Timeline for Bounds on $\int \log(1+x) g(x) \mathrm{d}x$?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 20, 2016 at 21:07 | comment | added | John Jiang | Why the negative vote if some good samaritan took the pain to redact my answer? | |
| S May 18, 2016 at 6:20 | history | suggested | Amir Sagiv | CC BY-SA 3.0 |
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| May 18, 2016 at 6:08 | review | Suggested edits | |||
| S May 18, 2016 at 6:20 | |||||
| Mar 25, 2016 at 23:10 | comment | added | Chee | Can you first start by assuming $(X,Y)$ is bivariance Gaussian? Then you can look at a bivariate Lancaster family; then your problem relates to extremal points in the corresponding family. Without a joint distribution on $(X,Y)$, how can one get exact expression? Think about it this way: you are tying to expand a function in terms of the moments of another random variable. | |
| Mar 16, 2016 at 23:32 | comment | added | John Jiang | My Y satisfies P(Y > X|X=x) < k for all little x. Now integrate over $x < \beta$. Same holds for the other inequality. What's your counterexample? | |
| Mar 16, 2016 at 20:45 | comment | added | Jeff | $Y=X \, \mathbb{1}_{Z<k} + W \, 1_{Z \geq k}$ will not necessarily ensure both $\Pr[Y\geq \beta \mid X<\beta] \leq k$ and $\Pr[Y< \beta \mid X \geq \beta] \leq k$. | |
| Mar 15, 2016 at 2:02 | comment | added | John Jiang | Define Y = X 1{Z < k} + W 1{Z>=k}, where Z is a uniform [0,1], X, W, Z mutually independent. By manipulating W you can get arbitrarily large moments for Y regardless of moments of X. | |
| Mar 14, 2016 at 22:59 | comment | added | Jeff | not clear to me how we can let $Y=X$ with probability higher $k$ for each $X$. | |
| Mar 14, 2016 at 15:11 | history | answered | John Jiang | CC BY-SA 3.0 |