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Given a set of random variables $X_1,\dots,X_n$ such that $${\rm cov}(X_i,X_j)=\rho$$ for $i\neq j$. What is the minimum value of $\rho$ possible? Can it be negative? Thanks!

EZLearner
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  • Expanding $\operatorname{var}(\sum X_i)\ge 0$ would give you an answer when the variables are identically distributed with finite variance (see https://math.stackexchange.com/questions/1032456/show-that-in-this-case-rho-geq-frac1n-1?rq=1 for example). – StubbornAtom Jun 21 '19 at 19:01
  • And what of the case when their not identically distributed? – EZLearner Jun 21 '19 at 19:05
  • Actually equal variance suffices for the bound in the linked question. Apart from that, you would need additional information. – StubbornAtom Jun 21 '19 at 19:10

1 Answers1

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For ease, suppose we normalize so that all random variables have mean $0$ and variance $1$. What you are asking is equivalent to for what values of $\rho$ is the supposed covariance matrix $\Sigma_{\rho}$ with diagonal $1$ and off-diagonals $\rho$ positive semi-definite. This is possible precisely when $\rho\in [\frac{-1}{n-1},1]$; when $\rho=\frac{-1}{n-1}$, one can easily see that $\Sigma_{\rho}$ is then diagonally dominant, and therefore p.s.d. Moreover, $\Sigma_{\rho}$ is easily seen to be p.s.d. when $\rho=1$, as this is easily seen to be rank-one with nonnegative eigenvalue (a more probabilistic argument is this is the all-ones matrix, and this is realizable as a covariance matrix by taking $X_1=\ldots=X_n$). For all other $\rho$ in that range, one may take convex combinations to show positive semi-definiteness.

For $\rho<\frac{-1}{n-1}$, simply observe $(1,\ldots,1)^T$ is an eigenvector with strictly negative eigenvalue.

J.G
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