# When is this linear combination of Normals exactly zero?

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.

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.
• 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. Mar 17, 2022 at 13:30