Timeline for Expected value of the quotient times quotient of the expected values
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 21, 2021 at 19:22 | comment | added | Floromidante | I see it now. Thank you very much! | |
| Dec 21, 2021 at 19:10 | comment | added | Martin Hairer | Sorry, the minus sign in $B$ was a typo. The expectations of $f$ and $g$ are then obviously equal but the expectation of the ratio is greater than $e^{\lambda/2}$ say for $\lambda$ large enough. (You get a lower bound by restricting the integral to a small cap around $(0,1)$.) | |
| Dec 21, 2021 at 18:08 | comment | added | Floromidante | @MartinHairer But in that case $$\int_{\mathbb{S}^1}\exp(x^\top[B-A]x)dx=\int_{\mathbb{S}^1}e^{-\lambda}dx=e^{-\lambda}$$ And \begin{align*} \frac{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Ax\right)dx}{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Bx\right)dx} &=\frac{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}(A-B)x\right)\exp\left(x^{\top}Bx\right)dx}{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Bx\right)dx}\\ &=\int_{\mathbb{S}^1}e^{\lambda}\frac{\exp(x^\top B x)}{\int_{\mathbb{S}^1}\exp(x^\top Bx)dx}dx\\ &=e^{\lambda} \end{align*} So the product is $1$. | |
| Dec 20, 2021 at 17:28 | comment | added | Martin Hairer | You probably can't do much better, consider $n=2$, $A = \mathrm{diag}(\lambda,0)$, and $B = -\mathrm{diag}(0,\lambda)$ for some very large $\lambda$. | |
| Dec 20, 2021 at 4:36 | history | edited | Floromidante | CC BY-SA 4.0 |
added 65 characters in body
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| Dec 20, 2021 at 4:23 | comment | added | Floromidante | @MattF. I am interested in "quotient times quotient". Sorry for the misleading title | |
| Dec 20, 2021 at 4:20 | history | edited | Floromidante | CC BY-SA 4.0 |
edited title
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| S Dec 20, 2021 at 1:26 | review | First questions | |||
| Dec 20, 2021 at 4:39 | |||||
| S Dec 20, 2021 at 1:26 | history | asked | Floromidante | CC BY-SA 4.0 |