The question goes like this: Use Jeffreys prior for Bernoulli distribution and find the prior for $\eta$ where: $$\eta(p) = \left(\frac{p}{1-p}\right) $$

So $\eta$ here is some kind of a transformation on the Bernoulli parameter $p$, and I know that Jeffreys prior $\pi(p)$, for Bernoulli distribution, is:

$$\pi(p) = \sqrt{I(p)} = \sqrt{\frac{1}{p(1-p)}}$$

The solution for the question is:

$$f_n(\eta) = f_p(g^{-1}(\eta))\cdot (g^{-1}(\eta))'$$ $$= \sqrt{\frac{1}{p(1-p)}} \cdot p(1-p)$$ $$= \sqrt{p(1-p)}$$

I'm trying to understand the solution:

$f_p(g^{-1}(\eta))$ is just $\pi(p)$, right? cause I see its equal to $\sqrt{\frac{1}{p(1-p)}}$.

And I didn't get what is $(g^{-1}(\eta))'$.. here its equal to $p(1-p)$ so it can't be the derevative of $\pi(p)$, so what it is then?

And basically they didn't use the fact that $\eta(p) = \left(\frac{p}{1-p}\right)$, they just used the prior for Bernoulli, right?

  • 2
    $\begingroup$ Please add the self-study tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. $\endgroup$ Commented Mar 11, 2023 at 1:22
  • $\begingroup$ Close-voters: OP is showing where they got stuck, so it's a well-posed self-study question. Voting to leave open, and upvoting. $\endgroup$ Commented Mar 12, 2023 at 9:49

1 Answer 1


You know the Jeffreys' prior density for $p$ is proportional to $\sqrt{\frac{1}{p(1-p)}}$ and that the corresponding odds are $\eta=\frac{p}{1-p}$.

This looks as if it is just a change of variables with $g(p)=\frac{p}{1-p}$ and so $g^{-1}(\eta)=\frac{\eta}{1+\eta}$ with derivative $\frac{1}{(1+\eta)^2}$ giving a density for $\eta$ proportional to $\sqrt{\frac{1}{\frac{\eta}{1+\eta}\left(1-\frac{\eta}{1+\eta}\right)}}\frac{1}{(1+\eta)^2} =\frac{1}{(1+\eta)\sqrt{\eta}}$. It is possible to do the integration, so I would have thought that the density of $\eta$ is $$f_\eta(\eta)=\frac{1}{\pi (1+\eta)\sqrt{\eta}}$$ for $\eta>0$.

I cannot see why you might want to substitute back to put this in terms of $p$; a proper change of variables would of course take you back to the Jeffreys' prior of $\frac1\pi\sqrt{\frac{1}{p(1-p)}}$ while a simple substitution would give the rather meaningless $\frac1\pi\sqrt{\frac{(1-p)^3}{p}}$, not what your quoted solution says.

Here is an R simulation demonstration of the density for $\eta$:

p <- rbeta(10^6, 1/2, 1/2) # Jefferys' prior
eta <- p / (1-p)
plot(density(eta, from=0, to=50), log="y")
curve(1 / (pi * (1+x) * sqrt(x)), from=0, to=50, add=TRUE, col="red") 

enter image description here


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