Lipschitz function of independent sub-Gaussian random variables

Question

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?

Answer 1

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

Answer 2

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 $\|\Phi\|_{Lip}<M+\infty$ 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.