Timeline for Bochner integral of stochastic process = path by path Lebesgue integral?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 30, 2012 at 11:49 | vote | accept | Hauke L. | ||
| Oct 29, 2012 at 22:40 | answer | added | George Lowther | timeline score: 17 | |
| Oct 29, 2012 at 19:45 | comment | added | Hauke L. | To apply dominated convergence, we need that a $\xi_n(\omega,t) \rightarrow X(\omega,t)$ for almost all $(\omega,t)$. I don't see how we can get this. | |
| Oct 29, 2012 at 14:26 | comment | added | George Lowther | You can show that the answer is yes by calculating along quickly converging sequence of partitions and using dominated convergence for the Lebesgue integral. | |
| Oct 29, 2012 at 14:07 | comment | added | Hauke L. | I do not understand. There is no stochastic integral here. For every measurable process, the integral against a finite variation process can be computed pathwise as a Lebesgue integral, and this is equivalent to computing it as a stochastic integral against a semimartingale. | |
| Oct 29, 2012 at 13:26 | comment | added | Gerald Edgar | Perhaps take one of the texts on stochastic integrals, look at their example of a stochastic integral that cannot be computed pathwise, and compare it to what you want. | |
| Oct 29, 2012 at 12:26 | history | edited | Hauke L. | CC BY-SA 3.0 |
added 403 characters in body; edited title; added 15 characters in body; added 8 characters in body
|
| Oct 29, 2012 at 6:31 | history | edited | Hauke L. | CC BY-SA 3.0 |
deleted 5 characters in body
|
| Oct 28, 2012 at 18:09 | history | edited | Hauke L. | CC BY-SA 3.0 |
added 141 characters in body
|
| Oct 28, 2012 at 18:05 | comment | added | Hauke L. | I did not know the Bochner integral before. And yes, the point is whether this integral is a.s. equal to the Lebesgue integral defined path by path. This does not seem obvious to me. | |
| Oct 28, 2012 at 17:22 | comment | added | Dan | Of course, of course, hence the "linear". $X :[0,T] \rightarrow \mathcal{H}$ is certainly Bochner integrable, so if by $\int_0^tX(s)ds$ you mean the Bochner integral, then the answer is yes. But I suppose the point is that the Bochner integral may disagree with the pathwise Lebesgue or Riemann integral. | |
| Oct 28, 2012 at 15:30 | comment | added | Hauke L. | *orthogonal projection onto $\mathcal{L}(X,t)$ i should say (1st sentence) | |
| Oct 28, 2012 at 15:15 | comment | added | Hauke L. | The assertion that the orthogonal projection equals the conditional expectation is only true for Gaussian processes. In general, the conditional expectation is the orthogonal projection onto $L_2(\Omega,\mathcal{X}(t),\mathbf{P})$, $\mathcal{X}$ the natrual filtration of $X$, which is a larger subspace than $\mathcal{L}(X,t)$. So in general, these two are not equal. | |
| Oct 28, 2012 at 14:45 | comment | added | Dan | Perhaps I'm missing something, but it seems that the answer to your question is "yes" simply because the projection of any $Y \in \mathcal{H}_0$ onto $\mathcal{L}(X,t)$ is exactly $\mathbb{E}[Y|\mathcal{F}_t]$, where $\mathcal{F}_t = \sigma(X_s : s \le t)$. And of course $\int_0^tX_sds$ is $\mathcal{F}_t$-measurable. | |
| Oct 28, 2012 at 13:26 | history | edited | Hauke L. | CC BY-SA 3.0 |
deleted 50 characters in body
|
| Oct 28, 2012 at 12:18 | history | edited | Hauke L. | CC BY-SA 3.0 |
edited body
|
| Oct 28, 2012 at 11:13 | history | edited | Hauke L. | CC BY-SA 3.0 |
added 54 characters in body; deleted 32 characters in body
|
| Oct 28, 2012 at 11:08 | history | asked | Hauke L. | CC BY-SA 3.0 |