I would like to ask that, there is a question asking to show that $\bar{X}$ is a minimum variance unbiased estimator of the mean $\mu$ of a normal distribution.
I am having difficulty understanding the solution, it took log of the normal p.d.f., and differentiate the natural log respect to $\mu$, and it wrote as: hence, $E[(\partial ln f(x)/\partial\mu)^2=(1/\sigma^2)\times E[((x-\mu)/\sigma)^2=(1/\sigma^2)\times1$ Now I am having trouble understand why the expected value of $((X-\mu)/\sigma)^2$ would be 1.
Thank you very much for your reading and any help would be highly appreciated! Thanks!
$$ E\left[\frac{(X-\mu)^2}{\sigma^2}\right]=\frac{1}{\sigma^2}E\left[(X-\mu)^2\right]=\frac{\sigma^2}{\sigma^2} $$