TITLE: How to find the covariance of sample mean and sample variance $Cov(M,S^2)$ for Poisson distribution?
QUESTION [7 upvotes]: Suppose that I have a Poisson distribution $P(\lambda)$.
Let $X_1,X_2,\ldots,X_n$ be independent random variables from the distribution mentioned above.
Let us define sample variance $S^2 = \frac{1}{n-1} \sum (X_i - M)^2 $ and sample mean as $M = \frac{1}{n}\sum X_i $.
I want to find the covariance, $Cov(M,S^2)$. I've seen from this answer that when the distribution is symmetric, they are uncorrelated, which makes it zero. I also know that for large $\lambda$ values Poisson distribution is very close to Gaussian distribution, thus becoming symmetric, and probably the covariance is close to zero.
However, I want to find the exact value of $Cov(M,S^2)$, as I am working with small values of $\lambda$

Currently I tried the following:
$Cov(M,S^2) =E( (M-\lambda) (S^2-\lambda) ) = E(MS^2)-\lambda E(M) - \lambda E(S^2) + \lambda^2$
Thus,
$Cov(M,S^2) = E(MS^2) - \lambda^2 - \lambda^2 + \lambda^2  = E(MS^2) -\lambda^2$
I am having trouble with calculating $E(MS^2)$.
Any idea?

REPLY [6 votes]: One can solve this more generally. For any random variable whose moments exist, the covariance between the sample mean $M$ and the sample variance $S^2$ is:
$$\text{Cov}(M, S^2) = \frac{\mu_3}{n}$$
where $\mu_3$ denotes the $3^{\text{rd}}$ central moment of the parent random variable $X$. 
If $X \sim \text{Poisson}(\lambda)$, then $\mu_3 = \lambda$, so the answer is $\large\frac{\lambda}{n}$.