Timeline for Time-derivative of integral over sub-level set $s(t) := \int_{f^{-1}((-\infty,t])}p(x)dx$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 22, 2022 at 22:37 | vote | accept | dohmatob | ||
| Mar 5, 2022 at 12:40 | answer | added | memorial | timeline score: 1 | |
| Mar 2, 2022 at 15:38 | comment | added | dohmatob | @JochenWengenroth Thanks again. It seems a general way to compute the density of that push-forward is via "Mallavian calculus". In the case of Example 1, I can confirm (see below) that this kind of calculus gives the correct answer as computed via "brute-force" in my comment above. Any further insights are welcome. | |
| Mar 2, 2022 at 15:33 | answer | added | dohmatob | timeline score: 1 | |
| Mar 2, 2022 at 10:06 | comment | added | Jochen Wengenroth | Yes, $f_\sharp\mu(A)=\mu(f^{-1}(A))$. Your calculation seems correct. | |
| Mar 2, 2022 at 9:07 | comment | added | dohmatob | @JochenWengenroth Thanks for the hint. Do you mean pushforward of $\mu$ under $f$, i.e $f_\# \mu$ as usually written in probability literature ? If so, then in the case of $f(x):=w^\top x-c$ and multi-variate Gaussian distribution $\mu = N(m,\sigma^2 I_d)$, your hint leads us to: $s_f'$ is the density of $N(w^\top m-c,\sigma^2\|w\|^2) = N(f(m),\sigma^2\|w\|^2)$, that is $s_f'(t) = \varphi((t-f(m))/(\sigma\|w\|))$, where $\varphi$ is standard Gaussian pdf. Does this computation look correct to you ? Thanks in advance. | |
| Mar 2, 2022 at 8:49 | comment | added | Jochen Wengenroth | If $s_f$ is, e.g., continuously differentiable its derivative is a Lebesque density of the push forward measure $\mu\circ f^{-1}$. A trivial situation is a constant function $f$ for which $\mu\circ f^{-1}$ does not have a density. | |
| Mar 2, 2022 at 8:40 | history | edited | dohmatob | CC BY-SA 4.0 |
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| Mar 2, 2022 at 8:26 | history | asked | dohmatob | CC BY-SA 4.0 |