# Replacing expectation by Lp norm

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.

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

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

Plug back should solve the original problem.