Is $\sum X_i + \sum X_i^2$ a sufficient statistics for $g(\mu,\sigma^2)$? If $(\sum X_i ,\sum X_i^2)$ is a sufficient statistics for $(\mu,\sigma^2)$ , is $\sum X_i + \sum X_i^2$ a sufficient statistics for $g(\mu,\sigma^2)$?
I am expected to find the MVUE of parameter $\gamma = E(X^3)$ = $\mu^{3}+3 \mu \sigma^{2}$ where $X\sim N(\mu,\sigma^2)$, the unbiased estimator at hand is $1/n\sum X_i^3$ and I want to refine this estimator to MVUE by using Rao–Blackwell theorem. However, $\gamma$ is a function of both $\mu$ and $\sigma^2$ and I am not sure how to find the sufficient statistics of $\gamma$. The factorization criterion only allows me to conclude  $(\sum X_i ,\sum X_i^2)$ is a sufficient statistics for $(\mu,\sigma^2)$.
Intuitively I know sufficient statistics basically says by having the information of $(\sum X_i ,\sum X_i^2)$, any additional information won't help us better predict the parameters... I am just wondering since $\gamma$ is a function of both $\mu$ and $\sigma^2$, I can simply deduce a linear combination of $\sum X $ and $\sum X^2$ is also a sufficient statistics for $\gamma$? If not, what should I do? If yes, why?
 A: Just because you have to estimate a function of $\theta=(\mu,\sigma^2)$, does not mean you have to look for a different sufficient statistic. The statistic $T=(\sum X_i,\sum X_i^2)$ is sufficient for the $N(\mu,\sigma^2)$ family of distributions where $\theta$ is the parameter of interest. So $T$ is sufficient for $\theta$ and also trivially sufficient for any function $g(\theta)$ of $\theta$. But I don't have an answer to the specific question in your title.
For your problem, you only need an unbiased estimator of $\mu^3+3\mu\sigma^2$ based on $T$. Because $T$ is complete, this unbiased estimator is guaranteed to be the MVUE by Lehmann-Scheffé theorem.
The sample mean $\overline X$ and sample variance $S^2$ with divisor $n-1$ are independently distributed for a Normal population. So we readily have the MVUE of $\mu\sigma^2$:
$$E(\overline XS^2)=E(\overline X)E(S^2)=\mu\sigma^2$$
Recall that odd-ordered central moments vanish for a symmetric distribution. In particular, $$E(\overline X-\mu)^3=0 \tag{$\star$}$$
Using the distribution of $\overline X$, an unbiased estimator of $\mu^3$ can be found by simplifying $(\star)$ and replacing $\sigma^2$ by its unbiased estimator $S^2$. The resulting MVUE would be a function of $(\overline X,S^2)$. Now you can combine both MVUEs to get the MVUE of $\mu^3+3\mu\sigma^2$.
