun 25 '18 at 6:34
  • $\begingroup$ @Yves Could you check my answer posted below? $\endgroup$ – bellcircle Jun 25 '18 at 10:31
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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:

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

  2. 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 $Z_1, Z_2, \ldots, Z_{n\alpha}.$

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

  4. 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 $Y.$

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

  6. 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 $\bar X = Y/n.$

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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.

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    $\begingroup$ (+1) For the paper on the characterization of gamma distribution. $\endgroup$ – StubbornAtom Jun 25 '18 at 17:54
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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 $\alpha$.

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  • $\begingroup$ Good. The theorem can be invoked with $\alpha$ regarded as fixed so considering a one-parameter statistical model. $\endgroup$ – Yves Jun 25 '18 at 17:07

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