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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.

a06e
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  • 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
  • @Glen_b What's the limit? Can you be more explicit (or point me to some reference/theorem)? Thanks! – a06e Sep 24 '18 at 13:51
  • For $n$ variables, if they're all equicorrelated that correlation cannot go below $-\frac{1}{n-1}$ so for 3 variables, $-\frac12$ is as negative as you get. – Glen_b Sep 24 '18 at 13:52
  • @Glen_b Can you give an example where all three correlations are negative? – a06e Sep 24 '18 at 13:55
  • This is coursework, right? Or at least an exercise from a text? Please see the help/on-topic in relation to homework-style questions. – Glen_b Sep 24 '18 at 13:57
  • Your reference to "zero means" indicates the notation "$\langle\cdot\rangle$" cannot possibly refer to correlation or covariance, because (obviously) neither of those depends on means. At a minimum, then, please explain what you mean by this notation. If indeed it is intended to be proportional to a correlation, then your question has an answer at https://stats.stackexchange.com/questions/137178/how-to-calculate-the-correlation-coefficient-from-minimal-distributional-assumpt/365267#365267. – whuber Sep 24 '18 at 14:20
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    (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
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    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
  • @Glen_b It's not homework. It's just something I found while doing some unrelated work. – a06e Sep 24 '18 at 14:24
  • @whuber It denotes expected value. – a06e Sep 24 '18 at 14:25
  • @MatthewGunn You can't have a triangle with all three angles larger than $\pi/2$. This proves that my intuition was right, all three expectations can't be negative? I am now confused with the other comments claiming that my assertion is false. – a06e Sep 24 '18 at 14:28
  • Since the question contains instructions, it's presumably from a textbook, assignment, exam or similar set of exercises, and in that situation the discussion at the above link applies. If it's nothing like that you'd need to explain how it comes to be phrased as a set of instructions. – Glen_b Sep 24 '18 at 14:29
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    @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
  • @Glen_b What instructions? You mean the sentence about assuming that the means of $X,Y,Z$ are zero? It's just my intuition! Which can be wrong, reading the links you posted. – a06e Sep 24 '18 at 14:30
  • As @Glen_b pointed out, this question is obviously from a textbook or some course material. – Matthew Gunn Sep 24 '18 at 14:31
  • @MatthewGunn Ah yes of course. The angles don't have to be in a triangle, sorry. – a06e Sep 24 '18 at 14:31
  • @Glen_b Yes, I see. Also the question you linked fully answers my question, and more. Thanks. – a06e Sep 24 '18 at 14:34
  • "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" is an instruction. It would seem an odd way to phrase a question, but it's a standard way to set a task for a student. It might be best to clarify how it arises. – Glen_b Sep 24 '18 at 14:35
  • @Glen_b I just phrased it that way, since that's what I wanted to know. In the future if I have a question that I come up with myself I'll try to phrase it differently then. In this particular case this just came out of a complicated replica computation (https://en.wikipedia.org/wiki/Replica_trick), and if this property held my life would have been much easier. It was just too complicated to post here the whole motivation. Would also have been a waste of time, since the answer was negative and trivial. – a06e Sep 24 '18 at 14:45
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    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

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