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Can anyone give me a clue of how to address this theorem?

Suppose that $T$ es a real-valued statistic. Suppose that $Q(t,\theta)$ is a monotone function of $t$ for each value of $\theta\in \Theta$. Show that if the pdf of $T$, $f(t|\theta)$, can be expressed in the form

\begin{equation} f(t|\theta)=g(Q(t, \theta))\left|\frac{d}{dt}Q(t,\theta)\right| \end{equation} for some function $g$, then $Q(t,\theta)$ is a pivot.

I know it is also useful to use the change of variable theorem, but i do not see how.

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  • $\begingroup$ Could you please add the self-study tag this post deserves? $\endgroup$ – Xi'an Feb 21 '16 at 10:35
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Here is a possible proof if the function $Q(t|\theta)$ is strictly monotone in $t$, i.e. if the function can be inverted. Let us agree that $Q$ is a function of $t$ only and consider $\theta$ as just a parameter. First, let us look at what the density of $Q(T,\theta)$ is: \begin{align*} f_{Q}(q|\theta) dq &= \text{Prob}[ q\leqslant Q(T,\theta) < q + dq]\\ & = \text{Prob}[ Q^{-1}(q,\theta) \leqslant T < Q^{-1}(q,\theta) + \frac{dq}{|Q'(Q^{-1}(q,\theta),\theta)|}] \end{align*} The size of the differential follows from straightforward differential calculus. The question to ask is: for a change $dq$, how large is $dt$? If $Q$ is differentiable and invertible, then $q=Q(t,\theta)$ means both $\frac{dq}{dt}=Q'(t,\theta)$ and $t=Q^{-1}(q,\theta)$. Combining both gives $dt=\frac{dq}{Q'(Q^{-1}(q,\theta),\theta)}$. The absolute value is there because in the above probability notation, only the size of the differentials matter, not their sign. So we have $$ f_{Q}(q|\theta) = f(Q^{-1}(q,\theta)|\theta) \frac{1}{|Q'(Q^{-1}(q,\theta),\theta)|} $$ or, equivalently, with $t=Q^{-1}(q,\theta)$, $$ f_{Q}(Q(t,\theta)|\theta) = f(t|\theta) \frac{1}{|Q'(t,\theta)|} $$ Now, if $Q(T,\theta)$ is a pivot, that means its distribution does not depend on $\theta$. So its density $f_Q(\cdot |\theta)$ should really be $f_Q(\cdot)$, or in other notation, some function $f_Q(\cdot|\theta) = g(\cdot)$ independent of $\theta$. If we plug that in, we get $$ g(Q(t,\theta)) |Q'(t,\theta)| = f(t|\theta) $$

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  • $\begingroup$ Could you explain steps 2 and 3? please. Why $Q^{-1}(q+dq)=Q^{-1}(q,\theta)+\frac{dq}{|Q'(Q^{-1}(q,\theta),\theta)|}$? $\endgroup$ – Héctor Feb 21 '16 at 14:09

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