Timeline for What is meant by a continuous-time white noise process?
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12 events
| when toggle format | what | by | license | comment | |
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| Dec 1, 2019 at 15:22 | comment | added | Ben | @the_src_dude You can't really choose $f(t)$ to get $X$ out of the integral. To get a physical intuition, you might consider using a sequence $f_n(t)$ of approximations of the delta function: in the limit the pairing $(X, f_n)$ will diverge, which reflects the fact that Brownian motion is not differentiable and $X$ is a very "rough" process. | |
| Dec 1, 2019 at 15:18 | comment | added | Ben | @the_src_dude I'm not sure I understand what you're asking about the pairing. When $X$ is fixed (as in this case, where $X$ is a white noise process), you can think of $(X, f)$ as defining a mapping on possible test functions $f$. So $f$ here is just a dummy variable and you don't have to choose it to be anything. In other words, $(X, f)$ should depend on $f$ and typically would have different values for different choices of $f$. The thing that stays constant is $X$. | |
| Nov 27, 2019 at 12:49 | comment | added | the_src_dude | @BenCW So for example if I want to define a white noise process with variance $\sigma^{2}$, how would I do that using this distributional definition? Would I define $f(t) = \sigma^{2}$? | |
| Nov 27, 2019 at 12:27 | comment | added | the_src_dude | @BenCW So, in your def $(X,f) = \int_{0}^{T} f(t) dB_{t}$, how should I choose the function $f(t)$? What if I choose it as $f(t) = t$ or $f(t) = 1$ or $f(t) = \delta(t)$? These give all different answers, but honestly, I can't see the physical intuition. I want to choose $f(t)$ such that I get the derivative of Brownian motion back out of integral -- ie I want to get white noise out. So what should I do? | |
| Oct 14, 2019 at 9:13 | history | edited | Ben | CC BY-SA 4.0 | fixed typo |
| Sep 5, 2019 at 13:02 | comment | added | Ben | Yes, because any cross-variation involving a process that is continuous and of finite total variation is 0. Typically, a test function $f$ would be smooth with bounded support, so would satisfy this requirement. | |
| Sep 4, 2019 at 23:43 | comment | added | alpastor | Is the cross variation term that pops up in the stochastic integration by parts equal $0$? And if so why? | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot | replaced http://math.stackexchange.com/ with https://math.stackexchange.com/ | |
| Dec 15, 2015 at 22:56 | history | edited | Ben | CC BY-SA 3.0 | added 215 characters in body |
| Dec 13, 2015 at 2:26 | history | edited | Ben | CC BY-SA 3.0 | added 24 characters in body |
| S Dec 13, 2015 at 0:00 | history | answered | Ben | CC BY-SA 3.0 | |
| S Dec 13, 2015 at 0:00 | history | made wiki | Post Made Community Wiki by Ben |