It is known that for a Lipschitz function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ that if $X \sim \mathcal{N}(0,I_n)$ then $$ \| f(X) - \mathbb{E}f(X)\|_{\psi_2} \leq C \|f\|_{Lip} $$ where $\| \|_{\psi_2}$ refers to the subgaussian norm (i.e. the smallest constant $C$ such that $\mathbb{E} \exp(X^2/C^2) \leq 2$). Is it possible to replace the $\mathbb{E}f(X)$ by $(\mathbb{E} f(X)^3)^{1/3}$ and still get a concentration inequality of the same form.

I have no problem replacing the expectation by the median as it is well known that the expectation and median are close for subgaussian random variables. I would like to prove the same phenomenon for Lp norms.

  • $\begingroup$ How is this norm defined $\| \cdot\|_{\psi_2}$? $\endgroup$
    – Guy Fsone
    Oct 24, 2017 at 14:52
  • $\begingroup$ I have defined it in the question. $\endgroup$
    – JohnKnoxV
    Oct 24, 2017 at 15:23

2 Answers 2


Yes, if your $f$ is nonnegative. Once you have a subgaussian tail, median, mean and $L_p$ norm are all within an additive constant from each other (depending on $p$).

Let $Z = f(X)\geq 0$, what you have is $\|Z-\mathbb{E} Z\|_{\psi_2}\leq C\|f\|_{Lip}$. It then follows that $$ \|Z-\mathbb{E}Z\|_{p} \leq C'\sqrt{p}\|Z-\mathbb{E} Z\|_{\psi_2} $$ for some absolute constant $C'$. On the other hand, $$ \|Z-\mathbb{E}Z\|_{p} \geq \|Z\|_p - \|\mathbb{E}Z\|_{p} = \|Z\|_p - \mathbb{E} Z. $$ This implies that $$ \mathbb{E}Z \leq (\mathbb{E} |Z|^p)^{1/p} \leq \mathbb{E}Z + C'\sqrt{p}\|Z-\mathbb{E} Z\|_{\psi_2}. $$


Source of this question is Exercise 5.1.4 from Vershynin's book. I rewrite user58955's solution here in a more digestable form.

Let $Z:=f(X)$

$\|Z-\|Z\|_p\|_{\psi_2}\le \|Z-EZ\|_{\psi_2}+\|EZ-|Z\|_p\|_{\psi_2}\le C\|f\|_{Lip}+C'|EZ-\|Z\|_p|$

But $|EZ-\|Z\|_p|\le \|EZ-Z\|_p<C''\sqrt{p} \|Z-EZ\|_{\psi_2}<C'C\sqrt{p} \|f\|_{Lip}.$

Plug back should solve the original problem.


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