# The bound for the conditional expectation related to a sub-gaussian random vector

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.

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
$$\mathbb{E}\left[\exp\left(\lambda\|Y\|_2 X^{T}\frac{Y}{\|Y\|_2}\right) \mid Y\right] \leq \exp(C\lambda^2 \|Y\|^2_2 \|X\|^2_{\phi_2}).$$
• " 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! Commented Feb 19, 2020 at 20:49