2
$\begingroup$

Scale model: $x\sim \frac{1}{\theta} f(\frac{x}{\theta})$.

How can I prove that $\pi(\theta)\propto \frac{1}{\theta}$ is also a Jeffreys prior?

I was trying to somehow change variables, using a log-transformation, but I don't seem to get it right.

Any help would be appreciated.

$\endgroup$
  • $\begingroup$ A usual answer to this question is "because the Jeffreys prior on location parameters is constant". When considering $\log(X)=Y$, $Y$ has indeed a location distribution with parameter $\xi=\log(\theta)$. (If $X$ has a sign, its sign is an ancillary statistic and one can instead consider $|X|$.) $\endgroup$ – Xi'an Feb 4 '18 at 17:35
3
$\begingroup$

A solution seems to make the change of variable $z=x/\theta$. Then the density of z writes (normalizing the differential areas) $p_Z(z)=f(z)$. Let then compute the information matrix from this parametrisation (assuming twice differentiation and regularity of $f$ and noting the logarithm of $f$ as $lf(z)$): $$ I(\theta)= - \int \frac{\partial^2 lf(z)}{\partial \theta^2} f(z) dz $$ $$ = - \int \frac{\partial^2 lf(z)}{\partial z^2} \frac{\partial^2 z}{\partial \theta^2} f(z) dz $$ $$ = - \int \frac{\partial^2 lf(z)}{\partial z^2} \frac{\partial^2 x/\theta}{\partial \theta^2} f(z) dz $$

$$ = - \int \frac{1}{\theta^2} \frac{\partial^2 lf(z)}{\partial z^2} 2 z f(z)dz $$

$$ = \frac{1}{\theta^2} K_f $$ where $K_f$ depends only on $f$.

Then the Jeffreys gives $p(\theta) \propto \theta^{-1}$.

$\endgroup$
  • $\begingroup$ I'm not quite sure I follow the second line here. What is the justification for that chain rule-like substitution? $\endgroup$ – gogurt Jan 25 at 0:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.