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A question with a solution that I don't quite get: asking for the Cramér-Rao lower bound of a random Poisson sample.

If we take the log of the function $f(x; \theta)$ and take its first derivative with respect to theta,it becomes $(x-\theta)/\theta$ (which is the score function $S(x;\theta)$) and if we find the fisher information of that, it's $E[S(X;\theta)^2]$ which then becomes $E[[X-\theta]^2]/\theta^2]$.

The solution says this leads to $1/\theta$. Can anyone please explain how $E[[X-\theta]^2]/\theta^2]$ leads to $1/\theta$?

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The variance and mean of a Poisson distribution are equal, so $E[(x-\theta)^2]=\theta$ and $$E\left[\frac{(x-\theta)^2}{\theta^2}\right]=\theta/\theta^2=1/\theta$$

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  • $\begingroup$ Thank you for your answer! Also, it says the Rao Cramer Lower Bound is theta/n, and since this is the variance of X-bar, hence X-bar is an efficient estimator of theta. Is this always the case ? (if the variance is equal to the RCLB, does theta always become an efficient estimator)? $\endgroup$ Commented Jun 16, 2020 at 6:58
  • $\begingroup$ Yes. The bound is the reciprocal of the Fisher information, divided by the sample size, so $(1/(1\theta))/n= \theta/n$. And we know $\mathrm{var}[X]/n=\theta/n$ is always the variance of the sample mean, so the sample mean attains the bound in this case. $\endgroup$ Commented Jun 16, 2020 at 7:01
  • $\begingroup$ Thank you, that just clarified everything ! :) $\endgroup$ Commented Jun 16, 2020 at 7:10

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