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.

  • $\begingroup$ Should the first sentence be "I have two random vectors $X$ and $Y$"? $\endgroup$
    – angryavian
    Nov 10 '20 at 22:30
  • $\begingroup$ @angryavian yes, corrected, thanks $\endgroup$
    – youpilat13
    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 $\text{Var}(W) = -f(a,b)$.

  • $\begingroup$ You mean $a=b=0$ is the only solution ? $\endgroup$
    – youpilat13
    Nov 10 '20 at 23:03
  • $\begingroup$ No. it's never the only solution. $\endgroup$ Nov 11 '20 at 0:56
  • $\begingroup$ Sorry, my equation above is not correct. I invite you to look at a new post about this ( math.stackexchange.com/questions/3904581/…) with more details. Regards $\endgroup$
    – youpilat13
    Nov 12 '20 at 14:30

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