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?

  • 1
    $\begingroup$ 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 $\endgroup$
    – Henry
    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 $θ$.

  • $\begingroup$ Brilliant! Thanks. $\endgroup$ Mar 4 '18 at 20:25

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.