Timeline for An inequality based on expectation of continuous random variables
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 6, 2015 at 22:20 | vote | accept | Aliveli | ||
| Feb 6, 2015 at 22:20 | vote | accept | Aliveli | ||
| Feb 6, 2015 at 22:20 | |||||
| Feb 6, 2015 at 3:12 | vote | accept | Aliveli | ||
| Feb 6, 2015 at 22:20 | |||||
| Feb 4, 2015 at 15:54 | answer | added | Nate Eldredge | timeline score: 3 | |
| Feb 4, 2015 at 15:07 | answer | added | Gerald Edgar | timeline score: 2 | |
| Feb 4, 2015 at 14:50 | answer | added | user3429697 | timeline score: 0 | |
| Feb 4, 2015 at 14:23 | comment | added | Aliveli | Yes, I should have clearly stated that $g(\cdot) \ge 0$. Otherwise, based on the counterexamples you suggested the assertion is not necessarily true. Yemon, can you elaborate more about using the Cauchy-Schwarz inequality? My understanding is that one can show $E[X^2g(X)]\ge(E[Xg(X)^{1/2}])^2$. However, this is not exactly what I need. | |
| Feb 4, 2015 at 14:22 | history | edited | Aliveli | CC BY-SA 3.0 |
added 35 characters in body
|
| Feb 4, 2015 at 4:19 | comment | added | Nate Eldredge | And if you don't require $g \ge 0$ then this is false, even in the discrete case. Let $X = \pm 1$ with probability 1/2 each (the Rademacher distribution) and let $g(x) = x$ on $[-1,1]$ (you can extend it to a smooth bounded function on $\mathbb{R}$ if you like). Then your equation reads $E[X] E[X^3] \ge E[X^2] E[X^2]$ which is false; the left side is 0 and the right side is 1. For an example with a continuous random variable, let $X$ be uniformly distributed on $[-1,1]$. | |
| Feb 4, 2015 at 3:27 | comment | added | Yemon Choi | If you assume that $g\geq 0$ in the general problem then the result follows immediately from the Cauchy-Schwarz inequality: $Xg(X)= Xg(X)^{1/2}\cdot g(X)^{1/2}$ | |
| Feb 4, 2015 at 3:26 | comment | added | Yemon Choi | In your "simple case" you seem to be assuming that $g$ takes positive values (because you want $g_1g_2\geq 0$ | |
| Feb 4, 2015 at 3:12 | review | First posts | |||
| Feb 4, 2015 at 3:22 | |||||
| Feb 4, 2015 at 3:10 | history | asked | Aliveli | CC BY-SA 3.0 |