Timeline for Is finding the CDF from the Laplace transform well-posed?
Current License: CC BY-SA 4.0
19 events
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| Oct 27 at 12:43 | comment | added | Ziv | Your example is essentially an instance of the following phenomenon: convergence in distribution can occur without convergence of density functions | |
| S Oct 27 at 9:53 | history | bounty ended | Riemann | ||
| S Oct 27 at 9:53 | history | notice removed | Riemann | ||
| Oct 27 at 9:53 | vote | accept | Riemann | ||
| Oct 24 at 22:02 | answer | added | Iosif Pinelis | timeline score: 1 | |
| Oct 24 at 11:12 | history | edited | Riemann | CC BY-SA 4.0 |
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| S Oct 24 at 6:08 | history | bounty started | Riemann | ||
| S Oct 24 at 6:08 | history | notice added | Riemann | Draw attention | |
| Oct 19 at 15:00 | comment | added | Riemann | Thanks for your comment@MichaelRenardy! I thought about this, and the situation for $C$ seems to be much nicer than $f$. The reason is that $C$ is a monotone function (being a CDF). Knowing that the output of the Laplace Inversion is monotone, should make it much more well-behaved. | |
| Oct 18 at 17:52 | comment | added | Michael Renardy | No chance. A simple integration by parts yields a simple relation between the Laplace transforms of C and f, so inverting one is basically as bad as inverting the other. | |
| Oct 18 at 11:52 | history | edited | Riemann | CC BY-SA 4.0 |
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| Oct 18 at 11:40 | history | edited | Riemann | CC BY-SA 4.0 |
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| Oct 18 at 11:40 | comment | added | Riemann | Hi David! So in my counterexample, I intended $f_n(t)$ to have finite support. When $\sigma$ is small, more than $99.99\%$ of the area under the lognormal distribution is inside of $[0,L]$. So to convert it to an actual probability distribution with finite support, we just need to multiply by a normalizing factor (like 1.0001). | |
| Oct 18 at 11:37 | comment | added | David | This depends on the assumptions you make on $f_n(t)$: in the beginning you speak about $f(t)$ having finite support -> this is not true for your counterexample with log normal distributions. In general, you need a criterion on the tails of your random variables. Note that that the tail of the log normal distribution decays quite slowly as $t\to \infty$. | |
| Oct 18 at 11:33 | history | edited | Riemann | CC BY-SA 4.0 |
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| Oct 18 at 11:26 | history | edited | Riemann | CC BY-SA 4.0 |
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| Oct 18 at 11:21 | history | edited | Riemann | CC BY-SA 4.0 |
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| Oct 18 at 11:14 | history | edited | Riemann | CC BY-SA 4.0 |
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| Oct 18 at 11:05 | history | asked | Riemann | CC BY-SA 4.0 |