4
$\begingroup$

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?

$\endgroup$
2
  • 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

4
$\begingroup$

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 $\eta$:

set.seed(2023)
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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.