Timeline for A normal distribution inequality
Current License: CC BY-SA 3.0
21 events
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| Jul 7, 2013 at 20:19 | comment | added | Hans | @juan: OK. Just emailed you. | |
| Jul 7, 2013 at 20:14 | comment | added | Hans | @juan: I don't know why you repeat your discriminant inequality in your last comment. I know perfectly that from this inequality you can get the range for $N$. However, what I am saying is that you need to convert the value from $N(x)$ to $x$. That needs some work. You can not just quote whatever comes out of the numerical quadrature result. You need to give it tight error bounds. Do you see what I am getting at? | |
| Jul 7, 2013 at 20:07 | comment | added | juan | @Hans can you send me a mail. To send you this privately. They want to pass the discussion to chat. I prefer mail. My address is easy to get. | |
| Jul 7, 2013 at 19:57 | comment | added | Hans | @juan: Could you show your work for getting $|x|<0.597$ from the bound on $N$ in detail, or at least with some sketch? It is not enough coming from numerical computation. The proof is not completely obvious. | |
| Jul 7, 2013 at 19:54 | comment | added | Hans | @juan: Also, numerical computation shows the discriminant $\Delta<0$ for $0<x<1.3$. If we can show this, together with my result for $x>1.16$, the proof will be complete. | |
| Jul 7, 2013 at 19:37 | comment | added | juan | For $x>0$ in the case of discriminant positive, the inequality will be true if $x>$ the greater of the two roots. I think this is also promising. | |
| Jul 7, 2013 at 18:16 | comment | added | Hans | @juan: I like your discriminant trick. | |
| Jul 7, 2013 at 16:39 | comment | added | Hans | @juan: I have noticed your error in reversing the inequality some time ago, but I was busy with other things and I knew you'd notice it yourself anyway.:-) It seems there is still some work to actually prove the desired inequality holds in $|x|<0.597$. One way to do it is through bounding the integral $N$. It seems your previously suggested method of maximal slope gives a valid region much smaller than $|x|<0.597$. Maybe my derivative bound is not tight enough. Could you show your work in detail? | |
| Jul 7, 2013 at 16:01 | history | edited | Did | CC BY-SA 3.0 |
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| Jul 7, 2013 at 15:51 | comment | added | juan | @Did yes I was wrong. This proves the inequality for $|x|<0.597. And the rest is not so easy, because now the intervals are infinite. I changed my solution correspondingly. | |
| Jul 7, 2013 at 15:49 | history | edited | juan | CC BY-SA 3.0 |
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| Jul 7, 2013 at 11:52 | comment | added | Did | I am not sure to follow: the good regime $4N^2(1-N)^2\gt1/2\pi$ is when $|x|\lt x_*$ for some $x_*$, not the other way round (as an aside, note that the inequality one is interested in holds at $x=0$ hence also in a neighborhood of $x=0$). | |
| Jul 7, 2013 at 11:20 | comment | added | juan | @Hans I had edited my answer because at the start I said $|x|<0.597$ instead of the intended $|x|>0.597$. | |
| Jul 7, 2013 at 11:19 | history | edited | juan | CC BY-SA 3.0 |
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| Jul 7, 2013 at 9:24 | comment | added | juan | @Hans It is realy easy. But now I am a little confused about who is interested. | |
| Jul 7, 2013 at 9:10 | comment | added | user64494 | @ juan: If this is very easy to get, please give it. | |
| Jul 7, 2013 at 8:18 | comment | added | juan | @user64494 My answer proof completely the inequality for $x<-0.597$ or $x>0.597$. For this you have no need of a bound of the derivative. Now to show $f(x)>0$ (my $f$ is different from yours) on the interval $|x|<0.597$ you may apply the maximal slope principle. This need a bound of the derivative on $|x|<0.597$ (a rough bound suffice). This is very easy to get. And you finish without difficulty the proof with a little computation (see the paper cited in my answer). | |
| Jul 7, 2013 at 8:17 | comment | added | user64494 | @ juan: You don't answer my request concerning the estimate of the derivative. So called slope principle is the next step in your answer. | |
| Jul 7, 2013 at 8:11 | comment | added | juan | @user64494 To apply the maximal slope principle you only need a rough bound of the derivative. For example substitute all exp(-x^2/2) by 1 and all N(x) by 1, all x by 0.597. All in absolute value and this bound will suffice. | |
| Jul 7, 2013 at 5:11 | comment | added | user64494 | @ juan : You wrote "Since the derivative of your function is easily bounded". Could you explain that place in detail? The expression for $f'(x)$ (which can be downloaded from rapidshare.com/files/3032281090/derivative.pdf ) is not so simple. | |
| Jul 6, 2013 at 21:21 | history | answered | juan | CC BY-SA 3.0 |