8
$\begingroup$

What is the Jeffreys prior for the geometric distribution?

$\endgroup$
13
$\begingroup$

The geometric distribution is given by:

$$p(X|\theta)=(1-\theta)^{X-1}\theta \;\;\; X=1,2,3,\dots$$

The log likelihood is thus given by:

$$\log[p(X|\theta)]=L=(X-1)\log(1-\theta)+\log(\theta)$$

Differentiate once:

$$\frac{\partial L}{\partial \theta}=\frac{1}{\theta}-\frac{X-1}{1-\theta}$$

And again:

$$\frac{\partial^{2} L}{\partial \theta^{2}}=-\frac{1}{\theta^{2}}-\frac{X-1}{(1-\theta)^{2}}$$

Take the negative expectation of this conditional on $\theta$ (called Fisher information), note that $E(X|\theta)=\frac{1}{\theta}$

And so we have:

$$I(\theta)=\frac{1}{\theta^{2}}+\frac{\theta^{-1}-1}{(1-\theta)^{2}}=\theta^{-2}\left(1+\frac{\theta}{1-\theta}\right)=\theta^{-2}(1-\theta)^{-1}$$

The Jeffreys prior is given by the square root of this:

$$p(\theta|I) \propto \sqrt{I(\theta)}=\theta^{-1}(1-\theta)^{-\frac{1}{2}}$$

$\endgroup$
  • 4
    $\begingroup$ It's worth adding that this prior is improper. $\endgroup$ – cardinal Mar 29 '11 at 0:09
  • 1
    $\begingroup$ But this is only the prior, the posterior is always proper, as $X\geq 1$ $\endgroup$ – probabilityislogic Mar 29 '11 at 8:05
  • $\begingroup$ the above answer is wrong because the likelihood of Geometric distribution is L(.)=(P^(n))*(1-p)^(summation(X) -n). it will be (1-p)^(-0.5). $\endgroup$ – user94054 Nov 4 '15 at 13:36
  • 1
    $\begingroup$ @Hamid - the likelihood is actually fine. For example, it adds up to 1, because of the well known geometric sum $\sum_{x=1}^{\infty }(1-\theta)^x=\frac {1-\theta}{\theta} $. So the only way it could be "wrong" was by mis-interpreting what distribution the OP meant. The answer was accepted - so I think it's fine on that count. Also, the likelihood you've written looks like a binomial one, not geometric (eg what is 'n'?) $\endgroup$ – probabilityislogic Nov 6 '15 at 1:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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