 Solution

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

S = \sum_{i} X_{i}

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

Using

E S = n θ

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

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