# Joint raw moments of multivariate normal

Let $X_1,X_2$ follow a bivariate standard normal distribution with some non-zero correlation coefficient, $\rho\neq 0$. Let the function $f(z) = z^k,\; k=1,...$. By Stein's lemma, we have that

$$\text{Cov}(f(X_1),X_2) = \text{Cov}(X_1,X_2)\cdot E[f'(X_1)]$$

and in our case

$$\text{Cov}(X_1^k,X_2) = E[X_1^k X_2] = \rho\cdot kE(X_1^{k-1})$$

Since $E(X_1^{k-1}) = 0$ when $k-1$ is odd, it follows that

$$\text{Cov}(X_1^k,X_2) = E[X_1^k X_2] = 0,\;\; \text{iff}\;\;\ k+1\;\;\text{is odd}$$

namely when the sum of the moment-orders is odd.

Moreover by Isserlis Theorem, if $X_1,...,X_n$ follow jointly a zero-mean multivariate normal distribution with dependence, we have for $m\leq n$

$$E[X_1\cdot ... \cdot X_m] = 0 \;\; \text{iff}\;\;\ m\;\;\text{is odd}$$

What I cannot seem to find anywhere is the generalized combination of the two results and its proof, namely that if $X_1,...,X_n$ follow jointly a zero-mean multivariate normal distribution with dependence, we have

$$E[X_1^{r_1}\cdot ... \cdot X^{r_m}_m] = 0 \;\; \text{iff}\;\;\sum_{j=1}^m r_j \;\;\text{is odd}$$

Do we?

• The symmetry $(X_1, \ldots, X_m)\to (-X_1, \ldots, -X_m)$ preserves the distribution but changes the moment to $(-1)^{r_1+\cdots+r_m},$ QED. Am I overlooking some subtlety? It doesn't seem any of the quoted theorems are needed.
– whuber
Jun 18, 2018 at 21:20
• @whuber Good lord, that was painfully simple. So, if I understand correctly, we have the result I am after for any multivariate distribution for which the symmetry you invoke holds, and not just for the multivariate normal? And that would be the spherical distributions that have moments? Jun 18, 2018 at 21:39
• That's right. I'm glad you added that final condition, Alecos, because the argument's conclusion relies on the moments being defined and not infinite.
– whuber
Jun 18, 2018 at 23:57
• I vote for Alecos and whuber to write a joint academic paper on this!
– Ben
Jun 19, 2018 at 0:37

To close this one, as whuber's comment pointed out, let a multivariate distribution be symmetric in the sense

$$(X_1, \ldots, X_m) \sim_d (-X_1, \ldots, -X_m)$$

Suppose also that moments exist. Then

$$E[X_1^{r_1}\cdot ... \cdot X^{r_m}_m] = E[(-X_1)^{r_1}\cdot ... \cdot (-X_m)^{r_m}]$$

$$\implies E[X_1^{r_1}\cdot ... \cdot X^{r_m}_m] = (-1)^{r_1+\cdots+r_m}\cdot E[X_1^{r_1}\cdot ... \cdot X^{r_m}_m]$$

If $r_1+\cdots+r_m$, the sum of moment-orders, is an even number, then $(-1)^{r_1+\cdots+r_m}=1$ and the relation holds always.

But if $r_1+\cdots+r_m$ is an odd number then we have

$$E[X_1^{r_1}\cdot ... \cdot X^{r_m}_m] = - E[X_1^{r_1}\cdot ... \cdot X^{r_m}_m]$$

and this can only hold if

$$E[X_1^{r_1}\cdot ... \cdot X^{r_m}_m] = 0$$