confusion on ancillary of gamma distribution

Question

Here is the question. I am concerned about part (ii).

I found out, $T$ is complete sufficient statistic for $\beta$. Now I need to show that $X_{(i)}$ is ancillary. But, for of all, I can not find a explicit form of pdf of $X_{(i)}$. On the other hand, if I don't have to find that, then what is the other way to show $X_{(i)}$ is an ancillary?

Thanks for any help.

Note: Basu's Theorem

Answer 1

You have \begin{align} \Pr(X_i\in A) & = \int_A \frac 1 {\Gamma(\alpha)} (\beta x)^{\alpha-1} e^{-\beta x} (\beta\, dx) \\[10pt] & = \int_{\beta A} \frac 1 {\Gamma(\alpha)} u^{\alpha-1} e^{-u} \, du \qquad \text{where } \beta A = \{\,\beta x:x\in A\,\}. \end{align}

Let $Y_i= \beta X_i.$ Then $$ \Pr(Y_i\in\beta A) = \Pr(X_i\in A), $$ so $Y_i\sim\operatorname{gamma}(\alpha,1).$

Let $U= Y_1+\cdots+Y_n.$

Then $\dfrac{Y_{(i)}} U = \dfrac{X_{(i)}} T,$ so $$ \Pr\left( \frac{X_{(i)}} T \in B \right) = \Pr\left( \frac{Y_{(i)}} U \in B \right) $$ and the latter probability clearly does not depend on $\beta.$ Hence $X_{(i)}/T$ is an ancillary statistic.