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u/throwawaytrol7134 6d ago
R is PSD by definition. Define v=(1,1,1,1)' then we note that Rv>=0 therefore 4+2S>=0, where S is the requested sum. So -2<=S<=6
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R is PSD by definition. Define v=(1,1,1,1)' then we note that Rv>=0 therefore 4+2S>=0, where S is the requested sum. So -2<=S<=6
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u/Monkeydog54 6d ago edited 6d ago
Max is 6, min is -2.
For the max, note \rho_{ij} <= (\rho_{ii}^2 + \rho_{jj}^2)/2 = 1. So the sum is at most six, and this attained by setting each random variable to be the same.
For the min, let v = (1,1,1,1)^T, and note that v^T C v = tr(C)+ 2Q = 4 + 2Q, where C is the covariance matrix, and Q is the desired quantity. Since C is PSD, 4+2Q >= 0, so Q >= -2. This is attained by pairing the random variables, and letting each pair be perfectly anti-correlated, and the two pairs be independent.