Timeline for Constructing the 'idealized white noise' stochastic process
Current License: CC BY-SA 4.0
24 events
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| Apr 5, 2021 at 12:49 | comment | added | Iosif Pinelis | @MateuszKwaśnicki : Thank you for the references to those two questions. | |
| Apr 5, 2021 at 6:57 | comment | added | Mateusz Kwaśnicki | @IosifPinelis: It is not very difficult to see that a jointly measurable resalisation of such a process $(X_t)$ is not possible. With your statement (that $t \mapsto X_t$ is measurable with probability one), however, this may get difficult. See, for example, these two questions: one and two (especially the latter one, I think). | |
| Apr 5, 2021 at 6:52 | comment | added | Mateusz Kwaśnicki | @UserA: The usual time-derivative of the Brownian motion would have infinite variance, I think this is what Ofer Zeitouni meant. | |
| Apr 3, 2021 at 7:01 | comment | added | UserA | @MartinHairer As you said in your comments and as mentioned in the answer, we can "use Kolmogorov's extension theorem to construct an $\mathbb{R}$-indexed i.i.d. collection of Gaussian random variables with some fixed finite variance". But I still don't understand, why did some of the commenters say that $\{\dot W_t\}$ will have infinite variance? (I'm referring to the comments by @oferzeitouni) | |
| Apr 2, 2021 at 20:43 | comment | added | Martin Hairer | @UserA If by random field you mean any $\mathbb{R}$-indexed family of random variables then yes. Most authors would however implicitly assume some measurability with respect to the underlying space when talking about a 'random field', so I would think that in most cases it wouldn't qualify... | |
| Apr 2, 2021 at 20:04 | comment | added | UserA | @MartinHairer Indeed, using the Kolmogorov extension theorem to come up with a family of iid Gaussian random variables with fixed finite variance is what I seek. However, the comment (under the question) by ofer zeitouni suggested that even with this construction, the variance is infinite, which was confusing for me. So to seal the deal: $\{\dot W_t\}$ is constructed using the Kolmogorov theorem (as $\otimes_t N(0,1)$), but is not a stochastic process but rather a random field. Is that correct? | |
| Apr 2, 2021 at 16:16 | comment | added | Iosif Pinelis | @MartinHairer : I guess I meant this:Do there exist a probability space and a family $X=(X_t)_{t\ge0}$ of independent standard normal variables defined on this probability space such that the inner probability of the set of all measurable realizations of $X$ is $1$? | |
| Apr 2, 2021 at 15:48 | comment | added | Martin Hairer | @IosifPinelis What exactly would you mean by such a statement given that the set of measurable functions isn’t itself measurable with respect to the product sigma-algebra? | |
| Apr 2, 2021 at 14:54 | comment | added | Iosif Pinelis | @MartinHairer : I think I have an idea how to prove the impossibility of the joint measurability. But do you know what can be said about the (lack of ?) measurability of almost all realizations of the "idealized white noise"? | |
| Apr 2, 2021 at 10:46 | comment | added | Martin Hairer | ... On the other hand, you can use Kolmogorov's extension theorem to construct an $\mathbb{R}$-indexed i.i.d. collection of Gaussian random variables with some fixed finite variance (say $1$). This is what most commenters here believe you mean when you talk about 'idealised white noise' and what I mentioned cannot be realised as a jointly measurable stochastic process. The relation between the two is like the relation between the 'delta function' and the indicator function of the set $\{0\}$, i.e. intuitively appealing but actually pretty much nonexistent. | |
| Apr 2, 2021 at 10:42 | comment | added | Martin Hairer | @UserA If you want your $\dot W$ to be actual white noise (which is rigorously defined in one of the two ways you mention in your question), then this can heuristically be thought of as i.i.d. Gaussians with infinite variance, just like the 'delta function' can be heuristically thought of as a function that vanishes everywhere except at the origin where it is infinite. Real white noise is of course not built by Kolmogorov's extension theorem in the way suggested. (It can as a test function-indexed process though.) ... | |
| Apr 2, 2021 at 6:35 | comment | added | UserA | How about the infinite variance? What I understand so far is that $\dot W$ has infinite variance, yet we can construct it using the Kolmogorov extension theorem. Can you clarify this part? Sorry for asking too much but this is a crucial point for me. | |
| Apr 2, 2021 at 3:49 | comment | added | Iosif Pinelis | @UserA : The comment about the non-measurability was made by Martin Hairer, not by me. I have not seen this fact, and I have not even seen this "idealized white noise" in publications. If I had to prove the non-measurability, I guess I'd try using, first, Lusin's theorem (en.wikipedia.org/wiki/Lusin%27s_theorem) and then maybe something like the Lebesgue's density theorem (en.wikipedia.org/wiki/Lebesgue%27s_density_theorem). | |
| Apr 1, 2021 at 11:22 | comment | added | UserA | Thank you for the answer. To answer comment 2, I do not need it for anything specific. First of all, it seemed a priori natural to have the 'continuous equivalent 'of a coin tossing process, which is how I heuristically think of white noise. Second, to clarify: I did not know that such a process (what I call idealized white noise) is not measurable w.r.t $t$ and has infinite variance (would be great if you can clarify why). So I looked at at how authors define white noise and said: "maybe I can use their definition of noise to define idealized white noise". | |
| Mar 31, 2021 at 23:26 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 31, 2021 at 20:27 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 31, 2021 at 20:22 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 31, 2021 at 20:20 | comment | added | Iosif Pinelis | @oferzeitouni : Yes, of course, because the "idealized white noise" is not an actual white noise. | |
| Mar 31, 2021 at 20:09 | comment | added | ofer zeitouni | Sure, I did not notice you divided by $\sqrt{\epsilon}$... Then, this object does not converge to $\dot{W}$.... what he wanted is white noise, and your scaling is inappropriate for that - if you integrate your $\dot{W}_t^{(\epsilon)}$ against a smooth test function, you will get $0$ in the limit. | |
| Mar 31, 2021 at 15:52 | comment | added | Iosif Pinelis | @oferzeitouni : The variance of $\dot W^{(\varepsilon)}_t$ is $1$. | |
| Mar 31, 2021 at 15:48 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 31, 2021 at 15:45 | comment | added | ofer zeitouni | The variables $\dot{W}^{(\epsilon)}_t$, for a fixed $t$, do not converge anywhere in distribution: it is a sequence of Gaussian variables whose variances blows up... | |
| Mar 31, 2021 at 15:41 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 31, 2021 at 15:35 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |