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?

  • 3
    $\begingroup$ 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. $\endgroup$
    – whuber
    Jun 18, 2018 at 21:20
  • 3
    $\begingroup$ @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? $\endgroup$ Jun 18, 2018 at 21:39
  • $\begingroup$ 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. $\endgroup$
    – whuber
    Jun 18, 2018 at 23:57
  • $\begingroup$ I vote for Alecos and whuber to write a joint academic paper on this! $\endgroup$
    – Ben
    Jun 19, 2018 at 0:37

1 Answer 1


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

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


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