# Show that distribution of statistic is independent of parameter

Consider a random sample of size $$n$$ froma gamma distribution, $$X_i\sim GAM(\theta, \kappa)$$, and let $$\bar X=\dfrac{1}{n}\sum X_i$$ and $$\tilde X=(\prod X_i)^{1/n}$$ be the sample mean and geometric mean, respectively.

Show that the distribution of $$T=\bar X/\tilde X$$ does not depend on $$\theta$$.

It can be shown easily that $$\bar X$$ and $$\tilde X$$ are jointly complete and sufficient statistics for $$\theta$$ and $$\kappa$$. I'm not necessarily sure if that will help me with this, but I don't really know how to go about this. Surely I don't need to use bivariate transformation, do I? There's got to be an easier way to show this than that. Any ideas?

• It may just be my eyesight or screen resolution, but in the question the tilde in $\tilde{X}$ looks very like the bar in $\bar{X}$, though in this comment they look different Mar 4 '18 at 14:18

$\def\deq{\stackrel{\mathrm{d}}{=}}$Suppose $Y_1, \cdots, Y_n$ are i.i.d. such that $Y_k \sim {\mit Γ}(1, κ)$, then$$(X_1, \cdots, X_n) \deq (θY_1, \cdots, θY_n).$$ So$$(\overline{X}, \widetilde{X}) = \left( \frac{1}{n} \sum_{k = 1}^n X_k, \left( \prod_{k = 1}^n X_k \right)^{\frac{1}{n}} \right) \deq \left( θ \cdot \frac{1}{n} \sum_{k = 1}^n Y_k, θ \cdot \left( \prod_{k = 1}^n Y_k \right)^{\frac{1}{n}} \right) = (θ \overline{Y},θ \widetilde{Y}),$$ where$$\overline{Y} = \frac{1}{n} \sum_{k = 1}^n Y_k,\ \widetilde{Y} = \left( \prod_{k = 1}^n Y_k \right)^{\frac{1}{n}}.$$ Thus$$\frac{\overline{X}}{\widetilde{X}} \deq \frac{\overline{Y}}{\widetilde{Y}},$$ which does not depend on $θ$.