How can I show $Y=\frac{\ln X_1}{\ln X_2}$ is an ancillary statistic?

Question

Suppose $X_1,X_2$ be random sample with probability density function $f(x)=\alpha x^{\alpha-1}e^{{-x}^\alpha}$, $x>\alpha$, $\alpha>0$. How can I show $Y=\frac{\ln X_1}{\ln X_2}$ is an ancillary statistic.

Answer 1

You have $f(x)\,dx = e^{-x^\alpha}\Big( \alpha x^{\alpha-1}\,dx\Big) = e^{-u}\,du$. This suggests we consider $U=X^\alpha$, so that \begin{align} f_U(u) & = \frac d {du} \Pr(U\le u) \\[10pt] & = \left(\frac d{dx} \Pr(U\le u)\right) \cdot \frac{dx}{du} \\[10pt] & = \left(\frac d{dx} \Pr(X \le x)\right) \cdot \frac 1 {\alpha x^{\alpha-1}} \\[10pt] & = \alpha x^{\alpha-1} e^{-x^\alpha}\cdot\frac 1 {\alpha x^{\alpha-1}} \\[10pt] & = e^{-x^\alpha} = e^{-u}\quad\text{for }u>0. \end{align}

Next, we have $\ln U=\alpha\ln X$, so $$ \frac{\ln X_1}{\ln X_2} = \frac{\ln U_1}{\ln U_2}. $$ Since the distribution of neither $U_1$ nor $U_2$ depends on $\alpha$, the distribution of this random variable does not depend on $\alpha$.