1
$\begingroup$

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 $1/\theta$?

$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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)? $\endgroup$ – mathmathmathmath Jun 16 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. $\endgroup$ – Thomas Lumley Jun 16 at 7:01
  • $\begingroup$ Thank you, that just clarified everything ! :) $\endgroup$ – mathmathmathmath Jun 16 at 7:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.