# For any three random variables $X,Y,Z$, prove or disprove that $\langle XY\rangle$, $\langle XZ\rangle$, $\langle YZ\rangle$ can't all be negative [duplicate]

For any three real random variables $$X,Y,Z$$, prove or disprove that $$\langle XY\rangle$$, $$\langle XZ\rangle$$, $$\langle YZ\rangle$$ can't all be negative.

Here $$\langle \cdot \rangle$$ denotes an expectation over the joint probability distribution of the three variables.

Intuitively I think they can't all be negative, because if $$X,Y$$ are anticorrelated, and $$X,Z$$ are anticorrelated, how can $$Y,Z$$ also be anticorrelated? But I am not sure how to prove this, or if there is a counterexample.

It might be necessary to assume that $$X,Y,Z$$ have zero means.

• You easily can have three variables all with negative pairwise correlation, though there's a limit on how negative you can make them. – Glen_b Sep 24 '18 at 13:47
• (a) If the inner product between two vectors is negative, this is equivalent to saying what about the angle $\theta$ between the vectors? Hint1: $\cos \theta_{xy} = \frac{\langle x, y \rangle}{\|x\| \|y\|}$ (b) Can you draw a picture where every angle $\theta_{xy}, \theta_{yz}, \theta_{xz}$ satisfies the conditions in (a)? – Matthew Gunn Sep 24 '18 at 14:21
• Hint2: $\langle x, y \rangle > 0$ implies $\cos \theta_{xy} > 0$ hence $\theta_{xy}$ is an acute angle. What does $\langle x, y \rangle < 0$ imply? – Matthew Gunn Sep 24 '18 at 14:22
• @Becko Can you draw three vectors that have obtuse angles between them? That point away from each other? – Matthew Gunn Sep 24 '18 at 14:30
• To generate a trivariate sample with population correlations all -1/2, try this (in R): n <- 10000; u <- rnorm(n,0,0.5); v <- rnorm(n,0,sqrt(0.75)); xyz <- data.frame(x=u+u,y=-u+v,z=-u-v); cor(xyz) – Glen_b Sep 24 '18 at 16:17