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Setup: Suppose I have a mean-zero multivariate normal $X = (X_1, X_2, X_3, X_4)$, where each component has strictly positive but possibly unequal variance. $(X_1, X_3)$ is independent of $(X_2, X_4)$. $X_1$ and $X_3$ are dependent, and $X_2$ and $X_4$ are dependent. Consider $a = (a_1,a_2,a_3,a_4)$ where the $a_i$ are either $1$ or $-1$ for each $i =1,\dots,4$.

Question: When is $ a^TX = 0$ with positive probability? Suppose $(a_1,a_2)\neq -(a_3,a_4)$ to rule out the trivial case where $X_1=X_3$ and $X_2 = X_4$. Are strictly positive variances somehow enough to make this a probability zero event?

Why I am asking: The variance of $X$ is $$\Sigma = \begin{pmatrix} \sigma_1^2 &0 & \sigma_{13} & 0\\ 0 &\sigma_2^2 &0 & \sigma_{24}\\ \sigma_{13} &0 & \sigma^2_{3} & 0\\ 0 &\sigma_{24} &0 & \sigma^2_{4} \end{pmatrix}$$ This looks a bit strange, which makes me believe that there is a better characterization of $P(a^TX \neq 0) =1$ than "$a^T \Sigma a$ has to be positive,'' especially because the choice of $a$ is restricted.

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As mentioned by @angryavian, $a^TX$ is univariate gaussian, so the event $a^TX=0$ has positive probability iff $a^TX$ is identically zero. In particular, $a^TX$ must have zero variance.

Assume each $a_i$ is either $+1$ or $-1$. Then the quantity $$\operatorname{Var}(a^TX)=a^T\Sigma a=\sigma_1^2+\sigma_2^2+\sigma_3^2+\sigma_4^2+2a_1a_3\sigma_{13}+2a_2a_4\sigma_{24}\tag{$\ast$}$$ simplifies to $$\sigma_1^2\pm2\sigma_{13}+\sigma_3^2+\sigma_2^2\pm2\sigma_{24}+\sigma_4^2,\tag1$$ so there are four cases to consider. In all cases we have the inequality $$|\sigma_{ab}|\le\sigma_a\sigma_b\tag2$$ which follows from Cauchy-Schwarz. This implies $$(1)\ge \sigma_1^2-2\sigma_1\sigma_3+\sigma_3^2 + \sigma_2^2-2\sigma_2\sigma_4+\sigma_4^2=(\sigma_1-\sigma_3)^2+(\sigma_2-\sigma_4)^2.\tag3$$ Note the RHS of (3) is nonnegative. So if we demand that $(\ast)$ be zero, then it follows that $\sigma_1=\sigma_3$ and $\sigma_2=\sigma_4$. Substituting into (1), conclude that $$(\sigma_1\sigma_3\pm\sigma_{13}) + (\sigma_2\sigma_4\pm\sigma_{24})=0.\tag4$$ Apply inequality (2) again to see that both parenthesized quantities in (4) are non-negative, hence $\sigma_{13}=\mp\sigma_1\sigma_3$ and $\sigma_{24}=\mp\sigma_2\sigma_4$. This means the correlation between $X_1$ and $X_3$ is either $+1$ or $-1$, and the same is true between $X_2$ and $X_4$. In other words, we have $X_1=\pm X_3$ and $X_2=\pm X_4$ are the only four possibilities for which $(\ast)$ equals zero.

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  • $\begingroup$ I figured it had to be CS to pin this down further but I was missing the key step (3). Thank you for writing up such a neat solution. This argument is quite general and extends nicely to higher dimensional $(X_1, X_3)$ and $(X_2,X_4)$. The reason for my question was the combinatorial analysis of certain Gaussian processes and this argument put me on the right track. $\endgroup$
    – Galton
    Mar 17, 2022 at 13:30

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