A uniform prior for a scale parameter (like the variance) is uniform on the logarithmic scale.

What functional form does this prior have on the linear scale? And why so?


3 Answers 3


It's just a standard change of variables; the (monotone & 1-1) transformation is $y = \exp(x)$ with inverse $x=\log(y)$ and Jacobian $\frac{dx}{dy} = \frac{1}{y}$.

With a uniform prior $p_y(y) \propto 1$ on $\mathbb{R}$ we get $p_x(x) = p_y(x(y)) |\frac{dx}{dy}| \propto \frac{1}{y}$ on $(0, \infty)$.

Edit: Wikipedia has a bit on transformations of random variables: http://en.wikipedia.org/wiki/Probability_density_function#Dependent_variables_and_change_of_variables. Similar material will be in any intro probability book. Jim Pitman's "Probability" presents the material in a pretty distinctive way as well IIRC.

  • 3
    $\begingroup$ just adding to @JMS answer, a great reference to check out about is Bayesian Inference in Statistical Analysis by Box and Tiao. It presents conceptual ideas behind it too. $\endgroup$
    – suncoolsu
    Feb 21, 2011 at 6:20
  • $\begingroup$ Can you explain why we want this transformation? $\endgroup$ Jan 21, 2021 at 18:52

We are told that the scale parameter is uniform on the logarithmic scale. That means that if x is the scale parameter, then $y=\log(𝑥)$ and the distribution function for $y$ is the one uniform on the logarithmic scale, $p_Y(y) \propto 1$.

Then, applying the Jacobian transformation which comes from the fact that the probability contained in a differential area must be invariant under change of variables, we must have that $p_X(𝑥)=p_Y(y(x))|\frac{dy}{dx}|$. Since $\frac{dy}{dx} \propto \frac{1}{x}$, we obtain $p_X(𝑥) \propto \frac{1}{x}$.

Note: I tried posting this as a comment but I do not have privileges to post comments because I am a new user. The currently accepted answer to the question (given by @JMS) has errors in it. I tried to edit the answer given by @JMS to make the minimum necessary changes but my edit was rejected because people wanted me to put this as a comment or as an answer. Firstly, $p_𝑋(𝑥)$ should end up being a function of $x$, not a function of y. The way @JMS's the answer is phrased right now gives $p_X(x)\propto\frac{1}{y}$. Secondly, there is an error in the Jacobian formulation, it should be $p_X(𝑥)=p_Y(y(x))|\frac{dy}{dx}|$; right now it is given as $p_X(𝑥)=p_Y(x(y))|\frac{dx}{dy}|$. Thirdly, $y=\log(𝑥)$, not $y=\exp(x)$, due to the reason explained in this answer.

  • $\begingroup$ Why are we told that the scale parameter is uniform on the logarithmic scale? Who decided this? $\endgroup$ Jan 21, 2021 at 18:51
  • $\begingroup$ It was part of the problem the poser of the question was trying to solve: i.e: if a scale parameter is uniform on the logarithmic scale,what functional form does this prior have on the linear scale? $\endgroup$ Jan 21, 2021 at 19:31

@JMS answer is adequate for the nuts and bolts of changing variables. However, This question may help you a bit with why it is uniform on that scale.

My answer to this question goes through a slightly longer derivation of the "jacobian rule" result given in @JMS's answer. It may help with understanding why the rule applies.

  • $\begingroup$ +1 for the additional references. My favorite derivation for the change of variables formula starts with the cdf, like in your other answer. $\endgroup$
    – JMS
    Mar 27, 2011 at 15:31
  • $\begingroup$ @JMS - the cdf rule is the only one I don't get confused with, I usually struggle to remember if its $\frac{dy}{dx}$ or $\frac{dx}{dy}$ with the jacobian $\endgroup$ Mar 27, 2011 at 22:40
  • $\begingroup$ same for me - Pitman gives a nice geometric explanation, which is why I referenced it in my answer, but I can't ever remember it when it counts :) When I TA'd a probability class we used this text and some students found it very helpful. $\endgroup$
    – JMS
    Mar 28, 2011 at 1:56
  • $\begingroup$ can you explain why we want this transformation? Why we want the variance invariant on the log scale? $\endgroup$ Jan 21, 2021 at 18:52

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