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Lets say we are given random (row) vector $x \in \mathbb{R}^n$, and a non-random matrix $M \in \mathbb{R}_{nxn}$

I came across a claim (which does not impose any assumptions on the distribution of $x$ a priori), that the following expectation equality holds:

$$ \mathop{\mathbb{E}}[||x^tM||] = \sigma_x||M||_F^2$$

Where $||M||_F$ is the Frobenius matrix norm and $\sigma_x$ is the standard deviation of $x$ (identically component-wise I assume although again this was not qualified in the original statement)

I'm trying to understand this statement and prove it for myself. To begin with, trying it out on a 1x1 case I think there's an error in the claim in that it should use $||M||_F$ rather than $||M||_F^2$, is that right?

After that, I am still unclear how to derive the full equality. I was able to derive it as an inequality using Jensen's inequality and assuming the mean of $x$ is zero, but thats as far as I got.

Any insight appreciated on whether this statement is generally true. Alternatively, does it hold if we impose additional requirements on the vector $v$? e.g. zero-mean or stronger claims (such as component-wise i.i.d $\mathcal{N}(0,\,\sigma^{2})$ etc?)

giorgio
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    What distribution is your random row vector $x$ drawn from? (And what is $\sigma_x$?) – TonyK Dec 05 '24 at 16:49
  • @TonyK I mention in the last paragraph, it is not stated in the original reference, so part of the exercise for me here is to figure out if by assuming certain distribution / properties the claim holds, e.g. zero mean, normality etc? updated the question body addressing your questions to make it clear in any case. – giorgio Dec 05 '24 at 16:50
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    An eyeball dimensional analysis shows $\Vert\cdot\Vert^2$ can’t be right. – A rural reader Dec 05 '24 at 16:54
  • If $\mathbb{E}[x_ix_j] = 0$ for all $i\ne j$ we could get $\mathbb{E}[|x^t M |^2] = \sigma_x^2 |M|_F^2$ by just expanding out (assuming each component of $x$ has the same variance $\sigma_x^2$ and mean $0$). – raj Dec 05 '24 at 17:01
  • @raj agreed, in that case though the original equality as stated is wrong yes? eg. we need the form as you state with the appropriate squarings in place – giorgio Dec 05 '24 at 17:09

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