First question: is it possible to have a set of $k$ random variables $\left\{X_i\right\}$ s.t. each $X_i \sim N(0,1)$ individually, and $\text{Corr}(X_i,X_j)=\rho$, $∀i\neq j$?

If those conditions are self-contradicting, then the rest of this post is just gibberish.

Assuming we could have such $\left\{X_i\right\}$, I tried to compute $\mathbb{E}\prod_i (\mu + \sigma X_i)$ as follows:

$\mathbb{E}\prod_i (\mu + \sigma X_i) = \sum_i \mu^{k-i} \sigma^i \mathbb{E} S(i)$, where $S(i) \equiv \sum_{A(i)_n \in \{A(i)\}} \prod_{a \in A(i)_n} X_a$, and $\{A(i)\} \equiv \{\text{all possible combinations of }i\text{ many indices chosen from }\{1,2,...,k\}\}$.

For example, if $k=3$, then $\{A(2)\}=\{\{1,2\},\{1,3\},\{2,3\}\}$

I'm not good at making the notations easy to understand, but here I'm just trying to state something like the binomial theorem: $\mathbb{E}\prod_i (\mu + \sigma X_i) = \mu^k+\mu^{k-1}\sigma\mathbb{E}\left(X_1+X_2+...+X_k\right)+\mu^{k-2}\sigma^2\mathbb{E}(X_1X_2+X_1X_3+...+X_1X_k+X_2X_3+...+X_2X_k+...)+...+\mu\sigma^{k-1}\mathbb{E}(X_2...X_k+X_1X_3...X_k+...)+\sigma^k\mathbb{E}X_1X_2...X_k$.

Assume $\mathbb{E}\prod_{a \in A(i)_n} X_a=\mathbb{E}\prod_{a \in A(i)_m} X_a=p(i)\in\mathbb{R}$ for any $A(i)_m,A(i)_m \in \{A(i)\}$, then

$\mathbb{E}S(i) = \text{C}(k,i)p(i)$, where $\text{C}(k,i)$ is the number of combinations of $i$ elements chosen from $k$ elements.

I assume that the expectation of the product of an odd number of 1-D Gaussians is $0$ (It's a shaky assumption! It's prompted by the fact that the article I'm going to cite only gives that of an even number of variables). And according to this article, the product of $2i$ (which is an even number of) 1-D Gaussians can be written as the sum of all possible products of pairwise correlations. But since we've assumed equal pairwise correlations, the assumption made about $p$ is true.

So, $\mathbb{E}\prod_{a \in A(2i)_n} X_a = M_{2i}\rho^i$, where $M_{2i}$ is the number of pair-partitions (as defined in that article cited) of $\{1,2,...,2i\}$.

$M_{2i}$ can be calculated as the number of ways to consecutively take out $2$ elements from a set of size $2i$ without replacement, i.e. $M_{2i} = \frac{\text{C}(2i,2)\text{C}(2i-2,2)...\text{C}(2,2)}{i!} = \frac{(2i)!}{i!2^i}$. Hence,

$\begin{align}\mathbb{E}\prod_i (\mu + \sigma X_i)& = \sum_i \mu^{k-i} \sigma^i \mathbb{E} S(i) \\ &= \sum_i \mu^{k-2i} \sigma^{2i} \text{C}(k,2i)\frac{(2i)!}{i!2^i}\rho^i \\ &= \sum_{i=0}^{\lfloor k/2\rfloor} \left(\frac{\rho\sigma^2}{\mu^2}\right)^i\mu^k\frac{k!}{2^i(k-2i)!i!} \end{align}$

Have I done anything wrong in these steps?

If not, why does the sign of this sum seem to oscillate dramatically at big $k$ for negative $\rho$?--I wrote a computer program to calculate this sum for $\mu=1$, $\sigma=0.2$, and $\rho=-0.1$, and found that the absolute value of this sum first approached $0$ for $k<30$, after which the absolute value increased roughly exponentially and the sign almost seemed random. (In my opinion, it's not likely to be a float number problem, because I designed a set of arithmetics under scientific notation; at least the order of magnitude is reliable.) Because of this, I strongly feel that the equation is wrong. In that case, did I miscalculated or applied the wrong theorem? Was the expectation of an odd number of uniformly pairwise correlated Gaussians in fact nonzero? Or, was my assumptions of $\{X_i\}$ impossible to begin with?

  • 3
    $\begingroup$ Since the determinant of the covariance matrix $\Sigma$ is$$\text{det}(\Sigma)=(1+k\rho/(1-\rho))(1-\rho)^k$$ the constraint on $\rho$ is $$1\ge\rho\ge-1\big/ k-1$$ $\endgroup$
    – Xi'an
    Jun 13, 2022 at 10:19
  • $\begingroup$ See stats.stackexchange.com/q/578090/7224 for an explanation of why the product increases in instability with $k$. $\endgroup$
    – Xi'an
    Jun 13, 2022 at 10:43
  • $\begingroup$ My post in the duplicate thread answers your first question. It also demonstrates the question has nothing to do with normality: it's purely about the first two moments of any multivariate distribution. Consequently, it is difficult to see why your subsequent computations of higher moments of Normal distributions would be relevant. $\endgroup$
    – whuber
    Jun 13, 2022 at 16:30
  • $\begingroup$ @whuber Thank you for pointing me to that thread. It also explains what I saw in the numerical computation--the case where $\rho=-0.1$ and $k>30$ is impossible. If I'm not mistaken, my subsequent computation is another question, which is in the title; I was trying to compute the expectation of the product of these random variables with the same pairwise correlation. Is that task meaningless to begin with? $\endgroup$ Jun 14, 2022 at 0:43
  • $\begingroup$ The task is not meaningless--but it's not relevant in the context you establish at the outset. Expectations of products are (multivariate) moments. See stats.stackexchange.com/questions/176702 for computing univariate Normal moments, stats.stackexchange.com/questions/155007 for more on multivariate moments, and stats.stackexchange.com/a/267021/919 for reducing the computation of multivariate moments to univariate moments. If these don't address your question, then by all means please edit your post to focus on what you still need to know. $\endgroup$
    – whuber
    Jun 14, 2022 at 13:57