Let $X_1,...,X_n$ be iid random variables with a common density function given by:

$f(x|\theta)=\theta x^{\theta-1}$

for $x\in[0,1]$ and $\theta>0$.

Put a prior distribution on $\theta$ which is $EXP(2)$, where $2$ is the mean of the exponential distribution. Obtain the posterior density function of $\theta$. Do you recognize this distribution?

Clearly $f(x|\theta)$ is a beta distribution. But now I need to determine the posterior distribution. So this implies that the posterior distribution is

$$f(\theta|x)\propto f(x|\theta)\pi(\theta)=\frac{\theta x^{\theta-1}e^{-\theta/2}}{2}$$

But I don't really recognize the distribution at all. Was my calculation incorrect? If not, what is the name of this distribution?


You have $$f_{\Theta|X=x}(\theta)\propto \frac\theta 2 x^{\theta-1}e^{-\tfrac \theta 2}=\frac\theta 2 e^{\log{x^{\theta-1}}}e^{-\tfrac \theta 2}\propto \theta e^{\theta \log x}e^{-\tfrac\theta 2}=\theta^{2-1} e^{-\left(\tfrac12-\log(x)\right)\theta}.$$

So, this is proportional to the density of a $\Gamma\left(2,\tfrac12-\log(x)\right)$, or a $\Gamma\left(2,\tfrac2{1-2\log(x)}\right)$, depending on the convention you are used to.

(If you can't see it, maybe try putting $y$ —or even $x$— instead of $\theta$, and some letter like $b$, $\eta$, etc. instead of $x$.)

  • $\begingroup$ I see it now. Those distributions really know how to hide sometimes. Thanks. $\endgroup$ – ereHsaWyhsipS Feb 14 '18 at 0:29
  • $\begingroup$ I mean it: if you feel lost, change the parameter to $x$ or $y$ and the $x$'s to a letter that you don't associate with a variable but a constant ($\alpha$, $c$, $\zeta$, whatever). It sounds silly, but it works many times. $\endgroup$ – Alejandro Nasif Salum Feb 14 '18 at 1:08
  • $\begingroup$ I appreciate the tip. $\endgroup$ – ereHsaWyhsipS Feb 14 '18 at 1:16

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