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 $.

  1. 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?
  2. 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.

  • 2
    $\begingroup$ Could you add what is meant by convex order? $Ef(X)\le Ef(Y)$ for every convex $f$? $\endgroup$
    – jlewk
    Oct 11, 2023 at 16:04
  • $\begingroup$ @jlewk yes, exactly as you say. $\endgroup$
    – πr8
    Oct 11, 2023 at 21:14

1 Answer 1


If $Z\sim N(0,1)$ and $X$ is subgaussian with subgaussian norm less than $1/\sqrt 2$ then $$P(|X|>t)\le 2 e^{-t^2} \le K P(|Z|>t) $$ for some numerical constant $K>1$, thanks to lower bound on $P(Z>t)$. For instance, $K=4$ is achievable thanks to Formula 7.1.13 in Handbook of Mathematical Functions with. Formulas, Graphs, and Mathematical Tables by Milton Abramowitz and Irene A. Stegun, which gives $P(|Z|>t)\ge 4e^{-t^2/2}/(\sqrt{2\pi}(t+\sqrt{4+t^2})$.

Once the above inequality holds for some $K>0$, the following is essentially a trick in Lemma 4.6 of Probability in Banach Spaces by Ledoux and Talagrand.

First, let us study the case where $X$ is symmetric in the sense that it is equal in distribution to $\epsilon |X|$ where $\epsilon$ is a Rademacher random variable independent of $|X|$.

Let $\delta\sim$Bernoulli$(1/K)$ independent of everything else so that $K E[\delta]=1$. Then by independence $$ P(|\delta X|>t) = P(|X|>t) / K \le P(|Z|>t) $$ for all $t>0$ which grants stochastic dominance of $|X|$ by $|\delta Z|$, i.e., there is a rich enough probability space such that $X,\delta,Z$ have the same distribution as before and $P(|X|\delta \le | Z|)=1$.

Let $F$ be convex. Denote by $E_\epsilon$ the conditional expectation given $(\delta,|X|,Z)$ (i.e., integration with respect to the law of $\epsilon$ only). The function $$a \mapsto E_\epsilon[ F(a \epsilon |Z|)]$$ is convex and it attains is maximum over $a\in[-1,1]$ at an extreme point, either $-1$ or $1$. With $a=X\delta/|Z|$ this gives almost surely $E_\epsilon[F(\epsilon|X|\delta)] \le E_\epsilon F(\epsilon |Z|)$. Taking expectation and using Jensen's inequality with respcet to $\delta$ noting that $E[\delta|Z,X]=1/K$, $$ E[F(X/K)] = E[F(\tfrac1K \epsilon|X|)] \le E[F(\delta \epsilon|X|)] \le E[F(\epsilon |Z|) = E[F(Z)]. $$ where the two equalities follow thanks to $X=^d \epsilon|X|$ and $Z=^d \epsilon |Z|$. This solves the question if $X$ is symmetric.

If $Y$ is subgaussian but not symmetric, the definition of subgaussianity used in the question implies $E[Y]$ so that if $Y'$ is an independent copy of $Y$, $E[F(Y-E[Y'])]\le E[F(Y-Y')]$. Now $X=Y-Y'$ is subgaussian (with twice the subgaussian norm) and we may apply the result to the symmetric random variable $X$. This gives $E[F(Y/8)]\le E[F(Z)]$ for any $1/\sqrt 2$-subgaussian random variable $Y$.

  • $\begingroup$ I will take a look at the reference in Ledoux-Talagrand, thanks. I'm not so clear about the final point; all of the variables which I consider are centred, and so $P(X>0)$ is never $1$. $\endgroup$
    – πr8
    Oct 11, 2023 at 21:19
  • $\begingroup$ I also wonder whether your assertion that $F(|X|\delta) \leq F(|Z|)$ secretly uses monotonicity of $F$ in some form? $\endgroup$
    – πr8
    Oct 11, 2023 at 21:35
  • $\begingroup$ If $F$ is twice differentiable, the first and second derivatives of $a\to F(|aZ|)$ are $|Z| F'(|aZ|)$ and $Z^2 F''(|aZ|)$ if $a>0$, and $-|Z| F'(|aZ|)$ and $Z^2 F''(|aZ|)$ if $a<0$. This proves convexity in this case and I do not see a secret use of monotonicity. You are right regarding the final point, I had in mind the definition of subgaussian rv that does not require zero-mean, but your definition with the MGT requires it. I will think about it. $\endgroup$
    – jlewk
    Oct 12, 2023 at 1:37
  • $\begingroup$ I modified the answer to obtain $E[F(Y/8)]\le E[F(Z)]$ for any 1-subgaussian random variable $Y$. $\endgroup$
    – jlewk
    Oct 12, 2023 at 5:37
  • 1
    $\begingroup$ Thanks again for correcting the first paragraph. Indeed the variance of $Z$ needs to be a little larger than the sub-gaussian norm of $X$ to compare the tails for all t, for instance $2e^{-t^2}\lesssim P(|Z|>t)\asymp \frac1t e^{-t^2/2}$ holds if there is no $1/2$ in the leftmost exponential. The answer is fixed. $\endgroup$
    – jlewk
    Oct 12, 2023 at 13:33

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