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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.

Analysis801
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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 $\alpha$.

drhab
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  • How does $U_i = X_i^\alpha$ follows from the same distribution? – Nikita Dezhic Dec 08 '19 at 16:57
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    The first displayed equation in the answer shows that the distribution functions of $U_i^{1/a}$ and $X_i$ are the same. Hence the distributions of the $a$th powers of them are the same, too. – kimchi lover Dec 08 '19 at 17:19
  • @kimchi lover But why does that mean that the variables are equal? I understand why they have the same distribution, but does that mean they are equal? I have examples like here in mind https://stats.stackexchange.com/questions/24938/can-two-random-variables-have-the-same-distribution-yet-be-almost-surely-differ – Nikita Dezhic Dec 08 '19 at 19:05
  • I would also add an example like this: $\Omega = {\omega_1,\omega_2}, P(\omega_1) = P(\omega_2) = \frac{1}{2}, X_1(\omega_1) = X_2(\omega_2) = 1, X_1(\omega_2) = X_2(\omega_1) = 0$. – Nikita Dezhic Dec 08 '19 at 19:07
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    @DeuzharNickens Which variables are claimed to be equal, by whom, and where? – kimchi lover Dec 08 '19 at 19:08
  • @kimchi lover "So $U_{i}:=X_{i}^{\alpha}$ are iid and have standard exponential distribution", right after the division line. – Nikita Dezhic Dec 08 '19 at 19:09
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    @DeuzharNickens Here drhab defined $U_i$ to be $X_i^a$, that's why they are equal. Since the $X_i$ are iid, the $U_i$ are too. (The symbol $\defeq$ means "is defined to equal".) – kimchi lover Dec 08 '19 at 19:15