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Suppose $X$ is a $k\times 1$ vector of random variables. Then please verify that $EX^{\prime}(EXX^{\prime})^{-1}EX\leq 1$.

When $K=1$ this is a well known result that $(EX)^{2}\leq EX^{2}$. But how shall I claim this in general?

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1 Answer 1

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Let $\mu = EX$ and $\Sigma = E(XX^T) - \mu\mu^T$. Then we need to show $$ \mu^T (\Sigma + \mu\mu^T)^{-1}\mu \leq 1. $$

Let $C = E(XX^T)$ so that $\Sigma = C - \mu\mu^T$. Using the matrix determinant lemma and the existence of $C^{-1}$ we can see that $\Sigma^{-1}$ exists exactly when $\mu^T C^{-1}\mu\neq 1$. If $\mu^T C^{-1} \mu = 1$ we are done, so WLOG we assume $\mu^T C^{-1}\mu\neq 1$.

Then by the Sherman–Morrison formula we have $$ \mu^T (\Sigma + \mu\mu^T)^{-1}\mu = \mu^T \Sigma^{-1}\mu - \mu^T \left(\frac{\Sigma^{-1}\mu\mu^T\Sigma^{-1}}{1 + \mu^T \Sigma^{-1}\mu}\right)\mu = c - \frac{c^2}{1+c} = \frac{c}{1+c} < 1 $$ since $\Sigma^{-1}$ is PSD so $c \geq 0$.

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