5
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3 random variables have equal pairwise correlation, what are the possible values of correlation?

I think 1 and 0 are the obvious answers, but are there any more, and how could I find them?

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  • $\begingroup$ Do you know the formula to calculate correlation? $\endgroup$
    – rocinante
    Commented May 15, 2014 at 20:30
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    $\begingroup$ A more general version of this question is addressed at stats.stackexchange.com/questions/5747. $\endgroup$
    – whuber
    Commented May 15, 2014 at 22:09

2 Answers 2

10
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Let $X_1$, $X_2$, $X_3$ be three random variables with common pairwise correlation coefficient $\rho$, that is $\mbox{corr}(X_i, X_j)= \rho$ for $i \neq j$ with $|\rho|\leq 1$. So, the correlation matrix of $X = (X_1, X_2, X_3)$ is $$ \left( \begin{array}{ccc} 1 & \rho & \rho \\ \rho & 1 & \rho \\ \rho & \rho & 1 \end{array} \right) . $$ Correlation matrices need to be positive-semidefinite, which implies that their leading principal minors are all positive. So $\rho$ must satisfy the following two conditions $$ \begin{cases} 1 - \rho^2 &\geq 0,\\ 1 - 3\rho^2 + 2\rho^3 &\geq 0 . \end{cases} $$ The first condition is always satisfied, and the second condition implies that $\rho \geq -0.5$.

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    $\begingroup$ In case it's not immediately obvious to everyone $1-3\rho^2+2\rho^3 = (1-\rho)^2(1+2\rho)$ $\endgroup$
    – Glen_b
    Commented May 16, 2014 at 2:59
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Just a side note for the otherwise correct answers already given (+1 both). The correlation/covariance matrix described is compound symmetric. This has some rather important theoretical implication on how one interpreters a model; in particular one would assume that the co-variance of the variables examined can be perfectly partitioned in a "shared" component and the "unshared" component between your variables. A (somewhat) common setting for such structures to be used is when the assumption for equal correlation of residuals is plausible; for example when one deals with repeated trials under the same condition in an experiment.

The CV link: "What is compound symmetry in plain english?" gives a more in-depth presentation of compound symmetry.

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