# Creating a uniform prior on the logarithmic scale

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?

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.

• 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. – suncoolsu Feb 21 '11 at 6:20

@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.

• +1 for the additional references. My favorite derivation for the change of variables formula starts with the cdf, like in your other answer. – JMS Mar 27 '11 at 15:31
• @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 – probabilityislogic Mar 27 '11 at 22:40
• 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. – JMS Mar 28 '11 at 1:56

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.