# Comparison Inequality for Moment Generating Function.

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}