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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 $b$?

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  • $\begingroup$ It's not obvious. stats.org.uk/priors/noninformative/YangBerger1998.pdf $\endgroup$
    – HStamper
    Commented 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
    Commented 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
    Commented Feb 18, 2016 at 17:27
  • $\begingroup$ @EricMittman I'll edit the question so that this is clear. $\endgroup$
    – ADB
    Commented Feb 18, 2016 at 22:39

1 Answer 1

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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.

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  • $\begingroup$ 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
    Commented Feb 18, 2016 at 7:45
  • $\begingroup$ Indeed, everything but $b$ is a constant for the proportionality $\propto$ sign. $\endgroup$
    – Xi'an
    Commented Feb 18, 2016 at 8:39

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