= Gaussian concentration inequality
{c}
For a <Gaussian random vector> $X\sim N(m,\Sigma)$ and an $L$-<Lipschitz function> $f$ in the <Euclidean norm>, put $v=L^2\|\Sigma\|_{\mathrm{op}}$. If $v>0$, the <Gaussian concentration inequality> gives
$$
\log\mathbb E\exp[\lambda(f(X)-\mathbb Ef(X))]\leq\frac{v\lambda^2}{2}\quad(\lambda\in\mathbb R),
$$
and therefore
$$
\mathbb P(f(X)-\mathbb Ef(X)\geq t)\leq e^{-t^2/(2v)},\qquad
\mathbb P(|f(X)-\mathbb Ef(X)|\geq t)\leq2e^{-t^2/(2v)}\quad(t\geq0).
$$
Write $X=m+\Sigma^{1/2}G$ with a standard <Gaussian random vector> $G$. The composed <function> has <Lipschitz constant> at most $L\|\Sigma\|_{\mathrm{op}}^{1/2}$. The <Gaussian logarithmic Sobolev inequality> and <Herbst argument> give the exponential bound, and the <Chernoff bound> supplies the tails.<Probability in High Dimension by Ramon van Handel>[Theorems 3.24–3.25 and their exponential-moment proof]{ref}
For example, $f(x)=\|x\|$ has $L=1$, so its deviations depend on the largest <covariance> <eigenvalue> rather than directly on the number of coordinates. The center $\mathbb E\|X\|$ can still grow with dimension. If $v=0$, $f(X)$ is constant <almost surely> and every positive deviation has <probability> zero.
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