We all know $E\left[\left(X_{i}-\overline{X}\right)^{2}\right] = \frac{n-1}{n}\sigma^{2} = \frac{n-1}{n}E\left[\left(X_{i}-\mu\right)^{2}\right]$ where $\overline{X}$ is the average of a sample with size $n$ that includes $X_{i}$.
Can $E\left[\left(X_{i}-\overline{X}\right)^{3}\right]$ be nicely written in terms of $E\left[\left(X_{i}-\mu\right)^{3}\right]$? If so, what is that formula?
This is not as easy as I thought it would be.
According to this answer which cites page 7 of this document, $$E[(X_i - \bar{X})^3] = \frac{(n-1)(n-2)}{n^2} E[(X_i-\mu)^3].$$ Proving this amounts to expanding $(X_i - \bar{X})^3$ and doing a bit of accounting for each term.