Can we conclude that $\bar{X}$ and $S^2$ are also uncorrelated? (Solution Verification)

58 Views Asked by At

Consider a simples random sample $ X_1, ..., X_n $ taken from a population with mean $\mu$ and variance $\sigma^2$.

(i) Show that $\bar{X}$ and the deviation $X_i - \bar{X}$ are uncorrelated statistics.

(ii) With this result, can we conclude that $\bar{X}$ and $S^2 = \sum_{i=1}^{n}(X_i - \bar{X})^2/(n-1)$ are also uncorrelated?

I did item (i), but I'm unsure if my answer to (ii) is entirely correct. I answered as follows:

"We cannot conclude, generally, that $ \text{Cov}(\bar{X}, X_i) = 0 $ for any population with mean $\mu$ and variance $\sigma^2 \Rightarrow \text{Cov}(\bar{X}, S^2) = 0$. A problem proposed previously proves that $\text{Cov}(\bar{X}, S^2) = \mu_3/n$, where $\mu_3$ is the third central moment of the population. So it depends on the population."

Question: Is this answer 100% correct?