Proof (at least name or source) of this probability inequality?

Question

I stumbled upon this beautiful piece of inequality : I'd appreciate very much if anyone can provide me a source where I can find the proof of this inequality. Does it have a name? (It must! But I don't know where to look.)

Answer 1

Probably no name. For a proof, first note that the hypothesis that $$c_1>0\qquad c_2>e^{-1}$$ should read $$c_1>e^{-1}\qquad c_2>0$$ for the conclusion to make sense and let us choose values of the parameters such that $$c_1=c\qquad c_2n=1$$ Replacing $X_n$ by $\sqrt{c_2n}X$, this is not a loss of generality, thus it suffices to consider the case when, for every $x>0$, $$P(X>x)\leqslant ce^{-x^2}$$ Then $$E(X)=\int_0^\infty P(X>x)dx$$ Thus, if $c<1$, $$E(X)\leqslant\int_0^\infty ce^{-x^2}dx=c\frac{\sqrt{\pi}}2$$ so the task in this case is to prove that, for every $c$ in $(e^{-1},1)$, $$c\frac{\sqrt{\pi}}2\leqslant\sqrt{1+\log c}$$ or, equivalently, that for every $a$ in $(0,1)$, $$e^{-a}\frac{\sqrt{\pi}}2\leqslant\sqrt{1-a}\tag{1}$$ On the other hand, if $c>1$, $$E(X)\leqslant\int_0^{\sqrt{\log c}}dx+\int_{\sqrt{\log c}}^\infty ce^{-x^2}dx$$ so the task in this case is to prove that, for every $c>1$, $$\sqrt{\log c}+\int_{\sqrt{\log c}}^\infty ce^{-x^2}dx\leqslant\sqrt{1+\log c}$$ or, equivalently, that, for every $a>0$, $$a+\int_a^\infty e^{a^2-x^2}dx\leqslant\sqrt{1+a^2}\tag{2}$$ The trouble with all this is that your problem is really equivalent to proving $(1)$ and $(2)$ and that $(1)$ is obviously wrong.

Edit: It appears that, while $(1)$ is wrong, the desired inequality holds when $c>1$. Since the technique used in the notes mentioned in a comment below works in this case, avoids completely the clumsy inequality $(2)$ above, and is, independently of the specific question at hand, worthy of being remembered, we outline it here.

Again, one starts from the hypothesis that $X\geqslant0$ almost surely and that, for some $c\geqslant1$, for every $x\geqslant0$, $$P(X>x)\leqslant ce^{-x^2}$$ Then one makes the apparent détour of considering $Y=X^2$ instead of $X$... Thus, for every $y\geqslant0$, $$P(Y> y)=P(X>\sqrt y)\leqslant ce^{-y}$$ hence $$E(X^2)=E(Y)=\int_0^\infty P(Y> y)dy\leqslant\int_0^{\log c}dy+\int_{\log c}^\infty ce^{-y}dy=(\log c)+1$$ hence, using Cauchy-Schwarz inequality, $$E(X)\leqslant\sqrt{E(X^2)}\leqslant\sqrt{1+\log c}$$ To sum up the above, passing by $X^2$, one loses a bit due to Cauchy-Schwarz but one gains a lot because the primitives of $e^{-y}$ are handier than those of $e^{-x^2}$. Cute...