What is the Jeffreys prior for the geometric distribution?


1 Answer 1


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:


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:


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

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

  • 5
    $\begingroup$ It's worth adding that this prior is improper. $\endgroup$
    – cardinal
    Mar 29, 2011 at 0:09
  • 1
    $\begingroup$ But this is only the prior, the posterior is always proper, as $X\geq 1$ $\endgroup$ Mar 29, 2011 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, 2015 at 13:36
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    $\begingroup$ @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'?) $\endgroup$ Nov 6, 2015 at 1:13

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