# Jeffreys prior for geometric distribution?

What is the Jeffreys prior for the geometric distribution?

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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:

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

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It's worth adding that this prior is improper. –  cardinal Mar 29 '11 at 0:09
But this is only the prior, the posterior is always proper, as $X\geq 1$ –  probabilityislogic Mar 29 '11 at 8:05