Let $X_1,X_2...X_n$ be a random sample from $N(\mu,\sigma^2)$. Find the umvue of $\mu^3$.

Question

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 $?

Answer 1

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 $S^2$.