I suspect the following relation is true for random variable in $\mathbb{R}^d$ can someone suggest a way of proving it?
$$E[xx'\otimes xx'] \succ E[xx']\otimes E[xx']$$
Using Loewner order symbol $\succ$ and Kronecker product symbol $\otimes$
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Rewrite $xx'$ as a $d^2$-vector $y$. The elements of $(xx') \otimes (xx')$ are the elements of $yy'$, with the same permutation $P$ applied to rows and columns.
Let $a$ be a unit $d^2$-vector. Then as scalars $$E[a'yy'a]\geq a'E[y]E[y]'a$$ regardless of $a$, and so $$E[yy']\succ E[y]E[y]'$$ in the Loewner order. Call the permutation $P$. We have $$E[P(xx') \otimes (xx')P']\succ PE[(xx')] \otimes E(xx')P']$$ and the permutation doesn't affect the Loewner ordering.
Thomas Lumley
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I'm confused about dimensions, xx'a is a vector, so how did you get from comparing scalars to comparing vectors? – Yaroslav Bulatov Sep 14 '20 at 23:54
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Ok, I was confused by your notation. The idea still holds, that the easiest way to do this is to show the inequality holds for all 1-d projections. – Thomas Lumley Sep 15 '20 at 00:11
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Got it. Actually, no permutation is even needed here since (xx')\otimes(xx')=yy' where y=vec(xx') – Yaroslav Bulatov Sep 15 '20 at 04:20
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1I guess it's even stronger than Loewner ordering -- every element of left matrix is larger than right, which implies Loewner – Yaroslav Bulatov Sep 15 '20 at 04:49
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I didn't want to go find a pen and paper to work out what the ordering would be, so I just put in a permutation in case ;-) – Thomas Lumley Sep 15 '20 at 04:55
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Actually how do you show that E[yy']>E[y]E[y]'? This is true for diagonal entries by Jensen's, but for non-diagonal, this is equivalent to claim that E[y_1 y_2]>E[y_1]E[y_2] – Yaroslav Bulatov Sep 16 '20 at 00:13
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1The LHS minus the RHS is a covariance matrix. – Thomas Lumley Sep 16 '20 at 00:46