If $Y$ is a sub-Gaussian random variable, must $Y-E(Y\mid X)$ be also sub-Gaussian? 
Question. If $Y$ is a sub-Gaussian random variable, must the regression residual $\varepsilon=Y-E(Y\mid X)$ be also sub-Gaussian? That is, whether there exists some constant $\sigma>0$ such that $E\big( e^{\lambda \varepsilon} \big) \leq e^{\lambda^2\sigma^2/2}$ for any $\lambda\in\mathbb R$.

In terms of the variance, it is well known that $\mathrm{var}(\varepsilon)\leq \mathrm{var}(Y)$. However, I can't find out a way to prove (or disprove) the sub-Gaussianity of $\varepsilon$ when $Y$ is sub-Gaussian. I wonder if $\varepsilon$ is always sub-Gaussian, by the sub-Gaussianity of $Y$, and maybe plus some additional requirements on $X$ (such as $X$ is also sub-Gaussian) or $E(Y\mid X)$ (such as $E(Y\mid X)$ is linear).
 A: If you're dealing with linear regression - the answer is yes, $\epsilon$ is indeed subG.
First and most obvious, $E[\epsilon]=E[Y]-E[E[Y|X]]=0$ by law of total expectation. Then, according to Vershynin, all you need to prove is the bound on the MGF of $\epsilon$.
Now, if you're modelling linear regression then $E[Y|X]=\hat{Y}\sim\mathcal{N}$, which means this is a subG variable as well and by Hoeffding's lemma we get that their sum is also subG.
This won't hold for Poisson regression, as the Poisson distribution is not subG.
Edit:
The thing with subG property is this: you need the distribution to fulfill that, from a point $\lambda>0$, the probability tail $P(|Y|>\lambda)$ is lighter than the Gaussian tail. This is usually the case when $Y$ is some variant of Gaussian or when it is bounded. If $Y$ is subG with variance proxy $\sigma^2$ then by definition  of regression (classic, not Bayesian) we get that $\hat{Y}$ has subG distribution with the same proxy, hence the residual is subG with this proxy as well. The only condition here is that $E[Y|X]=E[Y]$ - that is, $\hat{Y}$ should be an unbiased estimator.
