Jeffreys prior for geometric distribution?

What is the Jeffreys prior for the geometric distribution?

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:

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

• It's worth adding that this prior is improper. Mar 29, 2011 at 0:09
• But this is only the prior, the posterior is always proper, as $X\geq 1$ Mar 29, 2011 at 8:05
• 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).
– user94054
Nov 4, 2015 at 13:36
• @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'?) Nov 6, 2015 at 1:13