 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2012/o/o1/c/i/solution

For the stated geometric distribution, write

S = \sum_{i} X_{i}

. Then

ℓ_{n} (θ) = n lo g θ + (S - n) lo g (1 - θ), - ℓ_{n}^{''} (θ) = \frac{n}{θ ^{2}} + \frac{S - n}{( 1 - θ ) ^{2}} .

Since

E X_{i} = 1/ θ

, we have

E (S - n) = n (1 - θ) / θ

I_{n} (θ) = \frac{n}{θ ^{2}} + \frac{n}{θ ( 1 - θ )} = \frac{n}{θ ^{2} ( 1 - θ )} .

The support begins at one, so this expectation uses the trials-until-first-success convention of the geometric law.

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