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Suppose that we have $X_1,...,X_n$ iid observations from a Geometric($p$) distribution.

I found that the MLE of p is $\hat{p} = \frac{n}{n+\sum{X_i}}$.

I am trying to find the MLE of $\theta = \frac{p}{1-p}, \hat{\theta}$ and the Fisher Information of a single observation $I_1(\theta)$.

Now, by the functional invariance property of MLE estimators, I found that the MLE of $\hat{\theta}$ is just:

$\hat{\theta}$= $\frac{\hat{p}}{1-\hat{p}}$.

However, I have no idea how to find the Fisher Information $I_1(\theta)$.

I know of two ways of doing this, one is to exploit the asympotic efficiency and the lower bound princple of MLE's. That is, $Var(\hat{\theta})=\frac{1}{nI_1(\theta)}$.

Also, is the parametrization formula

${\mathcal I}_\eta(\eta) = {\mathcal I}_\theta(\theta(\eta)) \left( \frac{{\mathrm d} \theta}{{\mathrm d} \eta} \right)^2$ where ${\mathcal I}_\eta$ and ${\mathcal I}_\theta$ are the Fisher information measures of $η$ and $θ$, respectively.

Is there another easier way to do this? I have no idea how to find the variance above. Thank you!

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By the formula for the MLE, I understand that you are dealing with the variant of the Geometric distribution where the random variables can take the value $0$. In this case we have

$$E(X_1) = \frac {1-p}{p},\,\,\, \text {Var}(X_1) = \frac {1-p}{p^2}$$

The Fisher Information of a single observation can be derived by applying its definition :

$$I_1(p) = \operatorname{E} \left[\left. \left(\frac{\partial}{\partial p} \ln f(X_1;p)\right)^2\right|p\right] = \operatorname{E} \left[\left. \left(\frac{\partial}{\partial p} \ln(1-p)^{X_1}p \right)^2\right|p \right]$$

$$=\operatorname{E} \left(-\frac {X_1}{1-p}+\frac 1p \right)^2 = \operatorname{E} \left(\frac {X_1^2}{(1-p)^2}+\frac 1{p^2}-2\frac {X_1}{(1-p)p}\right)$$

$$=\frac 1{p^2} - \frac {2}{(1-p)p} E(X_1)+ \frac {1}{(1-p)^2}\left(\text {Var}(X_1) + (E[X_1])^2\right)$$ $$=\frac 1{p^2}- \frac {2}{(1-p)p}\cdot \frac {1-p}{p} + \frac {1}{(1-p)^2}\left( \frac {1-p}{p^2} + \frac {(1-p)^2}{p^2}\right)$$

$$=\frac 1{p^2}- \frac {2}{p^2}+\frac {1}{(1-p)p^2}+\frac 1{p^2} = \frac {1}{(1-p)p^2}$$

We also have $$\frac {d\theta}{dp} = \frac {1}{(1-p)^2}$$ So $$I_1(\theta) = I_1(p)\cdot \left(\frac {d\theta}{dp} \right)^{-2} = \frac {1}{(1-p)p^2}\cdot (1-p)^4 = \frac {(1-p)^3}{p^2}$$

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