Lipschitz function of independent sub-Gaussian random variables

If $X\sim \mathcal{N}(0,I)$ is a Gaussian random vector, then Lipschitz functions of $X$ are sub-Gaussian with variance parameter 1 by the Tsirelson-Ibragimov-Sudakov inequality (eg. Theorem 8 here).

Suppose $X = (X_1,X_2,\ldots, X_n)$ consisted of independent sub-Gaussian random variables themselves, which are not normally distributed. Does the above property still hold?

• Hi, any news on your question? I'm also interested. Commented Jan 22, 2017 at 20:02
• No, I have not been able to resolve this yet. Commented Jan 28, 2017 at 16:43

Here are two options that may suit your needs.

1. Concentration inequality for convex functions of bounded random variables. If $$X_1,...,X_n$$ are independent taking values in in $$[0,1]$$ and $$f$$ is a quasi-convex, then $P(f(X) > m+t) \le 2e^{-t^2/4}, P(f(X) < m - t) \le 2e^{-t^2/4} \qquad$ where $$m$$ is the median of $$f(X)$$. See Theorem 7.12 in the book Concentration Inequalities: A Nonasymptotic Theory of Independence by Gábor Lugosi, Pascal Massart, and Stéphane Boucheron. It follows from the convex distance inequality due to Talagrand.

2. View $$X_i$$ as a function of a standard normal. If $$X_i$$ can be written as $$\Phi(Z_i)$$ where $$Z_i$$ is standard normal, then $$f(X) = f\circ \Phi(Z)$$ where $$Z_1,...,Z_n$$ are iid standard normal. Here, the multivariate function $$\Phi:R^n\to R^n$$ applies $$\Phi$$ on every coordinate.
Then the Tsirelson-Ibragimov-Sudakov inequality applies to $$f\circ \Phi$$, and the Lipschitz norm of $$f\circ \Phi$$ is at most $$\|f\|_{Lip} \|\Phi\|_{Lip}$$. Now, the question is whether $$\|\Phi\|_{Lip}$$ is bounded by an absolute constant (and, in particular, whether $$\Phi$$ is Lipschitz at all, otherwise $$\|\Phi\|_{Lip}=+\infty$$ and we do not get anything). Inequality $$\|\Phi\|_{Lip} holds, for instance, if $$X_i$$ is uniformly distributed on $$[0,1]$$, see Theorem 5.2.10 in the book High Dimensional Probability by Roman Vershynin where this approach is described.

3. If $$X$$ has density $$e^{-U(x)}$$ for strongly convex $$U:R^n\to R^n$$. If $$U$$ is twice continuously differentiable and strongly convex in the sense that the Hessian $$H$$ of $$U$$ (i.e., $$H_{ij} = (\partial/\partial x_i)(\partial/\partial x_i) U$$ satisfies for all $$x\in R^n$$ that $$H(x) - \kappa I_{n\times n}$$ is positive semi-definite, then for any 1-Lipschitz function $$f$$ of $$X$$, $P( |f(X) - E[f(X)] | > t) \le 2 \exp(-\kappa c t^2)$ for some absolute constant $$c>0$$. This is Theorem 5.2.15 in the book High Dimensional Probability by Roman Vershynin.

• How do you derive 1 from the convex distance inequality?
– Meni
Commented Jul 18, 2019 at 10:03

Try the following extension of McDiarmid’s inequality for metric spaces with unbounded diameter: https://arxiv.org/pdf/1309.1007.pdf

• If $X$ is d-dimensional sub-gaussian vector with $Cov(X) = \sigma^2 I$ and $f$ is L-Lipschitz w.r.t Euclidean norm, can we use the result of this paper? If so, what would be the sub-gaussian diameter as defined in the paper?
– ie86
Commented May 25, 2017 at 6:30
• No, this result would work for the $\ell_1$ norm, but not for $\ell_2$. Commented Jun 18, 2019 at 10:24