Timeline for Trying to establish an inequality on probability
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12, 2020 at 10:38 | history | edited | CommunityBot | Commonmark migration | |
| Mar 18, 2015 at 23:21 | vote | accept | CommunityBot | ||
| Mar 15, 2015 at 21:05 | answer | added | Zilin J. | timeline score: 2 | |
| Mar 15, 2015 at 21:01 | comment | added | Did | Hint: Use $$ P( Y \geq y ) \leq \frac{ E(Y)}{y} $$ with $$Y=\alpha-f(X)\ge0\ \text{almost surely},\qquad y=\alpha-a>0.$$ | |
| Mar 15, 2015 at 20:56 | comment | added | Michael Hardy | If $Y=f(X)$ then you can say $Y:\Omega\to\mathbb R$ and $\Pr(0\le Y\le\alpha)=1$ and then say you want to prove that for $0\le a\le\alpha$ you have $\Pr(Y\ge a)\ge\dfrac{\operatorname{E}(Y)-a}{\alpha-a}$. There doesn't seem to be any point in writing it as $f(X)$ instead of just $Y$. ${}\qquad{}$ | |
| Mar 15, 2015 at 20:52 | comment | added | Michael Hardy | I doubt you need countability here. ${}\qquad{}$ | |
| Mar 15, 2015 at 20:50 | history | asked | user139708 | CC BY-SA 3.0 |