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?

  • 2
    $\begingroup$ 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

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 $\bar X = Y/n.$


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.

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

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 $\alpha$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.