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Can 3 random variables X, Y, Z have pairwise correlation -1?

This seems to be easy question, but I just got confused over this.

Confusian
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  • So, by @mookid's answer we see that the pairwise correlations $(r_{XY},r_{YZ},r_{XZ})$ cannot be $(-1,-1,-1)$, while obvious they can be $(1,1,1)$ if $X=Y=Z$. I think the generalized question is interesting: What is the space of possible values of the triple $(r_{XY},r_{YZ},r_{XZ})$? –  Mar 27 '14 at 05:59
  • Wild guess, from thinking of the correlation as the cosine of the angle between two vectors: if the correlation of $X$ and $Y$ is $\alpha$ and the correlation of $X$ and $Z$ is $\beta$, then the correlation of $Y$ and $Z$ must lie between $\alpha\beta-\sqrt{1-\alpha^2}\sqrt{1-\beta^2}$ and $\alpha\beta+\sqrt{1-\alpha^2}\sqrt{1-\beta^2}$...? – Greg Martin Mar 27 '14 at 07:15

1 Answers1

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No: if $X,Y,Z$ are your variables, then as $corr(X,Y)=-1$, first $Var \ X\neq 0$ and then

$$ Var\left(\frac X{\sqrt{Var \ X}}+ \frac Y{\sqrt{Var\ Y}}\right) = 2 + 2corr(X,Y)=0 $$ Then $$ \frac X{\sqrt{Var \ X}} = - \frac Y{\sqrt{Var\ Y}} = + \frac Z{\sqrt{Var\ Z}} = - \frac X{\sqrt{Var \ X}} $$by symetry, then $X=0$. But it is impossible because $Var \ X>0$.

If you replace correlation by covariance, then it is possible: for example, any gaussian vector with variance matrix $\bigl(\begin{smallmatrix} 8 &-1&-1\\ -1 & 8&-1\\ -1&-1&8 \end{smallmatrix} \bigr)$ (because such a matrix has $>0$ eigenvalues).

mookid
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