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?
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 $c>0$.