Upper bound on the expectation from Bernstein-type inequality

Question

Let $Z$ be a nonnegative random variable satisfying the following concentration inequality for all $t\ge 0$:

$$\mathbb{P}[Z\ge t]\le C e^{-\frac{t^2}{2(1+t)}}$$

(for some $C\ge 1$)

Then:

$$\mathbb{E}[Z]\le 2(\sqrt{\pi}+\sqrt{\log C})+4(1+\log C)$$

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This is basically exercise 2.8 from High-Dimensional Statistics: A Non-Asymptotic Viewpoint. I tried to use:

$$\mathbb{E}[Z]=\int_0^\infty \mathbb{P}[Z\ge t]\, dt\le C\int_0^\infty e^{-\frac{t^2}{2(1+t)}}\, dt$$

But I don't know how to upper bound the term in the integral (also, I don't see how I would get $\log C$ or $\sqrt{\log C}$ terms from this expression). Is there another way to get $\mathbb{E}[Z]$ from $\mathbb{P}[Z\ge t]$?

Answer 1

Thanks to the hint of @Saad in comments, I am now able to finish the proof.

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First I separate the summand in the integral into two parts (the first one will be relevant when $t$ is small, the second one when $t$ is large):

$$\exp\left(-\frac{t^2}{2(1+t)}\right)\le \exp\left(-\frac{t^2}{4\max(1,t)}\right)\le e^{-t^2/4}+e^{-t/4}$$

Now I can use the hint in the following way:

$$\mathbb{E} Z=\int_0^\infty\mathbb{P}[Z\ge t]\le \int_0^\infty \min\left(C \exp\left(-\frac{t^2}{2(1+t)}\right),1\right)\,dt\le A+B$$

with:

$$A=\int_0^\infty \min(C e^{-t^2/4},1)\,dt$$ $$B=\int_0^\infty \min(C e^{-t/4},1)\,dt$$

For $B$ the crossing point of the two regimes is for $t=4 \log C$, hence:

$$B\le 4\log C+\int_{4\log C}^\infty C e^{-t/4}\,dt=4(\log C+1)$$

For $A$ it happens at $t=2\sqrt{\log C}$, so that:

$$A\le 2\sqrt{\log C}+\int_{2\sqrt{\log C}}^\infty Ce^{-t^2/4}\,dt$$

And the integral can be upper bounded as follows:

$$\int_{2\sqrt{\log C}}^\infty Ce^{-t^2/4}\,dt=\sqrt{2} C\int_{\sqrt{2\log C}}^\infty e^{-t^2/2}\,dt\le 2\sqrt{\pi}$$

where I used the standard inequality:

$$\int_x^\infty e^{-t^2/2}\, dt\le \sqrt{2\pi} e^{-x^2/2}$$