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Let $X$ be an $d\times d$ random matrix satisfying $\mathbb{E}[X]=0$ and $\|X\|_2\leq 1$ almost everywhere. Let $\mathcal{F}$ be the $\sigma$-field generated by $X$. Now suppose we have another $\sigma$-field $\mathcal{G}$, it satisfies that \begin{equation*} \rho(\mathcal{F},\mathcal{G})=\sup_{A\in\mathcal{F},B\in\mathcal{G}}|\mathbb{P}(AB)-\mathbb{P}(A)\mathbb{P}(B)|\leq\phi. \end{equation*} Now I want to prove that \begin{equation}\label{eq:main} \mathbb{E}\big[\|\mathbb{E}[X|\mathcal{G}]\|_2\big]\leq Cd\phi, \end{equation} where $C$ is some constant and $d$ is the dimension.

From Lemma 4.4.1 of this paper, I already know that \begin{equation} \mathbb{E}\big[|\mathbb{E}[X|\mathcal{G}]|\big]\leq C\phi, \end{equation} hold if $X$ is a scalar random variable. Now I want to extend this result to matrix case.

I tried to use the discretization technique as in Proposition 5.17 of Wainwright's book, but then I can only prove it bounded by $C9^d\phi$, which is undesirable because it is exponentially related to the dimension $d$. So I hope someone can give me some idea about it.

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I have figured out how to prove a relatively weaker result.

By elementary inequality of matrix, we know that \begin{equation*} |X|_{\infty}\leq\|X\|\leq 1, \quad\text{ and }\quad\|X\|\leq \|X\|_F. \end{equation*} For every element $X_{i,j}$ of $X$, by Lemma 4.4.1 of this paper we have that \begin{equation*} \mathbb{E}\Big[\big|\mathbb{E}[X_{ij}|\mathcal{G}]\big|^2\Big]\leq \mathbb{E}\Big[\big|\mathbb{E}[X_{ij}|\mathcal{G}]\big|\Big]\leq 2\pi\phi. \end{equation*} Therefore, we have that \begin{align*} &\mathbb{E}\Big[\|\mathbb{E}[X|\mathcal{G}]\|\Big]\leq \mathbb{E}\Big[\|\mathbb{E}[X|\mathcal{G}]\|_F\Big],\\ \leq&\Big\{\mathbb{E}\Big[\|\mathbb{E}[X|\mathcal{G}]\|^2_F\Big]\Big\}^{1/2}\\ \leq&\Big\{\mathbb{E}\Big[\sum_{i,j=1}^d\big|\mathbb{E}[X_{ij}|\mathcal{G}]\big|^2\Big]\Big\}^{1/2}\leq\sqrt{d^22\pi\phi}=d\sqrt{2\pi\phi}. \end{align*}

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