Proof: Pivotal Quantity

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

Suppose that $$T$$ is 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:

$$$$f(t|\theta)=g(Q(t, \theta))\left|\frac{d}{dt}Q(t,\theta)\right|$$$$ for some function $$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.

1 Answer

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 $$g(Q(t,\theta)) |Q'(t,\theta)| = f(t|\theta)$$

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