# How to find the covariance of sample mean and sample variance $Cov(M,S^2)$ for Poisson distribution?

Suppose that I have a Poisson distribution $P(\lambda)$.

Let $X_1,X_2..,X_n$ 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?

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A square is missing in the definition of $S^2$, isn't it? – Davide Giraudo Nov 12 '12 at 13:00
@DavideGiraudo, you are right, I made a typo. Thanks for pointing that out. – Andrey Rubshtein Nov 12 '12 at 13:09
We can expand the square in the definition of $S^2$ to get a more simple expression, then express $E(MS^2)$ as a double sum. You will have to distinguish when the indexes are the same or not. – Davide Giraudo Nov 12 '12 at 13:13

Let us first center everything, using $\bar X_k=X_k-\lambda$ and $\bar M=M-\lambda$. Then $$\mathrm{Cov}(M,S^2)=\mathbb E(\bar MS^2)=\mathbb E(\bar X_1S^2)=\frac1{n-1}\mathbb E(\bar X_1U),$$ where $$U=\sum\limits_{k=1}^n(\bar X_k-\bar M)^2.$$ Note that $U$ is a linear combination of products $\bar X_k^2$ and $\bar X_k\bar X_i$ for $i\ne k$. Amongst these products, many will not contribute to the expectation of $\bar X_1U$ since $\mathbb E(\bar X_1\bar X_k\bar X_i)=0$ for every $k\ne i$ and $\mathbb E(\bar X_1\bar X_k^2)=0$ for every $k\ne1$.
Hence, one needs only the coefficient of $\bar X_1^2$ in $U$, which is $c_n=\left(\frac{n-1}n\right)^2+(n-1)\frac1{n^2}=\frac{n-1}n$. This yields $\mathbb E(\bar X_1U)=c_n\mathbb E(\bar X_1^3)$ and $\mathrm{Cov}(M,S^2)=\frac1{n-1}c_n\mathbb E(\bar X_1^3)=\frac1n\mathbb E(\bar X_1^3)$.
Finally, the third central moment of the Poisson distribution with parameter $\lambda$ is $\mathbb E(\bar X_1^3)=\lambda$ hence $$\mathrm{Cov}(M,S^2)=\frac\lambda{n}.$$
The argument needs some refining. When $\lambda\to\infty$, what happens is really that the scaled and centered versions are more and more gaussian-like, hence symmetric. The proper scaling is to center each $X_k$ and to divide it by $\sqrt{\lambda}$ hence one should divide $M$ by $\sqrt{\lambda}$ and $S^2$ by $\lambda$. Thus, $\mathrm{Cov}(M,S^2)/\lambda^{3/2}$ should go to zero when $\lambda\to\infty$ (and it does). – Did Nov 12 '12 at 13:59