Timeline for On the sum of uniform independent random variables
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| Jul 4, 2016 at 5:47 | vote | accept | CommunityBot | ||
| Jul 1, 2016 at 20:51 | answer | added | Iosif Pinelis | timeline score: 5 | |
| Jun 21, 2016 at 16:31 | comment | added | Christian Remling | @CarloBeenakker: Yes, the density is $P'_n$ in your notation and $p'_n$ in mine, and $P_n\to 1$ as $n\to\infty$ for fixed $c$, but that's clear and doesn't address the original question. | |
| Jun 21, 2016 at 6:37 | comment | added | Carlo Beenakker | @ChristianRemling --- correct me if I have misunderstood you, but isn't this "peaked density" just the derivative with respect to $\delta$ of the sinc integral in my posting below, so this distribution tends for $n\rightarrow\infty$ to $P_n(\delta)\rightarrow\sqrt{6n/\pi}\exp(-6n\delta^2)$ (with $\delta=c-1/2$) | |
| Jun 20, 2016 at 18:49 | comment | added | Christian Remling | @wolfies: The Bates and IH distributions are related by $B_n(x)=IH_n(nx)$, so either one is fine. In any event, I'd love to see a proof of the claims you're making (why does the density "become more peaked"), so if you have one, please post. | |
| Jun 20, 2016 at 17:28 | comment | added | wolfies | [ and mode at 1/2 ] | |
| Jun 20, 2016 at 16:52 | comment | added | wolfies | And the result should follow conceptually ... the $\text{Bates}(n)$ distribution has a constant mean at $\frac12$, and as $n$ increases, the pdf becomes more peaked ... it has variance of $\frac{1}{12n}$, so as $n$ increases, the distribution narrows and converges upon the mean , i.e. if $Z \sim \text{Bates}(n)$, then $P(Z < c)$, for $c > 1/2$ (the mean) must be increasing in $n$. | |
| Jun 20, 2016 at 16:25 | comment | added | wolfies | You want Bates distribution ... not Irwin-Hall. | |
| Jun 20, 2016 at 9:44 | comment | added | Carlo Beenakker | @BenCrowell --- the large-$n$ limit is actually reached very early, I didn't do a formal error bound analysis, but the numerics indicates a rapid convergence, see the plot at the end of my posting. | |
| Jun 20, 2016 at 1:01 | comment | added | Christian Remling | @BenCrowell: We're not in the regime of the CLT though, our probabilities are close to $1$ (we're many standard deviations away from the mean). | |
| Jun 19, 2016 at 22:30 | history | edited | user31317 | CC BY-SA 3.0 |
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| Jun 19, 2016 at 21:26 | comment | added | user21349 | Maybe it's possible to prove this by putting error bounds on the approximation of the distribution as a normal distribution. Related: math.stackexchange.com/questions/31365/… stats.stackexchange.com/questions/30468/… | |
| Jun 18, 2016 at 20:24 | answer | added | Carlo Beenakker | timeline score: 3 | |
| Jun 18, 2016 at 20:20 | comment | added | Christian Remling | The fact that $p_n''<0$ actually works against me, and I need the asymmetry of the interval I'm averaging over to at least compensate for this. | |
| Jun 18, 2016 at 20:18 | comment | added | user31317 | I saw the post, where was the mistake? | |
| Jun 18, 2016 at 20:16 | comment | added | Christian Remling | I have a feeling this might be rather subtle, at least for $c$ close to $1/2$. I just posted and deleted an incorrect answer, and the summary of that is that there are competing effects. | |
| Jun 18, 2016 at 20:02 | comment | added | user31317 | I did simulations too. That is why I asked the question. The proof is all that matters... | |
| Jun 18, 2016 at 20:01 | answer | added | Christian Remling | timeline score: 3 | |
| Jun 18, 2016 at 19:16 | comment | added | Mark L. Stone | Based om simulation, I think the result is true, presuming "increasing" is interpreted as meaning "non-decreasing" (in order to cover $c \ge 1$). I leave the proof to you. | |
| Jun 18, 2016 at 18:55 | history | edited | user31317 | CC BY-SA 3.0 |
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| Jun 18, 2016 at 18:39 | history | asked | user31317 | CC BY-SA 3.0 |