Timeline for How to integrate the multinomial over a ball in $\mathbb{R}^{n}$?
Current License: CC BY-SA 4.0
10 events
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| Oct 21 at 14:13 | comment | added | xiangsha | @CarloBeenakker Could you explain a little more about the concentration of the measure in your answer? or give some reference? Thanks! | |
| Jan 30, 2022 at 1:54 | comment | added | Iosif Pinelis | @CarloBeenakker : Even when $m$ is fixed (and $\ne1$), the asymptotics depends on how much the $a_j$'s differ from one another -- see Addendum 3 in my answer. | |
| Jan 28, 2022 at 2:18 | comment | added | xiangsha | Thanks a lot! The question is for both m→∞ and n →∞, however, I found your estimate held for m that not so large, for instance m=O(lnn) and n →∞. Thank you for the details which help me a lot! | |
| Jan 23, 2022 at 18:20 | comment | added | Carlo Beenakker | my estimate is for $n\rightarrow\infty$ at fixed $m$, the double limit $m,n\rightarrow\infty$ is different. | |
| Jan 23, 2022 at 18:18 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 23, 2022 at 18:18 | comment | added | Iosif Pinelis | But here $m\to\infty$. | |
| Jan 23, 2022 at 18:16 | comment | added | Carlo Beenakker | @IosifPinelis --- I tried this simple case for $m=1$ and actually find that the exact result equals the large-$n$ estimate... | |
| Jan 23, 2022 at 18:16 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 23, 2022 at 17:23 | comment | added | Iosif Pinelis | I think the asymptotics will very much depend on how much the $a_j$'s differ from one another. E.g., if $a_1=1$ and $a_2=\cdots=a_n=0$, then it seems clear that the main contribution to the integral will be from $\mathbf x$ with $x_1^2\approx1$ and $x_j^2\approx0$ for $j\ge2$. So, then $\sum_{j=1}^{n}a_{j}x_{j}^2$ will be close to $1$, rather than to $\sum_{j=1}^{n}a_{j}r^2/n\approx1/n$. | |
| Jan 23, 2022 at 16:42 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |