Suppose that $X$ is a $1$-sub-Gaussian real-valued random variable, i.e. for all $t \in \mathbf{R}$, it holds that $\log \mathbf{E} \exp \left( t X \right) \leqslant \frac{1}{2} t^2 $.

- Does there always exist a constant $\sigma = \sigma_X \geqslant 1$ such that (writing $G$ for a standard Gaussian variable), $X$ is dominated by $\sigma_X \cdot G$ in the convex ordering?
- If so, can $\sigma_X$ be taken independent of $X$?

An example, to show that there is hope: consider the case of (normalised) Bernoulli random variables, and write $X \overset{\mathrm{d}}{=} \mathsf{Ber}(p) - p $.

The Kearns-Saul inequality shows that

\begin{align} \log\mathbf{E}\exp\left(t \cdot X \right)&\leqslant\frac{1}{2}\cdot \left(\frac{1-2\cdot p}{2\cdot\log\left(\frac{1-p}{p}\right)}\right) \cdot t^{2}. \end{align}

With some work, one can show that $X$ is majorised in the convex order by $\sigma(p) \cdot G$, where $\sigma (p) =\frac{p\cdot\left(1-p\right)}{I_{\gamma}\left(p\right)}$ and $I_\gamma = \phi_\gamma \circ \Phi_\gamma^{-1}$ is the Gaussian isoperimetric function. This implies the (weaker) estimate on the MGF

\begin{align} \log\mathbf{E}\exp\left(t \cdot X \right)&\leqslant\frac{1}{2}\cdot\left(\frac{p\cdot\left(1-p\right)}{I_{\gamma}\left(p\right)}\right)^{2}\cdot t^{2}, \end{align}

which only loses at most a constant factor of at most $\sqrt{\frac{\pi}{2}} \approx 1.25$ in the implied $\sigma$, which happens when $p = 1/2$. So, in this (very) limited setting, the result holds.