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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?

Jeromy Anglim
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Vass
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3 Answers3

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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.

JMS
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We are told that the scale parameter is uniform on the logarithmic scale. That means that if x is the scale parameter, then $y=\log()$ 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 $p_X()=p_Y(y(x))|\frac{dy}{dx}|$. Since $\frac{dy}{dx} \propto \frac{1}{x}$, we obtain $p_X() \propto \frac{1}{x}$.

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, $p_()$ 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 $p_X()=p_Y(y(x))|\frac{dy}{dx}|$; right now it is given as $p_X()=p_Y(x(y))|\frac{dx}{dy}|$. Thirdly, $y=\log()$, not $y=\exp(x)$, due to the reason explained in this answer.

Ioana Zelko
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  • Why are we told that the scale parameter is uniform on the logarithmic scale? Who decided this? – Maverick Meerkat Jan 21 '21 at 18:51
  • 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? – Ioana Zelko Jan 21 '21 at 19:31
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@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.

Community
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probabilityislogic
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  • +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 01:56
  • can you explain why we want this transformation? Why we want the variance invariant on the log scale? – Maverick Meerkat Jan 21 '21 at 18:52