Timeline for Kac-Rice formula and Borell-TIS inequalities for gradient-flow of centered gaussian random field
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 30, 2020 at 9:31 | comment | added | Liviu Nicolaescu | $U(t)$ is a trajectory of the random gradient flow $\dot{U}(t)=-\nabla g\big(U(t)\big)$ and $g\big(U(0)\big)$ is the value of $g$ at the initial point on the trajectory. | |
| Sep 30, 2020 at 9:29 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| Sep 30, 2020 at 8:18 | comment | added | dohmatob | In the posted answer, what do you mean by $g(U(0))$, etc. ? Did intend to say $U(0)$ ? Also, there is a stray dollar-dollar sign in your post. | |
| Sep 30, 2020 at 8:15 | comment | added | dohmatob | Thanks for the details. | |
| Sep 23, 2020 at 19:56 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| Sep 23, 2020 at 19:16 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| Sep 23, 2020 at 10:20 | comment | added | Liviu Nicolaescu | You need to solve for $t$ the equation $$\sum_{i=1}^n\lambda_i e^{-2\lambda_i} x_i^0=0,$$ and view it as a function of $x_i^0,\lambda_i$. Note that this equation has no solution if $\lambda_i >0$ for any $i$. For example if $n=2$, $a,b>0$ the equation $ae^{-at}x-be^{-2bt} y=0$ so $$1-\frac{b}{a}e^{2(a-b)t}\frac{y}{x}=0$$, $$t=\log\left(\frac{xa}{by}\right)/2(a-b).$$ For higher $n$ analyzing $t(\lambda_i,x_j^0)$ is more involved. | |
| Sep 23, 2020 at 8:29 | comment | added | dohmatob | Maybe one can conjecture $E[N_T(U) \mid g(x_0) > 0] \ge 1 - \exp(-\alpha(T))$ where $\alpha$ is a nonnegative increasing function of $T$ ? (or something like that) | |
| Sep 23, 2020 at 8:27 | comment | added | Liviu Nicolaescu | I don't know how to compute that conditional expectation. Note that $\mathbb{E}[N_T(U)|g(x_0)<0]=0$ and $\lim_{T\to\infty} \mathbb{E}[N_T(U)|g(x_0)>0]=1$ a.s.. | |
| Sep 23, 2020 at 8:26 | comment | added | dohmatob | Concerning last but one comment, if we consider the even simpler example where $g(x):= \Lambda^Tx$, then $U(t) = g(x(t)) = g(x(0) + t\Lambda) = \Lambda^Tx(0) + t\|\Lambda\|^2$, so that $U'(t) = \|\Lambda\|^2$. Thus the formula would give $E[N_T(U)] = \int_0^TE[\|\Lambda\|^2 \mid \Lambda^Tx(0) + t\|\Lambda\|^2] p_{U(t)}(0)dt$, but I don't quite see how to proceed to get a concrete number. | |
| Sep 23, 2020 at 8:18 | comment | added | dohmatob | What about the limiting case when $n \to \infty$, does anything simplify (e.g via some LLNs / CLT-type argment) ? | |
| Sep 23, 2020 at 8:12 | comment | added | dohmatob | Thanks for the worked example (upvoted). Can this integral be evaluated explicity to get $E[N_T(U)] =$ some explicit function of $T$ ? I've expected that such a "simple" model would produce an explicit answer down the line :) | |
| Sep 23, 2020 at 8:04 | history | answered | Liviu Nicolaescu | CC BY-SA 4.0 |