Let $X_1,...,X_n$ be iid and follow $Gamma(\alpha, \beta)$, where $$f(x,\alpha, \beta)=\frac{x^{\alpha-1}e^{-x/\beta}}{\Gamma(\alpha)\beta^\alpha}$$ I already showed that $\overline{X}$ and $X^*=\left[ \prod_{i=1}^n X_i \right]^\frac{1}{n}$ are complete and sufficient statistics for $(\alpha, \beta)$. I am to show that the distribution of the statistic $T=\frac{\overline{X}}{X^*}$ does not depend on $\beta$ but am unsure how to do so. Any help would be greatly appreciated.
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1$\begingroup$ What do you mean by "does not contain $\beta$"? When I look at the formulae, I see no $\beta$... $\endgroup$– jbowmanCommented Apr 12, 2020 at 2:16
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$\begingroup$ I edited the wording of the problem a little bit; hopefully that clears it up. The $X_i$'s follow $Gamma(\alpha, \beta)$ and I need to show that the statistic $T$ is ancillary for $\beta$. $\endgroup$– Michael Devin SmithCommented Apr 12, 2020 at 2:25
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1$\begingroup$ Please mention the parameterization you are using, i.e. the pdf of Gamma(a,b). $\endgroup$– StubbornAtomCommented Apr 12, 2020 at 7:29
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4$\begingroup$ As a hint: try a change of variable from $x$ to $y=x/\beta$. If you solve for the distribution of $y$, you will see that it doesn't have $\beta$ as a parameter. Then show that $\bar{Y} / Y^* = \bar{X}/X^*$, and you're essentially done. $\endgroup$– jbowmanCommented Apr 12, 2020 at 14:35
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$\begingroup$ @jbowman that was a giant help. Thank you so much! $\endgroup$– Michael Devin SmithCommented Apr 12, 2020 at 15:54
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1 Answer
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Let $Y=X/\beta$. Then, $$f_Y(y)=f_X(y\beta)\beta=\frac{(y\beta)^{\alpha-1}e^{-(\beta y/\beta)}}{\Gamma(\alpha)\beta^\alpha}$$ $$=\frac{y^{\alpha-1}e^{-y}}{\Gamma(\alpha)}$$ Note that the distribution of Y does not depend on $\beta$. Now consider $$\overline{Y}=\frac{\sum Y_i}{n}=\frac{\sum X_i/\beta}{n}=\frac{1}{\beta}\frac{\sum X_i}{n}$$ $$Y*=\left[\prod_{i=1}^n Y_i\right]^{1/n}=\left[\prod_{i=1}^n X_i/\beta \right]^{1/n}=\frac{1}{\beta}\left[\prod_{i=1}^n X_i\right]^{1/n}$$ Then, when taking the ratio $\frac{\overline{Y}}{Y*}$, note that the $\frac{1}{\beta}$ pieces of both statistics cancel, and you're left with $\frac{\overline{X}}{X*}$. This tells us that the statistic $T=\frac{\overline{X}}{X*}$ does not depend on $\beta$.