5
$\begingroup$

What is the posterior distribution for parameter $b$ with $X \sim Gamma(a,b)$, under the Jeffreys prior? We can assume that $a$ is known.

The Jeffreys prior is the square of the Fisher information of $b$:

$p(b)=\frac{\sqrt(a)}{b}$.

Then using Bayes' rule we have

$p(b|x) \propto p(x|b) \,p(b) = \dfrac{b^a}{\Gamma (a)}x^{a-1}e^{-xb}\cdot\frac{\sqrt(a)}{b}$

Next we look for the kernel of a Gamma distribution. But this is where I am stuck.

What is the next step for deriving the posterior distribution for $b$?

Improve this question
$\endgroup$
4
  • $\begingroup$ It's not obvious. stats.org.uk/priors/noninformative/YangBerger1998.pdf $\endgroup$
    – HStamper
    Feb 18, 2016 at 6:55
  • $\begingroup$ @EricMittman: I do not understand the link since the authors write therein (p.13) that $b$ being a scale parameter the natural choice is $\pi(b)=1/b$. $\endgroup$
    – Xi'an
    Feb 18, 2016 at 7:34
  • $\begingroup$ @Xi'an: I didn't realize that $a$ was considered to be known. I was thinking of the Jeffrey's prior for $(a,b)$. $\endgroup$
    – HStamper
    Feb 18, 2016 at 17:27
  • $\begingroup$ @EricMittman I'll edit the question so that this is clear. $\endgroup$
    – ADB
    Feb 18, 2016 at 22:39

1 Answer 1

Reset to default
5
$\begingroup$

It is difficult to perceive where you get stuck: $$\begin{align*}p(b|x)&\propto \dfrac{b^a}{\Gamma (a)}x^{a-1}e^{-xb}\cdot\dfrac{\sqrt{a}}{b} \\ &\propto b^{a-1} e^{-xb} \\ &\propto \dfrac{x^a\,b^{a-1}}{\Gamma(a)}\,e^{-xb}\end{align*}$$ which shows the posterior is a Gamma $\mathcal{G}(a,x)$ distribution.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I didn't see that $\sqrt{a}$ could be absorbed into the proportionality. Later it seems that $x^a$ appears from nowhere... Is this allowed because in the posterior we are conditioning on $x$, which means that it is fixed (constant)? $\endgroup$
    – ADB
    Feb 18, 2016 at 7:45
  • $\begingroup$ Indeed, everything but $b$ is a constant for the proportionality $\propto$ sign. $\endgroup$
    – Xi'an
    Feb 18, 2016 at 8:39

Your Answer

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

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