# How to solve a system of equations Cov(X,Z) + Cov(Y,Z) + Cov(X,Y) = -Var(Z)/2?

Please Don't vote down here, I can't delete the first post about my initial questions since there are errors in the equations below.

I have 2 covariance matrices known $$X$$ and $$Y$$ . I am looking for a way to find a combination of vectors random variables $$Z$$ (not null) that could verify :

$$2\,\text{Cov}(X,Z) + 2\,\text{Cov}(Y,Z) + 2\,\text{Cov}(X,Y) = -\text{Var}(Z)$$

If I can manage to find these combinations of vectors, I could write :

$$2\,\text{Cov}(X+Y,Z) + 2\,\text{Cov}(X,Y) = - \text{Var}(Z)$$ and so :

$$\text{Var}(X+Y+Z) = \text{Var}(X) + \text{Var}(Y) + 2\,\text{Cov}(X,Y) + \text{Var}(Z) + 2\,\text{Cov}(X+Y, Z) = \text{Var}(X) + \text{Var}(Y)$$

Any help/track/suggestion is welcome.

EDIT 1: Sorry, I think that I have made an error in my initial post, A further checking if the question is well formulated is welcome.

I am going to inspire me from the first answer to solve this problem but I would be glad to get also any help.

• Should the first sentence be "I have two random vectors $X$ and $Y$"? Nov 10 '20 at 22:30
• @angryavian yes, corrected, thanks Nov 10 '20 at 22:30

If $$Z = a X + b Y$$, \eqalign{\text{Cov}(X+Y,Z) &= a \text{Var}(X) + (a+b) \text{Cov}(X,Y) + b \text{Var}(Y) \cr &= a (\text{Var}(X) + \text{Cov}(X,Y)) + b (\text{Cov}(X,Y) + \text{Var}(Y))} which you want to be $$0$$.
EDIT: For your new equation, if $$Z = a X + b Y + W$$ (where $$W$$ is independent of $$X$$ and $$Y$$, you want $$a^2 \text{Var}(X) + 2 a b \text{Cov}(X,Y) + b^2 \text{Var}(Y) + 2( \text{Cov}(X,Y) + \text{Var}(X)) a + 2 (\text{Cov(X,Y)} + \text{Var}(Y)) b + 2 \text{Cov}(X,Y) + \text{Var}(W)= 0$$ Call the left side $$f(a,b) + \text{Var}(W)$$. It turns out that $$f(a,b)$$ is minimized at $$a=-1$$, $$b=-1$$, with $$f(-1,-1) = - \text{Var}(X) - \text{Var}(Y)$$.
So one solution is $$a=-1, b=-1$$, $$\text{Var}(W) = \text{Var}(X) + \text{Var}(Y)$$. Or you can change $$a$$ and $$b$$ as long as $$f(a,b) \le 0$$ and take $$\text{Var}(W) = -f(a,b)$$.
• You mean $a=b=0$ is the only solution ? Nov 10 '20 at 23:03