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A set contains random variables where any two random variables in the set have the same correlation $\rho$. Then what is the maximum number of the elements of this set?

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The answer depends on what $\rho$ is. For $-1 \leq \rho < -\frac 12$, the answer is two random variables. More generally, the maximum number of random variables that can have common correlation $\rho$ is $n$ for $\rho$ in the range $\left[-\frac{1}{n-1}, -\frac{1}{n}\right)$. For $\rho \geq 0$, the number of random variables is unbounded. See the answers to this question for some results.

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  • $\begingroup$ If the random variable is restricted to a finite number of values, then I don't think the number of random variables can exceed that number of values. For instance, if you have a bunch of weighted dice, you can't have more than six of them with the same correlation. $\endgroup$ – Acccumulation Jan 28 at 7:08
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    $\begingroup$ you can if that correlation is 0 or 1 — a bunch of independent dice or a bunch of dice glued together (All facing the same way) $\endgroup$ – JDL Jan 28 at 13:43
  • $\begingroup$ @Acccumulation I don't think that what you are saying can be true. Consider $n$ independent discrete random variables $X_1i$ each taking on values $\pm 1$ with equal probability $\frac 12$, and define $Y_i = X_i - \bar{X}$ where $\bar{X}$ is the sample mean $\frac 1n \sum_i $X_i$. Then, as shown in this answer to the question cited in my answer above, the $Y_i$ are zero-mean random variables with common correlation $-\frac{1}{n-1}$ exactly as claimed above. $\endgroup$ – Dilip Sarwate Jan 28 at 22:18
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To supplement Dilip Sarwate's answer, if you take $I_0 \sim Bernoulli(p_0)$ and any number of independent $I_i \sim Bernoulli(p)$, all independent then $$cor(I_0 + I_i, I_0 + I_j) = \frac 1 {1 + \frac {p (1-p)} {p_0(1-p_0)}},$$ so you can choose $p_0$ and $p$ to get any $\rho$ in the interval $(0,1)$.

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