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.

  • $\begingroup$ 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
    Commented Feb 4, 2018 at 17:35

1 Answer 1


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 $p(\theta) \propto \theta^{-1}$.

  • $\begingroup$ I'm not quite sure I follow the second line here. What is the justification for that chain rule-like substitution? $\endgroup$
    – gogurt
    Commented Jan 25, 2019 at 0:47

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