Timeline for Time integrals of diffusion processes
Current License: CC BY-SA 2.5
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| Jan 20, 2011 at 12:24 | comment | added | Did | @jzadeh: Did I? Where? As regards mathematical accuracy, you might wish to reread slowly what I (and others) wrote, and to revise your post accordingly. | |
| Jan 20, 2011 at 11:42 | comment | added | jzadeh | And here is my mistake. I was trying to figure out why $Y$ would be Gaussian in general but it is not. My argument breaks down I should have said that $tX_t - \int_{0}^{t}X_sds = \int_{0}^{t}sdX_s$ is a Gaussian process. | |
| Jan 20, 2011 at 11:38 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 20, 2011 at 11:23 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 20, 2011 at 11:19 | comment | added | Simon Lyons | "The (Riemann) integral of an ito process is equal to the difference of two Gaussian processes which should again be Gaussian." Have you assumed that all diffusion processes are Gaussian? This is certainly not the case when $X_t$ has nonlinear drift. | |
| Jan 20, 2011 at 11:12 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 20, 2011 at 11:02 | comment | added | jzadeh | Since the processes are Gaussian they will have equivalent laws (as opposed to equal) for all time if the covariance functions equal. That is if $g_1(t,u) = g_2(t,u)$ for all $t,u >0$ the laws of the processes will be equivalent (as opposed to equal). Do you disagree with this fact Didier? | |
| Jan 20, 2011 at 10:28 | comment | added | Did | +1 for The Bridge's question. Re jzadeh's second edit: indeed, $E(Y^{(1)}_tY^{(1)}_s)=E(Y^{(2)}_tY^{(2)}_s)$ for every $(t,s)$ iff $E(X^{(1)}_tX^{(1)}_s)=E(X^{(2)}_tX^{(2)}_s)$ for every $(t,s)$. This is obvious and general--but not the point. The OP asks for cases when the laws of $Y^{(1)}$ and $Y^{(2)}$ are equivalent (as opposed to equal). | |
| Jan 20, 2011 at 10:22 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 20, 2011 at 10:16 | comment | added | jzadeh | Thanks for the downvote The Bridge.... I refer you to my above passage "it should then suffice to characterize the processes covariance structure in order to have a complete understanding of the law of the processes". Since that is obviously to vague I have elaborated a little more in edit 2 and I refer you to one of the excellent texts by Robert Adler for the theorems I am citing on Gaussian processes. # R.J. Adler, (1990), , An Introduction to Continuity, Extrema, and Related Topics for General Gaussian Processes, IMS Lecture Notes-Monograph Series, Vol 12, vii + 160 – jzadeh 0 secs ago | |
| Jan 20, 2011 at 10:02 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 20, 2011 at 9:49 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 20, 2011 at 9:35 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 20, 2011 at 7:16 | comment | added | The Bridge | I don't see the connection with the initial question. What is your point exactly ? | |
| Jan 19, 2011 at 22:58 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 19, 2011 at 22:51 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 19, 2011 at 22:48 | comment | added | jzadeh | Thank you for the help I will make the appropriate edits. | |
| Jan 19, 2011 at 21:31 | comment | added | Shai Covo | I have partially went over your answer. It should be noted that ${\rm E}[X_t \int_0^t {s\,{\rm d}X_s } ] = t^2 /2$, and ${\rm E}[(\int_0^t {X_s \,{\rm d}s} )^2 ] = t^3 /3$ (as is well known, $\int_0^t {X_s \,{\rm d}s} \sim {\rm N}(0,t^3/3)$). | |
| Jan 19, 2011 at 21:20 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 19, 2011 at 19:48 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 19, 2011 at 19:28 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 19, 2011 at 19:10 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 19, 2011 at 19:01 | history | undeleted | jzadeh | ||
| Jan 19, 2011 at 19:00 | history | edited | jzadeh | CC BY-SA 2.5 |
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| Jan 19, 2011 at 18:55 | history | deleted | jzadeh | ||
| Jan 19, 2011 at 18:54 | history | answered | jzadeh | CC BY-SA 2.5 |