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This is not assuming a perfect correlation between variables. I believe that the answer is yes as long as r<1.0. (r = Pearson product-moment correlation coefficient.)

Bonus: if the question is false at certain correlation coefficients, at what correlation coefficient does this begin to occur?

BigMistake
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  • I'm not sure. I recall a paper in the 2000s showed that $r_{xy}^2 + r_{yz}^2 > 1 \implies r_{xz}^2 > 0$ but that is a very weak condition and doesn't answer your question. – Galen Aug 23 '23 at 15:18
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    (1) Are you assuming positive correlations between $X$ and $Y$ and between $Y$ and $Z$? (2) Which correlation coefficient are you looking at? – Stephan Kolassa Aug 23 '23 at 15:24
  • @StephanKolassa (1) Good catch; thank you; fixed. (2) Pearson's r. – BigMistake Aug 23 '23 at 15:28
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    I am sure this has been addressed in several threads. See https://stats.stackexchange.com/questions/72790/bound-for-the-correlation-of-three-random-variables for one account where the correlation coefficients $\rho_{XY}$ and $\rho_{YZ}$ are equal and positive. Likely most of the posts I remember are in a regression context, such as https://stats.stackexchange.com/a/223948/919. A glance at the figures in the latter will reveal what's going on: vectors $X$ and $Z$ can make acute angles with $Y,$ yet make an obtuse angle with each other. Examples are easy to construct. – whuber Aug 23 '23 at 15:31
  • @whuber I don't see how that question is answered in the linked threads. Do you think those other threads say "yes" or "no" to the present question? – BigMistake Aug 23 '23 at 15:35
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    I assume you're interested in mathematical proof but if not, you can try this empirically yourself by generating data with different correlation structures (e.g. using R MASS package function mvrnorm). I tried some and rs=0.5 / -0.5 worked, rs 0.6 / -0.6 didn't. Of course, there are different correlation strength combinations you can try. – Sointu Aug 23 '23 at 15:36
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    Start simulating $X$, $Y$ and $Z$ until you find a triplet that provides a counterexample. Easiest to do for small vectors, e.g., of length 3. I was writing up the answer with the counterexample here when @whuber closed the thread... – Stephan Kolassa Aug 23 '23 at 15:37
  • My post in the duplicate thread explicitly constructs counterexamples, so yes, it provides a definitive yes/no answer. – whuber Aug 23 '23 at 15:38
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    @sointu A very general solution is given at https://stats.stackexchange.com/questions/305441. The present question is resolved quantitatively with the determinant-derived bound given at https://stats.stackexchange.com/questions/5747. In that post I also specifically answer the present question negatively in terms of the Spearman correlation which, because it is the Pearson correlation of ranks. also provides an answer here. – whuber Aug 23 '23 at 15:56
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    @whuber Thanks. That thread (https://stats.stackexchange.com/questions/5747/if-a-and-b-are-correlated-with-c-why-are-a-and-b-not-necessarily-correlated) has information that can be used to answer the question. (The answer is "yes.") Then, the formula to determine this is (r^2 between X and Y)+(r^2 between Y and Z) must be < 1 for X and Z to have a negative correlation, I think. – BigMistake Aug 23 '23 at 16:15
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    Right. Assuming the XY and YZ correlations are positive, that criterion is equivalent to the angle criterion I gave in my first comment. – whuber Aug 23 '23 at 16:27
  • The answer to the bonus is r = 0.70710678118 (assuming the same correlation). 1/sqrt(2). (50% variance explained). I wonder how to calculate this if I want a certain negative correlation, say, r = 0.15, between X and Z instead of r = 0. – BigMistake Aug 23 '23 at 17:53
  • The bonus question is unclear, because there are two original correlation coefficients: $\rho_{XY}$ and $\rho_{YZ}.$ But almost any version of it is easy to answer using the angle interpretation. For instance, with $\rho_{XZ}=-0.15,$ the necessary criterion for $\rho_{XZ}\le -0.15$ is $\cos(\arccos(\rho_{XY})+\arccos(\rho_{YZ}))\le -0.15.$ – whuber Aug 23 '23 at 18:42

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