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.

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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