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Here is the problem:

let $X$ be a random variable such that: $$P\{X > c(m+t) \}<2e^{-t^2} \ \ \ \forall t >0$$ where $c>0,m>0$ are constant.

Then I was asked to prove that : $$ \mathbb{E}[X] \leq cm$$

however I can't get rid of constant, here is my attempt:

\begin{align} \mathbb{E}[X] &= \int_0^\infty P\{X>u\}\mathrm{d}u\\ &= \int_{cm}^\infty P\{X>u\}\mathrm{d}u +\int_{0}^{cm} P\{X>u\} \mathrm{d}u \\ &\leq \int_{cm}^\infty P\{X>u\}\mathrm{d}u+ cm \\&= \int_{cm}^\infty P\{X>cm+c\delta\}\mathrm{d}u+ cm \ \ \ \ (\delta>0) \\&= c\int_{0}^\infty P\{X>cm+c\delta\}\mathrm{d}\delta \ \ \ (Change \ of \ vairible) \\& \leq c\int_{0}^\infty e^{-\delta^2} \mathrm{d}\delta +cm \ \ (assumption) \end{align}

what I have been missing here?

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  • $\begingroup$ In light of the answer, this begs the question; who asked you to prove that? $\endgroup$ – Clement C. Dec 30 '18 at 18:11
  • $\begingroup$ @ClementC. math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf exercise 4.4.6 $\endgroup$ – ShaoyuPei Dec 30 '18 at 23:07
  • $\begingroup$ Doesn't Exercise 4.4.6 include another constant $C$? (I.e., I wonder if the author doesn't use the "notation" that $C$ is an absolute constant, which may not take the same value across results). $\endgroup$ – Clement C. Jan 1 at 17:57
  • $\begingroup$ @ClementC. you are right , C is an absolute constant ,so Constant C is free to choose ? For small c in this question, I think if constant part is $\frac{1}{c}\int_{0}^\infty e^{-\delta^2}$,we could take c to be large , then the constant term can be neglect, or I did integration wrong ... $\endgroup$ – ShaoyuPei Jan 2 at 13:19
  • $\begingroup$ Well, with what your wrote in your OP, you get $c\cdot(\frac{\sqrt{\pi}}{2}+m)$. so since $m\geq 1$ in the book ($m,n$ are integers, right, and "your" $m$ is their $\sqrt{n}+\sqrt{m}$), you do have $c\cdot(\frac{\sqrt{\pi}}{2}+m) \leq c\cdot(\frac{\sqrt{\pi}}{2}+1)m = c'm$, setting $c'\stackrel{\rm def}{=} c\cdot(\frac{\sqrt{\pi}}{2}+1)$. $\endgroup$ – Clement C. Jan 2 at 13:52
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The inequality is false. For example if $Z$ has uniform distribution on $(0,1)$ and $X=cZ+cm$ then the hypothesis reduces to $P(Z>t) \leq 2e^{-t^{2}}$ or $1-t \leq 2e^{-t^{2}}$ for $t\in (0,1)$ which can be proved easily by writing the inequality as $e^{t^{2}}(1-t)-2\leq 0$. [ The left side is decreasing and it has the value $-1$ at $t=0$]. Note that $EX=\frac c 2 +cm>cm$ if $c>0$.

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