Timeline for Representing uncertainty as a noise process in a numerical simulation
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| Jan 12, 2017 at 20:41 | answer | added | Strum | timeline score: 1 | |
| Jan 9, 2017 at 15:54 | comment | added | user130529 | BTW, why don't you compute your numerical integration without noise and add the noise after? You can add it in such a way to have any desired velocity deviation, and you avoid the accumulation in the scheme. | |
| Jan 9, 2017 at 15:24 | comment | added | user130529 | And the mean square displacement doesn't change when you divide by $\Delta t$? This is strange, I would expect the contrary (imagine what happens to your velocity when $\Delta t=10^{-20}$), and also that it would rather not change if you multiply by $\sqrt{\Delta t}$ (knowing that the deviation for $n$ steps behaves like $\sqrt{n}$). | |
| Jan 9, 2017 at 14:55 | comment | added | user12800 | I have an easier time following your way of thought here. However running the simulation without the scaling gives different behavior depending on the time step (eg. the mean square displacement changes depending on the value of dt), which shouldn't be the case. | |
| Jan 9, 2017 at 13:56 | comment | added | user130529 | If you are using standard uncertainty (probability of the true value falling within the uncertainty range being roughly 68.3%) and want it to be equal to 2, then your standard deviation should be equal to your wanted uncertainty (for a normal distribution), that is, simply use $\alpha_s=A{\xi}$ with $A = 2$ and no time step scaling. You scaling looks BTW questionable as the smaller the time step, the higher the deviation, which you surely don't want to have. | |
| Jan 9, 2017 at 11:47 | history | asked | user12800 | CC BY-SA 3.0 |