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I'd like to ask if the below computation of the information number for the CRLB is correct:

Consider $x_1, x_2$ as iid $\sim Geo(\theta)$

Since $x_1, x_2$ ae iid and the geometric distribution is part of the exponential family, the information number is

$-nE_\theta(\frac{d^2}{d\theta^2}lnf(x|\theta))=-2E_\theta[\frac{d^2}{d\theta^2}ln(\theta(1-\theta)^{x-1})]=\frac{\theta^2(1-\theta)}{2}$

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I'll check, it equals: \begin{align*} -2 E \left[ \frac{d^2}{d \theta^2} \left( \ln ( \theta) + (x-1) \ln(1- \theta) \right) \right] &= -2 E \left[ \frac{d}{ d \theta} \left( \frac{1}{\theta} + (x-1) \frac{-1}{1-\theta} \right) \right] \\ &= -2 E \left[ - \frac{1}{\theta^2} + (x-1) \left(- \frac{1}{(1-\theta)^2} \right) \right] \\ &= \frac{2}{\theta^2} - \frac{2}{(1-\theta)^2} + \frac{2}{(1-\theta)^2} E[X] \end{align*}

Assuming the Geometric paramaterization you're using with $E[X] = \frac{1}{\theta}$, this simplifies to $- \frac{2}{\theta^2(\theta-1)}$.

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