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Let $\sigma_1,\ldots,\sigma_n>0$ and $X_i\sim \mathcal N(0,\sigma_i^2)$, $i=1,\ldots,n$. Write $\sigma_{ij}=\mathrm{Cov}(X_i,X_j)$, $i,j=1,\ldots,n$, and $\Sigma=(\sigma_{ij})_{i,j}\in \Bbb R^{n\times n}$ (with $\sigma_{ii}=\sigma_i^2$). I am looking for a condition on $\sigma_1,\ldots,\sigma_n$, under which $\sigma_{ij}\neq 0$ can be chosen such that

$$\tag{1} S:=X_1+\ldots+X_n=0 \quad \text{almost surely.} $$

Here are my thoughts so far:

  1. For $n=2$, (1) is equivalent to $X_2=-X_1$. Since $-X_1\sim \mathcal N(0,\sigma_1^2)$, this means that my problem has a solution if and only if $\sigma_1=\sigma_2$, and this solution is given by $\sigma_{12}=-\sigma_1^2$.

  2. In general, note that since $\Bbb E(S)=0$, the condition (1) is fulfilled if and only if we also have $\mathrm{Var}(S)=0$, which is equivalent to $$\tag{2} \sum_{i\neq j}\sigma_{ij}=-\sum_i\sigma_i^2. $$ This looks like we have a lot of choice for the $\sigma_{ij}$, but recall that $\Sigma$ needs to be a proper covariance matrix, so it has to be non-negative definite.

  3. If $\sigma_i=\sigma_1$ for all $i$, we can take $\sigma_{ij}=-\sigma_1^2/(n-1)$ for all $i\neq j$, as then (2) is obviously fulfilled and it is easy to check that the corresponding $\Sigma$ is in fact non-negative definite. So the one case that works for $n=2$ actually works similarly for any $n$. But what can we do if not all of the $\sigma_i$ coincide?

  4. If $J\subset\{1,\ldots,n\}$, then (1) implies $$ \mathrm{Var}(\sum_{i\in J} X_i) = \mathrm{Var}(\sum_{i\in J^c} X_i) $$ and hence $$\tag{3} \sum_{i\in J} \sigma_i^2 - \sum_{i\in J^c} \sigma_i^2 = \sum_{i,j\in J^c,i\neq j}\sigma_{ij} - \sum_{i,j\in J,i\neq j}\sigma_{ij}, $$ which includes (2) (for $J=\{1,\ldots,n\}$ or $J=\emptyset$). Note that (3) gives us the same equations for $J$ and $J^c$, so we get a total of $2^{n-1}$ linear equations in the $n(n-1)/2$ variables $\sigma_{ij}$ (recall that we consider the variances $\sigma_i^2$ as given).

  5. For $n=3$, we can check that (3) holds if and only if $$ \sigma_{12}=-\frac12 (\sigma_1^2+\sigma_2^2-\sigma_3^2), \; \sigma_{13}=-\frac12 (\sigma_1^2-\sigma_2^2+\sigma_3^2), \; \sigma_{23}=-\frac12 (-\sigma_1^2+\sigma_2^2+\sigma_3^2). $$ Checking non-negative definiteness of $\Sigma$ becomes kind of messy, though. One can show that necessarily $$ (\sigma_1-\sigma_2)^2\le \sigma_3^2 \le (\sigma_1+\sigma_2)^2, $$ and then another rather ugly condition will need to hold.

This is where I decided to stop for now. Does anybody know of more general conditions on $\Sigma$, under which (1) holds? I cannot get rid of the feeling that I'm missing something obvious...

Mars Plastic
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  • I don't think there is a way of describing all non-negative definite matrices with $\sum \sigma_{ij}-0$. – Kavi Rama Murthy Apr 25 '25 at 12:25
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    If you diagonalize $\Sigma = P^T D P$ , you get $0 = \sum_{i, j} \sigma_{ij}$ iff $\sum d_i^{1/2} (P 1)_i = 0$. So if you choose $(d_i)$ and $P$ such that $(P 1)_i = 0$ if $d_i > 0$ you get a matrix that satisfies your condition. – jvc Apr 25 '25 at 15:25

1 Answers1

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You can consider a space of all linear combination of some standard Gaussian variables and look for the solution there.

Then this becomes a purely geometric problem: when does a closed broken line consisting of segments $\sigma_1,...,\sigma_n$ exist? The necessary and sufficient condition for that is the triangle inequality: $\sigma_k\le \sum_{i\neq k}\sigma_i$ for all $k$, see e.g. here.

zhoraster
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