Let $\mathcal{F}$ denote a function class. A classic result by Talagrand states that

\begin{align*} \mathbb{P}\bigg\{\sup_{f\in\mathcal{F}}\big|\sum_{i=1}^nf(X_i)-\mathbb{E}\big[\sum_{i=1}^nf(X_i)\big]\big|\ge t\bigg\}\le K\cdot\text{exp}\bigg\{-\frac{1}{K}\frac{t}{U}\log\left(1+\frac{tU}{V}\right)\bigg\}, \end{align*} holds for all $t>0$, where $K$ is a universal constant, $U$ is a uniform bound for the functions in $\mathcal{F}$ and $V$ is any number satisfying $\mathbb{E}\sup_{f\in\mathcal{F}}\sum_{i=1}^n f^2(X_i)\le V$.

I am wondering about a one-sided version of this result. That is assume $f\ge -U$ holds for all $f\in\mathcal{F}$. Does \begin{align*} \mathbb{P}\bigg\{\inf_{f\in\mathcal{F}}\left(\sum_{i=1}^nf(X_i)-\mathbb{E}\big[\sum_{i=1}^nf(X_i)\big]\right)\le -t\bigg\}\le K\cdot\text{exp}\bigg\{-\frac{1}{K}\frac{t}{U}\log\left(1+\frac{tU}{V}\right)\bigg\}, \end{align*} hold?

  • $\begingroup$ I might get something wrong, but isn't your second probability upper-bounded by your first probability? Apparently inf a_n <-t => sup |a_n| > t $\endgroup$ – Koltchinskii Jan 8 '18 at 18:46
  • $\begingroup$ Thank you for your response. Yes I agree with your statement. However, please note that we now only have a one sided bound on f in the assumption rather than the two sided version. so one can not conclude the second identity from the first one due to the different assumptions. $\endgroup$ – mohi Jan 9 '18 at 19:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.