Bounding integral arising from expectation of a random variable satisfying Bernstein's inequality

Question

Let $X$ be a random variable s.t. for $v, b > 0$ and $C \geq 1$:

$$ P(X \geq t) \leq C\exp\left(-\frac{t^2}{2(v^2 + bt)} \right) $$

I am trying to show that $\mathbb{E}X \leq 2v(\sqrt{\pi} + \sqrt{\log C}) + 4b(1 + \log C)$.

In terms of progress I have made so far, I set $X = \int_0^\infty P(X \geq t)dt$, and have also made cases for when $v^2 > bt$ and otherwise (i.e. simplifying the exponentiated value to be $-t^2/4v^2$ in one case and $-t^2/4bt$ in the other).

This obtainis all the non $C$ values in the final inequality; However, I can't figure out how to squash the $C$ into the $\log C$ and $\sqrt{\log C}$ terms.

Answer 1

For $t>0$, we have $v^2+bt\le\max(2bt,2v^2)$, whence \begin{equation} P(X\ge t)\le1\wedge\max[Q_1(t),Q_2(t)]\le[1\wedge Q_1(t)]+[1\wedge Q_2(t)], \end{equation} where $x\wedge y:=\min(x,y)$, \begin{equation} Q_1(t):=C\exp\Big(-\frac{t^2}{4bt}\Big),\quad Q_2(t):=C\exp\Big(-\frac{t^2}{4v^2}\Big). \end{equation} So, \begin{equation} EX=\int_0^\infty P(X\ge t)\,dt\le I_1+I_2, \end{equation} where \begin{multline} I_1:=\int_0^\infty[1\wedge Q_1(t)]\,dt =4b\ln C+C\int_{4b\ln C}^\infty \exp\Big(-\frac{t}{4b}\Big)\,dt=4b(1+\ln C) \end{multline} and \begin{multline} I_2:=\int_0^\infty[1\wedge Q_2(t)]\,dt =v\sqrt 2\int_0^\infty\min(1,Ce^{-s^2/2})\,ds \\ =v\sqrt 2\Big(\sqrt{2\ln C}+C\int_{\sqrt{2\ln C}}^\infty e^{-s^2/2}\,ds\Big) \le 2v\big(\sqrt{\ln C}+\sqrt\pi/2\big); \end{multline} the latter inequality follows because $\int_t^\infty e^{-s^2/2}\,ds\le\sqrt{\pi/2}\,e^{-t^2/2}$ for $t\ge0$.

Thus, \begin{equation} EX\le 4b(1+\ln C)+2v\big(\sqrt{\ln C}+\sqrt\pi/2\big), \end{equation} which is a bit better than the bound requested in the OP, where we see $\sqrt\pi$ in place of $\sqrt\pi/2$.