Sub-gaussianity of product of a bounded random variable and sub-gaussian random variable

Question

Let $Z \sim \mathcal{N}(0,1)$ be a $1$-dimensional standard Gaussian random variable. Let $f(z)=\frac{1}{1+e^{-z}}$ be the logistic sigmoid function. For some $\alpha>0$, I want to show that the random variable $$ Y \triangleq f(\alpha Z) Z- \mathbb{E}[f(\alpha Z)Z] $$ is also sub-gaussian. Can anyone provide me some hints regarding how to prove this?

Answer 1

It may be slightly hand-wavy, but what about this direction? Firstly, $$-Z \leq f(\alpha Z)Z \leq Z$$ due to range of sigmoid. Then, using Cauchy-Scwartz, $$\mathbb{E}[f(\alpha Z) Z] \leq \sqrt{\mathbb{E}[Z^2] \mathbb{E}[f^2(\alpha Z)]} = \sqrt{\mathbb{E}[f^2(\alpha Z)]} \leq 1 $$ So, $-Z - 1 \leq Y \leq Z+1$ and this means $Y$ is sub-Gaussian.

Answer 2

I will use some lemmas and definitions given here, in particular Lemma 2.7.6 which tells you that a random variable $Z$ is subgaussian if and only if $Z^2$ is subexponential.

Let $\alpha>0$, $f(x)\triangleq 1/(1+e^{-x})$, $Z \sim \mathcal{N}(0,1)$, $\widetilde{Y} \triangleq f(\alpha Z)Z$ and $Y \triangleq f(\alpha Z)Z - \mathbb{E}[f(\alpha Z)Z] = \widetilde{Y} - E\widetilde{Y}$.

By Lemma 2.6.8 it suffices to show that $\widetilde{Y}$ is subgaussian. By Lemma 2.7.6, it suffices to show that $\widetilde{Y}^2$ is subexponential (see Definition 2.7.5). Since $|f|$ is bounded by 1, we have $$\widetilde{Y}^2 \leq Z^2 \quad a.s..$$ Because $\mathcal{A}\subseteq \mathcal{B} \implies P(\mathcal{A}) \leq P(\mathcal{B})$, we have $$\forall t\geq 0, \quad P(\widetilde{Y}^2\geq t) \leq P(Z^2\geq t).$$ To show that $\widetilde{Y}^2$ is subexponential, it remains to obtain an exponential decreasing upper-bound on $P(Z^2\geq t)$. Since $Z\sim\mathcal{N}(0,1)$ is 1-subgaussian (see Exemple 2.5.8), Lemma 2.7.6 implies that $Z^2$ is subexponential. This means that there exists some constant $C>0$ such that $$ \forall t\geq 0, \quad P(|Z^2| \geq t) \leq 2\exp(-tC).$$ Combining both inequalities yields $$\forall t \geq 0, \quad P(|\widetilde{Y}^2| \geq t) \leq 2\exp(-tC),$$

and this actually shows that $\widetilde{Y}^2$ is subexponential.