Timeline for Problem with making an estimate when values of many variables are unknown?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 2, 2012 at 16:31 | vote | accept | drewbarbs | ||
| Feb 1, 2012 at 18:54 | answer | added | magnetohydrodynamic | timeline score: 0 | |
| Jan 25, 2012 at 21:01 | comment | added | drewbarbs | Ok, I know that $p$ is greater than $\frac{1}{3}$, so I'm all set. Thanks for the help guys! | |
| Jan 25, 2012 at 20:20 | comment | added | Anthony Quas | $|\beta A-\beta B| \ge |\alpha A-\beta B| - |(\alpha-\beta)A|$ so you have to show $|(\alpha-\beta)A|<1/p$ and then you're done. Using your inequality for $|\alpha-\beta|$, it suffices to show that $A<3p$. Of course this is fine if $p>1/3$ since $A$ is a probability. | |
| Jan 25, 2012 at 20:15 | comment | added | Gerhard Paseman | In the above, I assume p < 3p^2, which would hold for all sensible values of p, namely positive integers. Gerhard "Ask Me About System Design" Paseman, 2012.01.25 | |
| Jan 25, 2012 at 20:01 | comment | added | Gerhard Paseman | If you call the first quantity C and note it is bounded away from 0 by 2/p, and call the second quantity D, how close are C and D? With A at most 1the and the step size less than 1/p, I can see how D is still far from 0. Gerhard "Ask Me About System Design" Paseman, 2012.01.25 | |
| Jan 25, 2012 at 19:57 | comment | added | drewbarbs | $\alpha$ and $\beta$ are also probabilities, but I'm not sure that is too important | |
| Jan 25, 2012 at 19:57 | comment | added | Gerhard Paseman | Actually, Charles's suggestion is better than mine. Having A integral would work for a different system of inequalities, similar but not identical to what is given. Gerhard "Ask Me About System Design" Paseman, 2012.01.25 | |
| Jan 25, 2012 at 19:53 | comment | added | drewbarbs | I'm sorry, probably an important detail is that A and B are probabilities, so they are bounded above by 1. I'm having trouble understanding how we could replace $\alpha$ in the 3rd to last inequality by $\beta$ (giving us the last inequality) | |
| Jan 25, 2012 at 19:53 | comment | added | Gerhard Paseman | You need some property of A. Knowing that A is an integer would help. Gerhard "Ask Me About System Design" Paseman, 2012.01.25 | |
| Jan 25, 2012 at 19:48 | comment | added | Charles Matthews | I think there must be an upper bound for A somewhere. The step troubling you should be the triangle inequality applied, but to conclude that the term you discard is at most 1/*p* you do need something. | |
| Jan 25, 2012 at 19:33 | history | asked | drewbarbs | CC BY-SA 3.0 |