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Let $X_1,....,X_n$ be some positive random variables.

Is $A_{ij} = \frac{E[X_i*X_j]}{E[X_i]E[X_j]}$ a positive (semi) definite matrix?

(note that $B_{ij} = Cov(X_i,X_j) = E[X_i*X_j] - E[X_i]E[X_j]$ is PSD because it is a covariance matrix.)

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That's an unusual formula; bad things happen when one of the $E[X_i]$ is 0. –  Rahul Oct 14 '10 at 22:38
    
thanks for pointing out. $X_i$ are strictly positive, and their expectation is never 0. does that help? –  normvector Oct 14 '10 at 23:26
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up vote 3 down vote accepted

Yes.

Suppose we choose arbitrary $\lambda_1,\ldots,\lambda_n$ and form a new random variable $$Z={{\lambda_1} \over {E[X_1]}}X_1+\cdots+{{\lambda_n} \over {E[X_n]}}X_n$$ Then $$E[Z^2]=\sum_{i=1}^n\sum_{j=1}^n A_{ij} \lambda_i \lambda_j$$ But being the expectation of a non-negative variable, $E[Z^2]$ is non-negative for any choice of $\lambda_1,\ldots,\lambda_n$. Which proves that your matrix $A$ is non-negative semi-definite.

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thanks. that is correct. –  normvector Oct 15 '10 at 0:34
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