Timeline for Let $a\in S^d$, $b\in S^{d-1}$ be uniform on the spheres. How to show $\mathbb E[\frac{||a||_1}{\sqrt {d+1}}] \le\mathbb E[\frac{||b||_1}{\sqrt d}]$?
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| Apr 29, 2021 at 10:30 | comment | added | A B | Thank you both for the answer! I followed your suggestion and separated the last question, which is now here: mathoverflow.net/questions/391461/…. | |
| Apr 29, 2021 at 9:47 | vote | accept | A B | ||
| Apr 29, 2021 at 3:22 | comment | added | Iosif Pinelis | @PierrePC : Thank you for your further comments. Hopefully, now my mistakes are corrected. :-) | |
| Apr 29, 2021 at 3:20 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Apr 29, 2021 at 1:56 | comment | added | Pierre PC | Using the same method, I have a slightly different result for $\mathbb E[Y]$, though. I get $\Gamma(d/2)/(\sqrt\pi\times\Gamma((d+1)/2))$. This differs from your result by a factor $\sqrt{d/(d-1)}$, or something similar. This gives $\rho_n^2 = (n(n+2)^3)/((n+3)(n+1)^3)$, and since $n(n+2)^3 - (n+3)(n+1)^3 = -2n-3$, it actually shows that $\rho_n<1$. This is coherent with the computations of the OP since, as you mentioned, $f_n$ converges to a constant (I think $\sqrt{2/\pi}$). | |
| Apr 29, 2021 at 1:55 | comment | added | Pierre PC | Thank you! :) I see you also corrected a mistake of mine, I agree with your formula although I think the first expression you give should be divided by $n$, which yields the result you give for $\mathbb E[\|X_n\|_1^2]/n$. | |
| Apr 28, 2021 at 22:18 | comment | added | Iosif Pinelis | @PierrePC : I see now -- a clever use of independence. | |
| Apr 28, 2021 at 22:18 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Apr 28, 2021 at 20:10 | comment | added | Pierre PC | I may be wrong, but this is the way I was thinking about it. Since $\|G\|_1^2=(|G_1|+\cdots+|G_n|)^2$, we can expand the square on the right, and we find $(*)=d\mathbb E[G_1^2]+d(d-1)\mathbb E[|G_1|]^2$ for its expectation. Now $\|G\|_1^2 = |G|^2\cdot\|X_n\|_1^2$, and since $|G|$ and $X_n$ are independent, $\mathbb E[\|X_n\|_1^2]$ is given by the quotient of $(*)$ by $\mathbb E[|G|^2]=d\mathbb E[G_1^2]$. Since $\mathbb E[G_1^2]=1$ (expectation of a $\chi^2(1)$) and $\mathbb E[|G_1|]=\sqrt{2/\pi}$ (expectation of a $\chi(1)$), this should the expression above. | |
| Apr 28, 2021 at 19:56 | comment | added | Iosif Pinelis | @PierrePC : To find $E\|X_n\|_1^2$, you will need to find the expectation of the product of the absolute values of two coordinates of $X_n$, and the joint distribution of the squares of such two coordinates is the Dirichlet distribution with parameters $1/2,1/2,(n-2)/2$. So, how did you do it without using the Dirichlet distribution? | |
| Apr 28, 2021 at 19:51 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Apr 28, 2021 at 19:42 | comment | added | Pierre PC | I think this approach shows that $\mathbb E[\|b\|_1^2]=\frac2\pi(d-1) + \sqrt{\frac2\pi}$. I am not familiar with Dirichlet distributions, maybe it is just another way of saying the same thing. | |
| Apr 28, 2021 at 19:37 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Apr 28, 2021 at 19:26 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Apr 28, 2021 at 18:57 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Apr 28, 2021 at 18:50 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |