Let $X$ be an $m\times n$ random real matrix with integrable entries. Can I show that
$$\|E[X]\|\leq E[\|X\|]$$
where $\|\cdot\|$ denotes the Frobenius norm?
I tried different combinations of Cauchy-Schwarz and Jensen, but did not succeed.
Any help is greatly appreciated.
Best Answer
By Jensen's inequality, $\mathbb E[X_{ij}]^2 \le \mathbb E[X_{ij}^2]$ for all $i,j$, so $$\begin{align}\Vert \mathbb E[X]\Vert^2 &= \sum_{1\le i,j\le n}\mathbb E[X_{ij}]^2\\ &\le \sum_{1\le i,j\le n}\mathbb E[X_{ij}^2]\\ &=\mathbb E\left[\sum_{1\le i,j\le n}X_{ij}^2\right]\ \text{ (by linearity)}\\ &=\mathbb E\left[\Vert X\Vert^2\right]\end{align} $$
Applying square root on both sides yields the desired inequality.Update : My above attempt does not work, and the result is in fact a consequence of the measure-theoretic formulation of Jensen's inequality which states that $$\varphi\left(\int_\Omega fd\mu\right) \le \int_\Omega\varphi\circ f\ d\mu $$ For any $\mu$-measurable $f : \Omega \to A$ and convex $\varphi : A \to \mathbb R$. In particular, the desired result follows from applying it to $f\equiv X : \Omega \to \mathcal M_n(\mathbb R)$ and $\varphi \equiv \Vert\cdot\Vert : \mathcal M_n(\mathbb R) \to \mathbb R$.
To give a more satisfactory explanation, I write down the proof of the result for the specific $\mathcal M_n(\mathbb R)$ case, but the argument is basically the same as the one given in the Wikipedia article :
For $\varphi : \mathcal M_n(\mathbb R) \to \mathbb R$ convex, there is for every $x\in\mathcal M_n(\mathbb R)$ a non-empty set of subgradients at $x$. Therefore, for any $x_0\in \mathcal M_n(\mathbb R)$, we can find $a\in\mathbb R$ and $b\in\mathcal M_n(\mathbb R)$ such that for all $x\in\mathcal M_n(\mathbb R)$, $$ \varphi(x) \ge a + \langle b,x\rangle\tag1$$ $$ \varphi(x_0) = a + \langle b,x_0\rangle\tag2$$ Where $\langle \cdot,\cdot\rangle $ is the inner product of $\mathcal M_n(\mathbb R)$ i.e. $\langle A,B\rangle = \mathrm{Tr}(A^TB)$. In particular, if I substitute in $(1)$ $x$ by the random variable $X$ and let $x_0\equiv \mathbb E[X]$, these two equations become $$ \varphi(X) \ge a + \langle b,X\rangle\tag{1'}$$ $$ \varphi(\mathbb E[X]) = a + \langle b,\mathbb E[X]\rangle\tag{2'}$$ Now, by applying expectation to line $(1')$ we get $$\begin{align}\mathbb E[\varphi(X)] &\ge a + \mathbb E[\langle b,X\rangle]\\ &=a + \langle b,\mathbb E[X]\rangle \ \text{ (by linearity) }\\ &=\varphi(\mathbb E[X]) \ \text{ (by equation [2']) }\\ &\square\end{align} $$