Sub-Gaussian concentration without the sub-Gaussian norm

Question

A random variable $X$ is said to have sub-Gaussian tails with parameter $\sigma>0$ if $$\Pr[|X|\geq t] \leq 2\exp(-t^2/(2\sigma^2))$$

I am interested if $X_0, X_1$ are independent, and have sub-Gaussian tails with parameters $\sigma_0, \sigma_1$, if $X_0+X_1$ has sub-Gaussian tails with parameter $\sqrt{\sigma_0^2+\sigma_1^2}$.

It is well-known by the theory of Sub-Gaussian random variables that they have sub-Gaussian tails with parameter $O(\sqrt{\sigma_0^2+\sigma_1^2})$. It is straightforward to show that they have parameter at most $\color{red}{10}\sqrt{\sigma_0^2+\sigma_1^2}$ (combine that $\lVert X_0+X_1\rVert_{\psi_2}^2 \leq \lVert X_0\rVert_{\psi_2}^2+\lVert X_1\rVert_{\psi_2}^2$ with the equivalence of the Orlicz norm $\lVert\cdot\rVert_{\psi_2}$ and having sub-Gaussian tails, see for example Prop. 2.5.2 of Vershynin). This leads to a loss in parameters that I want to avoid though, hence my question.

Answer 1

$\newcommand\si\sigma$The answer is no.

E.g., suppose that $P(X_i=1)=2/e=1-(X_i=0)$ for $i=0,1$. Then $X_0$ and $X_1$ are sub-Gaussian with parameter $\si=1/\sqrt2$, so that we can take $\si_0=\si_1=1/\sqrt2$.

On the other hand (assuming that $X_0$ and $X_1$ are independent), $$P(|X_0+X_1|\ge2)=\frac4{e^2}\not\le\frac2{e^2}=2\exp\Big(-\frac{2^2}{2(\si_0^2+\si_1^2)}\Big).$$ So, $X_0+X_1$ is not sub-Gaussian with parameter $\sqrt{\si_0^2+\si_1^2}$. $\quad\Box$

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Working slightly harder, one can even make $X_0$ and $X_1$ zero-mean. Indeed, suppose that $P(X_i=-p)=q$ and $P(X_i=q)=p$ for $i=0,1$, where $q:=1-p$ and $p\downarrow0$. Then $X_0$ and $X_1$ are sub-Gaussian with parameter $\si=q/\sqrt{2\ln\frac2p}$ if $p$ is small enough, so that we can take $\si_0=\si_1=q/\sqrt{2\ln\frac2p}$.

On the other hand (assuming that $X_0$ and $X_1$ are independent), $$P(|X_0+X_1|\ge2q)=p^2\not\le\frac{p^2}2=2\exp\Big(-\frac{(2q)^2}{2(\si_0^2+\si_1^2)}\Big)$$ if $p$ is small enough. So, $X_0+X_1$ is not sub-Gaussian with parameter $\sqrt{\si_0^2+\si_1^2}$. $\quad\Box$

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The idea of these examples is that, while all bounded random variables (r.v.'s) are sub-Gaussian, they differ much from Gaussian r.v.'s if those bounded r.v.'s are heavy tailed -- say in the sense that their skewness coefficients (and hence kurtoses) are large.