# Uniformly minimum variance unbiased estimator

How to prove $\overline{X}=\frac{1}{n}\sum_{i=1}^nX_i$ is the uniformly minimum variance unbiased estimator of $\mu$ when $X_i\sim N(\mu,\sigma^2),$ and $\sigma$ is known.

Idea: Let $X=(X_1,X_2,...,X_n)$, then we need to prove $E(\overline{X}-\mu)^2\leq E(f(X)-\mu)^2$ for any $f(X)$, an unbiased estimator of $\mu.$ Since $\overline{X}$ is sufficient and complete statistics, then by Lehmann-Scheffe theorem, we can easily get $\overline{X}$ is uniformly minimum variance unbiased estimator of $\mu$. Can someone directly prove this statement without applying the theorem?

• Have you thought about using the Cramer-Rao lower bound? Jul 22, 2017 at 17:55

Assuming $$X_i$$'s are independently distributed, joint density of $$X_1,\ldots,X_n$$ for $$\mu\in\mathbb R,\sigma>0$$ is
$$f_{\mu}(x_1,\ldots,x_n)\propto \exp\left\{-\frac1{2\sigma^2}\sum_{i=1}^n (x_i-\mu)^2\right\}\quad,\,(x_1,\ldots,x_n)\in\mathbb R^n$$
Or, $$\ln f_{\mu}(x_1,\ldots,x_n)=\text{constant }-\frac1{2\sigma^2}\sum_{i=1}^n (x_i-\mu)^2$$
$$\frac{\partial}{\partial \mu}\ln f_{\mu}(x_1,\ldots,x_n)=\frac1{\sigma^2}\sum_{i=1}^n (x_i-\mu)=\frac{n}{\sigma^2}\left(\frac1n\sum_{i=1}^nx_i-\mu\right)$$
In other words, the score function $$\frac{\partial}{\partial \mu}\ln f_\mu( X_1,\ldots,X_n)$$ is proportional to $$T(X_1,\ldots,X_n)-\mu$$ for some statistic $$T$$. By the equality condition of Cramér-Rao inequality, variance of $$T$$ attains the Cramér-Rao lower bound for $$\mu$$. Here of course $$T=\frac1n\sum\limits_{i=1}^n X_i$$ with $$E_{\mu}(T)=\mu$$ for every $$\mu$$, so that $$T$$ is the uniformly minimum variance unbiased estimator of $$\mu$$.