Suppose $X_1$ and $X_2$ are iid observations from the pdf $f(x|\alpha)=\alpha x^{\alpha-1}e^{-x^\alpha}$, $x>0$, $\alpha>0$. Show that $\frac{\log X_1}{\log X_2}$ is an ancillary statistic.
I guess I need to show that the distribution of this statistic is independent of $\alpha$, i.e. it is the same as if $\alpha = 1$. However, this concept is a little bit confusing to me and I am not sure how to approach this problem.
If $U\approx\mathsf{Exp}(1)$ then:$$P(U^{\frac1{\alpha}}>x)=P(U>x^{\alpha})=e^{-x^{\alpha}}$$ indicating that $U^{\frac1{\alpha}}$ has the distribution of $X_1,X_2$.
So $U_{i}:=X_{i}^{\alpha}$ are iid and have standard exponential distribution.
Then it follows easily that: $$\frac{\ln X_{1}}{\ln X_{2}}=\frac{\ln U_{1}}{\ln U_{2}}$$and distribution of RHS does not depend on $\alpha$ because the distribution of $U_1$ and $U_2$ does not depend on $\alpha$.