# Question on Rao-Cramer Lower Bound

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 $$1/\theta$$?

The variance and mean of a Poisson distribution are equal, so $$E[(x-\theta)^2]=\theta$$ and $$E\left[\frac{(x-\theta)^2}{\theta^2}\right]=\theta/\theta^2=1/\theta$$
• 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. – Thomas Lumley Jun 16 at 7:01