= Solution
For the stated <geometric distribution>, write $S=\sum_iX_i$. Then
$$
\ell_n(\theta)=n\log\theta+(S-n)\log(1-\theta),
\qquad
-\ell_n''(\theta)=\frac n{\theta^2}+\frac{S-n}{(1-\theta)^2}.
$$
Since $\mathbb E X_i=1/\theta$, we have $\mathbb E(S-n)=n(1-\theta)/\theta$. Taking <expectation> gives the sample <Fisher information>
$$
I_n(\theta)=\frac n{\theta^2}+\frac n{\theta(1-\theta)}
=\boxed{\frac n{\theta^2(1-\theta)}}.
$$
The support begins at one, so this <expectation> uses the trials-until-first-success convention of the geometric law.
Back to article page
