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Year 2 of the mathematics course of the University of Oxford

Options I course of the University of Oxford

Past exam of the options I course of the University of Oxford

2012

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.

Solution

Past exam of the options I course of the University of Oxford / 2012 / O / O1 / c / i / Solution

Past exam of the options I course of the University of Oxford / 2012 / O / O1 / c / i / Solution

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