# Know the concentration estimate expectation？

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?

• In light of the answer, this begs the question; who asked you to prove that? – Clement C. Dec 30 '18 at 18:11
• @ClementC. math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf exercise 4.4.6 – ShaoyuPei Dec 30 '18 at 23:07
• 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). – Clement C. Jan 1 at 17:57
• @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 ... – ShaoyuPei Jan 2 at 13:19
• 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 $c'\stackrel{\rm def}{=} c\cdot(\frac{\sqrt{\pi}}{2}+1)$. – Clement C. Jan 2 at 13:52

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 $$c>0$$.