This is a low-priority question, but it has been bugging me so I thought I'd ask. In my stats homework I have the following exercise:
Exercise. Suppose $X_1$ and $X_2$ are iid observations from the pdf $f(x\mid\alpha)=\alpha x^{\alpha-1}e^{-x^\alpha}$, $x>0$, $\alpha>0$. Show that $(\log X_1)/(\log X_2)$ is an ancillary statistic.
This is easy to do by showing that if $X\sim X_i$ then $(\log X)$ is a scale family and then applying the Theorem which states that if $U$ is a scale family and $U_1,\cdots,U_n$ are iid observations from $U$ then each $U_i/U_j$, $i\neq j$, is ancillary. However I attempted a different method which led to a curious problem. Observe.
Let $U=\frac{\log X_1}{\log X_2}$ and $V=\log X_2$, then consider the transformation $(X_1,X_2)\mapsto(U,V)$. Then $x_1=e^{uv}$ and $x_2=e^v$, giving us the Jacobian
\begin{equation*}J=\left|\begin{array}{cc}\frac{\partial x_1}{\partial u}&\frac{\partial x_1}{\partial v}\\\\\frac{\partial x_2}{\partial u}&\frac{\partial x_2}{\partial v}\end{array}\right|=e^{(u+1)v}.\end{equation*}
Thus by independence of $X_1$ and $X_2$ we have
\begin{multline*}f_{U,V}(u,v)=f_{X_1,X_2}(e^{uv},e^v)|J|\\\\=[\alpha (e^{uv})^{\alpha-1}e^{-(e^{uv})^\alpha}][\alpha (e^v)^{\alpha-1}e^{-(e^v)^\alpha}]e^{(u+1)v}\\\\=\alpha^2\exp[{(u+1)v\alpha}-e^{uv\alpha}-e^{v\alpha}].\end{multline*}
Hence by changing variables $z=v\alpha$ we get $dv=dz/\alpha$, and then
\begin{multline*}f_U(u)=\int_{-\infty}^\infty\alpha^2\exp[{(u+1)v\alpha}-e^{uv\alpha}-e^{v\alpha}]\;dv\\\\=\alpha\int_{-\infty}^\infty\exp[{(u+1)z}-e^{uz}-e^z]\;dz=:\alpha g(u),\end{multline*}
where $g(u)$ is independent of $\alpha$. However this is impossible!
This leads to my question: Where is my error?
