One sided bound for Lipschitz functions for gaussian variables

Question

Let $(X_1, . . . , X_n)$ be a vector of i.i.d. standard Gaussian variables, and let $f : R^n → R$ be L-Lipschitz with respect to the Euclidean norm. Then the variable $f(X) − E[f(X)]$ is sub-Gaussian with parameter at most L, and hence

$$ P\left[|f(X)-E[f(X)]| \geq t \right] \leq 2e^{\frac{t^2}{2L^2}}$$

Is there an equivalent bound for the one sided equivalent? $$ P\left[f(X)-E[f(X)] \geq t \right]$$

Answer 1

Not sure if this is what you're asking for but here's a simple lower bound:

\begin{align} 2e^{\frac{t^2}{2L^2}} &\ge P(|f(X)-E[f(X)]|\ge t) \\ \iff &\\ 1-2e^{\frac{t^2}{2L^2}} &\le P(|f(X)-E[f(X)]|\le t)\\ &=P(\{X\in\Omega:f(X)-E[f(X)]\le t\}\cap\{X\in\Omega:f(X)-E[f(X)]\ge -t\})\\ &\le \min \{ P(f(X)-E[f(X)]\ge t) , P(f(X)-E[f(X)] \le -t)\}\\ &\le P(f(X)-E[f(X)]\ge t) \end{align}