Let $X_1,X_2...X_n$ be a random sample from $N(\mu,\sigma^2)$. Find the umvue of $\mu^3$. Note: I know lehman scheffe theorem and that sample mean is umvue of $\mu$. But how can we find the UMVUE of hiher powers of $\mu $?
 A: We know the distribution of the sample mean, namely $$\bar X\sim\mathcal N\left(\mu,\frac{\sigma^2}{n}\right)$$
If $\sigma$ is known, a complete sufficient statistic for $\mu$ is $$\sum_{i=1}^n X_i \quad(\text{ and hence }\bar X)$$
Now, 
\begin{align}
\operatorname{Var}(\bar X)&=\frac{\sigma^2}{n}
\\\implies E(\bar X^2)&=\frac{\sigma^2}{n}+\mu^2
\end{align}
That is, $$E\left(\bar X^2-\frac{\sigma^2}{n}\right)=\mu^2 $$
Odd-ordered central moments of a symmetric distribution is zero.
So, 
\begin{align}
&E(\bar X-\mu)^3=0
\\\implies &E(\bar X^3)-\mu^3-3\mu E(\bar X^2)+3\mu^2E(\bar X)=0
\\\implies & E(\bar X^3)-\mu^3-3\mu \left(\mu^2+\frac{\sigma^2}{n}\right)+3\mu^3=0
\\\implies & E(\bar X^3)-3\mu\frac{\sigma^2}{n}=\mu^3
\end{align}
That is, $$E\left(\bar X^3-3\frac{\sigma^2}{n}\bar X\right)=\mu^3$$
As this unbiased estimator of $\mu^3$ is a function of the complete sufficient statistic $\bar X$, by the Lehmann-Scheffe theorem, the UMVUE of $\mu^3$ for known $\sigma$ is $$\bar X^3-3\frac{\sigma^2}{n}\bar X$$
For UMVUE of $\mu^r$ ( for some $r\in\mathbb N$, say ), one could try to find an unbiased estimator of $\mu^r$ starting from $E(\bar X-\mu)^r$ (separate cases for even and odd values of $r$), or find $E(\bar X^r)$ directly. To make it an 'estimator', one would have to substitute the lower powers of $\mu$ by their respective UMVUEs as done here.
If $\sigma$ is unknown, then a complete sufficient statistic for the family is $(\bar X,S^2)$ where $S^2$ is the sample variance. 
So by the above calculation, one would arrive at $$E\left(\bar X^3-\frac{3}{n}\bar X S^2\right)=\mu^3$$
By Lehmann-Scheffe, $$\bar X^3-\frac{3}{n}\bar X S^2$$ is the UMVUE of $\mu^3$ when $\sigma$ is unknown. Note the independence of $\bar X$ and $S^2$.
