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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?

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

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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, $$\ln f_{\mu}(x_1,\ldots,x_n)=\text{constant }-\frac1{2\sigma^2}\sum_{i=1}^n (x_i-\mu)^2$$

Therefore,

$$\frac{\partial}{\partial \mu}\ln f_{\mu}(x_1,\ldots,x_n)=\frac1{\sigma^2}\sum_{i=1}^n (x_i-\mu)=\frac{n}{\sigma^2}\left(\frac1n\sum_{i=1}^nx_i-\mu\right)$$

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 $\mu$.

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