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Let $X,Y$ be two mean-zero, independent subgaussian random vectors and assume their sub-gaussian norms are both bounded by a positive number $K$.

Recall that for a mean-zero real-valued sub-gaussian random variable $Z$, we have the bound
$$\mathbb E\exp(\lambda Z)\le \exp(C\lambda^2\|Z\|^2_{\phi_2}), \forall \lambda\in \mathbb R.$$

I want to prove the following inequality

$$\mathbb E[\exp(\lambda X^TY)\mid Y]\le \exp(C'\lambda^2 K^2 \|Y\|^2_2), \forall \lambda\in \mathbb R.$$

One of the major difficulties for this generalization is that the conditional expectation of two independent random variables may not be uncorrelated and thus the multiplicity rule of the $\exp$ may not work.

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Note that $$\mathbb{E}[\exp(\lambda X^{T}Y) \mid Y] = \mathbb{E}\left[\exp\left(\lambda\|Y\|_2 X^{T}\frac{Y}{\|Y\|_2}\right) \mid Y\right].$$

Because we condition on $Y$ and $X$ is independent of $Y$, we can simply take the terms involving $Y$ within the expectation to be constant (you can make this rigorous, e.g., by invoking the law of the unconscious statistician). Using the definition of a sub-Gaussian random vector in these notes (see Definition 1.2 therein and the discussion after it), we then have

$$\mathbb{E}\left[\exp\left(\lambda\|Y\|_2 X^{T}\frac{Y}{\|Y\|_2}\right) \mid Y\right] \leq \exp(C\lambda^2 \|Y\|^2_2 \|X\|^2_{\phi_2}).$$

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  • $\begingroup$ " we can simply take the terms involving Y within the expectation to be constant (you can make this rigorous, e.g., by invoking the law of the unconscious statistician)" I think actually this is preciously what I am asking.... how to make it rigorous s.t. I can take $Y$ to be a constant. Thanks anyway! $\endgroup$
    – No One
    Commented Feb 19, 2020 at 20:49
  • $\begingroup$ Do Propositions 10.16 and 10.19 here answer your question? $\endgroup$
    – ProAmateur
    Commented Feb 19, 2020 at 21:38
  • $\begingroup$ very good notes, thanks! $\endgroup$
    – No One
    Commented Feb 20, 2020 at 13:32

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