 Gaussian concentration inequality

For a Gaussian random vector $X\sim N(m,\Sigma )$ and an $L$-Lipschitz function $f$ in the Euclidean norm, put $v=L^2\Vert \Sigma {\Vert }_{\mathrm{op}}$. If $v>0$, the Gaussian concentration inequality givesand thereforeWrite $X=m+{\Sigma }^{1/2}G$ with a standard Gaussian random vector $G$. The composed function has Lipschitz constant at most $L\Vert \Sigma {\Vert }_{\mathrm{op}}^{1/2}$. The Gaussian logarithmic Sobolev inequality and Herbst argument give the exponential bound, and the Chernoff bound supplies the tails.^[ref]

For example, $f(x)=\Vert x\Vert$ has $L=1$, so its deviations depend on the largest covariance eigenvalue rather than directly on the number of coordinates. The center $\mathbb{E}\Vert X\Vert$ can still grow with dimension. If $v=0$, $f(X)$ is constant almost surely and every positive deviation has probability zero.