In Vershynin's book High Dimensional Probability, Exercise 6.2.6 (a) asks to show that if $X$ is a mean zero, subgaussian vector in $\mathbf{R}^n$ with $\| X \|_{\psi_2} \leq K$, and $B$ is an $m \times n$ matrix, then

$$ \mathbf{E} \exp(\lambda^2 |BX|^2) \leq \mathbf{E} \exp(C K^2 \lambda^2 |B^T g|^2) $$

where $g \sim N(0,I_m)$.

I'm having difficulty proving this. So far, I've noticed that for any $y$, we have $\| y \cdot BX \|_{\psi_2} \leq |B^T y| \| X \|_{\psi_2}$. Thus

$$ \mathbf{E} \exp(\lambda^2 y \cdot BX) \leq \mathbf{E} \exp(C K^2 \lambda^2 |B^T y|^2). $$

In particular, this means that we can prove that if $g$ is independent of $X$, then

$$ \mathbf{E} \exp(\lambda^2 g \cdot BX) \leq \mathbf{E} \exp(CK^2 \lambda^2 |B^T g|^2). $$

This seems like the right way to go, but I don't think the left hand side upper bounds the $\mathbf{E} \exp(\lambda^2 |BX|^2)$? Is this close to the correct way to prove this exercise, or am I missing a different path?


If $X$ and $g$ are independent, the conditional distribution of $g^TBX$ given $X=x$ is $\mathcal N(0,\|Bx\|_2^2)$, and $E(\exp(\mu g^TBX)) = E(E(\exp(\mu g^TBX)|X)=E\left(\exp(\frac{\mu^2 \|BX\|_2^2}{2})\right)$.

Letting $\mu=\sqrt 2 \lambda$ yields $\begin{aligned}[t]E(\exp(\lambda^2\|Bx\|_2^2)) &= E(\exp(\sqrt 2 \lambda g^TBX))=E(\exp(\sqrt 2 \lambda \langle X,B^Tg\rangle))\\ &=E(E(\exp(\sqrt 2 \lambda \langle X,B^Tg\rangle)|g))\\ &\leq E(\exp(C2\lambda^2 K^2\|Bx\|_2^2))\\ &= E(\exp(C'\lambda^2 K^2\|Bx\|_2^2)) \end{aligned}$

which proves the claim.


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