# Sub-Gaussian random variables and convex ordering

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.

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

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 $$Y$$.

• 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 $1$.
– πr8
Oct 11, 2023 at 21:19
• I also wonder whether your assertion that $F(|X|\delta) \leq F(|Z|)$ secretly uses monotonicity of $F$ in some form?
– πr8
Oct 11, 2023 at 21:35
• 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. Oct 12, 2023 at 1:37
• I modified the answer to obtain $E[F(Y/8)]\le E[F(Z)]$ for any 1-subgaussian random variable $Y$. Oct 12, 2023 at 5:37
• 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. Oct 12, 2023 at 13:33