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I was posed this problem where we know: $$ cov(x_1,x_2) \gg 0 \\ cov(y_1, y_2)\gg 0 \\ cov(x_1+y_1, x_2+y_2) = 0 \\ $$

What does this tell us about the structure of $x_1, x_2, y_1, y_2$?

Is there any significance to the $\gg$ here? I don't think $\gg$ tells us any additional detail than what $>$ would tell us since covariance isn't normalized. The magnitude of covariance depends on the domain of the random variables, so whether the covariance is much greater than zero or just slightly greater than zero doesn't give us any additional information. On the other hand, if it was instead said that correlation is much greater than zero, which means it's closer and closer to 1, then this could tell us that the 2 variables have a near-perfect linear relationship.

So the best that I think I can say about this is that $x_1$ and $x_2$ are positively correlated and have some degree of linearity, but the strength of their correlation isn't clear. Same can be said for $y_1$ and $y_2$.

$cov(x_1+y_1, x_2+y_2) = 0$ tells us that the newly formed random variables $x_1+y_1$ and $x_2+y_2$ are uncorrelated. They are linearly independent, but not necessarily independent. The way the problem was posed to me seems to be suggesting that $cov(x_1+y_1, x_2+y_2) = 0$ should tell us something about $x_1, x_2, y_1, y_2$, but I can't seem to think what it could be telling us. Any hints?

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  • $\begingroup$ Can it have anything to do with highly correlated parameters with opposite signs? They may be correlated but their summation may cancel the relationship between pairs? $\endgroup$
    – Atakan
    May 6, 2020 at 1:55
  • $\begingroup$ Hmmm I'm not sure, but let me think about that. Why do you say "high correlated?" I don't think we are given that information. I was verbally posed this problem, and I got the vibe that there's something to be said about the original variables $x_1, x_2, y_1, y_2$, when we know that $cov(x_1+y_1, x_2+y_2) = 0$. $\endgroup$
    – 24n8
    May 6, 2020 at 2:00
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    $\begingroup$ $cov(x_1+y_1, x_2+y_2) = cov(x_1,x_2) + cov(x_1,y_2)+cov(y_1,x_2)+cov(y_1,y_2)$. If $cov(x_1,x_2)>>0$ and $cov(y_1,y_2) >> 0$, then we have $cov(x_1,y_2)+cov(y_1,x_2)<<0$. $\endgroup$
    – Tim Mak
    May 6, 2020 at 2:46
  • $\begingroup$ @TimMak Oh I like the use of bilinearity. Although, I'm not sure what information $cov(x_1, y_2) + (y_2, x_2) << 0$ tells us about the 4 random vars. $\endgroup$
    – 24n8
    May 6, 2020 at 3:01

2 Answers 2

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$$ cov(x_1+y_1, x_2+y_2) = cov(x_1,x_2) + cov(x_1,y_2)+cov(y_1,x_2)+cov(y_1,y_2) $$

If $cov(x_1, x_2) \gg 0$ and $cov(y_1, y_2) \gg 0$, then we have $cov(x_1,y_2)+cov(y_1,x_2) \ll 0$.

If $x_1,x_2,y_1,y_2$ are all standardized, however, then supposing both $corr(x_1, x_2)$ and $corr(y_1, y_2)$ are close to 1, then it follows that $corr(x_1,y_2)$ and $corr(y_1,x_2)$ are both close to -1. Moreover, to ensure that the correlation matrix of $(x_1,x_2,y_1,y_2)$ is semi-positive definite, then $corr(x_1,y_1)$ and $corr(x_2, y_2)$ must be close to -1 also. An example is the following, $$ Corr \begin{pmatrix} x_1\\ x_2\\ y_1\\ y_2 \end{pmatrix} = \begin{pmatrix} 1 & 0.99 & -0.99 & -0.99 \\ 0.99 & 1 & -0.99 & -0.99 \\ -0.99 & -0.99 & 1 & 0.99 \\ -0.99 & -0.99 & 0.99 & 1 \end{pmatrix} $$

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One thing we learn is that the $y$-variables cannot be deterministic linear functions of their respective $x$- variables (and vice versa).

Assume, towards a contradiction that $y_1 = ax_1$, $y_2 = bx_2$ with $a\neq -1, b\neq -1$ (to avoid trivialities). To satisfy the second pair-wise positive covariance assumption we have also to assume that $a,b$ have the same sign (both positive or both negative), to obtain ${\rm Cov}(y_1, y_2) > 0$.

But then

$$0= {\rm Cov}(x_1+ y_1,\, x_2+y_2) = (1+a)(1+b){\rm Cov}(x_1,\, x_2) \neq0\;: {\rm contradiction}$$

Realistically I would extend this negative result (as "in general we cannot say that...") to stochastic linear functions (i.e. $y_1 = ax_1 + v$, $y_2 = bx_2 + u$) because then, in order to obtain the zero-covariance premise we would have to assume special structure for the newly arrived variables.

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