# UMVUE of $g(\mu)=2\mu$ for a normal distribution

Let $$X_1,X_2,...,X_n$$ be iid observations from a normal distribution with mean $$\mu$$ and variance $$\sigma^2$$, $$\sigma^2>0$$ is known and $$\mu$$ is an unknown real number. Let $$g(\mu)=2\mu$$ be the parameter of interest. I know that $$(\bar X,S^2)$$ is a complete sufficient statistics for ($$\mu$$,$$\sigma^2$$) where $$\bar X$$ is the sample mean and $$S^2$$ is the sample variance. I am a little confused here. How do I find the UMVUE of $$g(\mu)$$?

• It is unclear where exactly you are stuck. – StubbornAtom Apr 17 at 21:53
• UMVUE of $\mu$ is $\bar X$ so UMVUE of $2\mu$ is $2\bar X.$ – BruceET Apr 18 at 0:12

Using Lehmann-Shceffé Theorem, if $$T$$ is a Sufficient and Complete statistics for $$\mu$$ and $$T^{*}$$ $$=$$ $$h(T)$$ is unbiased for $$g(\mu)$$, then $$T^{*}$$ is UMVUE for $$g(\mu)$$.
The theorem applies to your case since $$T$$ $$=$$ $$\overline{X}$$ is Sufficient and Complete for $$\mu$$, hence $$T^{*}$$ = $$2\overline{X}$$ is UMVUE for $$g(\mu)$$ = $$2\mu$$.