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Suppose $X$ is an $n\times n$ symmetric PSD matrix. If $X$ concentrates around its expected value, i.e. $\|X-\mathbb{E}[X]\|_2\leq \epsilon$ w.h.p, can we conclude that $X^{1/2}$ concentrates around $\mathbb{E}[X^{1/2}]$?

In general, does $X^p$ concentrate around anything?

If not, are there any extra assumptions needed to make this concentration happen?

mathlover
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1 Answers1

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The answer is yes when $\epsilon>0$ is sufficiently small. More specifically, let $E(X)=M$. Using the uniform continuity of the matrix square root function on the cone of all positive definite matrices (the function is also differentiable on the PSD matrix cone, but this is irrelevant here), $\|X^{1/2}-M^{1/2}\|$ can be made arbitrarily and uniformly smaller than any $\delta>0$ whenever $\|X-M\|\le\epsilon$ for a sufficiently small $\epsilon>0$. Hence $$ \begin{align} \|X^{1/2}-E(X^{1/2})\| &= \|X^{1/2}-M^{1/2} + E(M^{1/2}-X^{1/2})\|\\ &\le \|X^{1/2}-M^{1/2}\| + E\|M^{1/2}-X^{1/2}\|\\ &\le \delta(1+O(\epsilon)). \end{align} $$

user1551
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  • What happens to the relative error? i.e. $|X^{1/2}-E(X^{1/2})|/|E(X^{1/2})|$. Could you also give a bound for that? – mathlover Sep 15 '17 at 21:16