Deriving covariance of sample mean and sample variance $\newcommand{\cov}{\operatorname{cov}}$I want to find the covariance between the sample mean and the sample variance, I think I am along the right track but am not sure.

Suppose we have i.i.d. random variables $X_1, X_2, \ldots, X_n$.

Define

$$\overline{X} = \frac{1}{n} \sum_{i=1}^n X_i \text{ and } S^2 = \frac{1}{2n(n-1)} \sum_{i=1}^n \sum_{j=1}^n (X_i - X_j)^2 $$

I want to find their covariance $\cov(\overline{X}, S^2)$. Here is what I have done:

\begin{align}
&amp; \cov(\overline{X}, S^2) = \cov\left(\frac{1}{n}\sum_{i=1}^n X_i, \frac{1}{2n(n-1)} \sum_{i=1}^n\sum_{j=1}^n (X_i - X_j)^2 \right) \\
= {} &amp; \frac{1}{2n^2(n-1)} \sum_{i=1}^n \cov(X_i,\sum_{k=1}^{n}\sum_{j=1}^{n} (X_k - X_j)^2) \\
= {} &amp; \frac{1}{2n^2(n-1)} \sum_{i=1}^n \cov(X_i,2\sum_{k=1}^{n} (X_i - X_k)^2)
\end{align}

(because $\cov(X_i, (X_a - X_b)^2)$ is zero if $i$ isn't $a$ or $b$)

$$= \frac{1}{n^2(n-1)} \sum_{i=1}^{n} \sum_{k=1}^n \cov(X_i, (X_i - X_k)^2).$$

Now, I compute $\cov(X_i, (X_i - X_k)^2)$:

for $ i = k, \cov(X_i, (X_i - X_k)^2) = 0$

for $ i \neq k,$

\begin{align}
&amp; \cov(X_i, (X_i - X_k)^2) = \cov(X_i, X_i^2 - 2X_iX_k + X_k^2) = \cov(X_i, X_i^2) - 2\cov(X_i, X_iX_k) \\[8pt]
= {} &amp; E(X^3) - E(X)E(X^2) -2E(X)E(X^2) + 2(E(X))^3 \\[8pt]
= {} &amp; E(X^3) -3E(X)E(X^2) + 2(E(X))^3.
\end{align}

When I plug this in, I do not get the answer that I want which is $E(X-E(X))^3$, can anyone please tell me where I went wrong?

UPDATE: I plugged it in again and it is actually correct! Sorry for being so careless. Thanks to anyone who has read the question. Should I delete this post?
I hope this will help you to find a mistake. 

https://www.tandfonline.com/doi/pdf/10.1198/000313007X188379
I hope this will help you to find a mistake. 

https://www.tandfonline.com/doi/pdf/10.1198/000313007X188379