Bound correlation coefficient for three random variables

[Master1022]

Homework Statement

Given three random variables ## X##, ##Y##, and ## Z ## such that corr(X, Y) = corr(Y, Z) = corr(Z, X) = r, provide an upper and lower bound on ##r##.

Relevant Equations

Correlation

Hi,

I just found this problem and was wondering how I might go about approaching the solution.

Question:
Given three random variables ## X##, ##Y##, and ## Z ## such that ##\text{corr}(X, Y) = \text{corr}(Y, Z) = \text{corr}(Z, X) = r ##, provide an upper and lower bound on ##r##

Attempt:
I don't quite know how to attempt this rigorously, but I started by thinking about the problem geometrically. I don't know if this is incorrect, but I thought of the variables as 'vectors' and the correlations as the cosine of the angle between them.

Then, let us fix two vectors, ## X ## and ##Y##, with the angle between them ##\theta = \cos(r)##. Now we need to find the vector ##Z## such that it has the same angle between itself and both of ##X## and ##Y##. After doing some visualisation, I thought that the maximum angle possible between each 'vector' was 120 degrees (when the vectors are all planar), thus the correlation is ## cos(120^{o}) = -0.5 ##. Thus, my final answer would be: ## -0.5 \leq r \leq 1 ##: is that correct?

I suppose I could imagine this by imagining taking three pencils/pieces of spaghetti and standing them all up in a bunch. The angle between them all is 0 degrees, so correlation = 1. Then I could start increasing the angle between them all (as the pieces become less inclined to the table), so correlation is decreasing between them. The largest angle occurs (I think!) when they are all flat on the table, and the angle between them all must be equal to satisfy the constraints of the problem. Thus, that led me to the 120 degrees. Apologies for the silly explanation, but that is how I thought about the problem.

Thanks in advance

 

[andrewkirk]

Your guess is correct, and I like your geometric way of thinking about it - very intuitive!

To prove it formally, note that the determinant of the correlation matrix must be non-negative. The correlation matrix has entries of r everywhere except on the main diagonal, where it is 1. The determinant is ##1 - 3r^2+2r^3##, which equals zero at ##r=1## and ##r =-0.5## and is positive between those points. It's negative for ##r<-0.5## so 0.5 is the lower bound.

 

[Master1022]

Your guess is correct, and I like your geometric way of thinking about it - very intuitive!

To prove it formally, note that the determinant of the correlation matrix must be non-negative. The correlation matrix has entries of r everywhere except on the main diagonal, where it is 1. The determinant is ##1 - 3r^2+2r^3##, which equals zero at ##r=1## and ##r =-0.5## and is positive between those points. It's negative for ##r<-0.5## so 0.5 is the lower bound.

Thanks @andrewkirk ! With the correlation matrix method, why must its determinant be non-negative? I might be missing something obvious... Otherwise, that arithmetic does make sense.

 

[andrewkirk]

Thanks @andrewkirk ! With the correlation matrix method, why must its determinant be non-negative? I might be missing something obvious... Otherwise, that arithmetic does make sense.

It is a theorem of probability theory that a correlation matrix of real random variables must be positive definite - which means the determinants of the upper-left 1x1, 2x2 and 3x3 submatrices must all be non-negative. See for instance here. There will be various ways of proving it, most involving linear algebra. It may relate to the fact that it must be possible to perform Cholesky decomposition of a correlation matrix.

The requirement turns out to be equivalent to the more intuitive notion of disallowing impossible correlations (such as random variables A, B and C all having pairwise correlations close to -1.).