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Will the following process $$r(t)=\int_0^ta(s,t)dW(s)$$ be adapted to the Brownian motion $W(s)$? Will $r(t)$ be an Ito process?

Edit: Maybe I should rephrase it a bit. The question is: does dependence $a(s,t)$ on $t$ (which is not the variable for intergration) interferes somehow with the usuall requirement for Ito integral that $a(s,t)$ must be adapted to the filtration generated by W?

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I am a it confused. Is this from a class in stochastic calculus? Much of the beginning of any such class starts with regularity conditions necessary on the $\alpha (s, t)$ to make this an Ito Process. Time dependence is close to a non-issue; it is usually covered. If it is from an application, maybe start by having a look at the wiki entry. From memory, you need something a bit weaker than square-integrability on $\alpha$, plus some adaptation conditions. –  gnometorule Feb 7 '13 at 22:18
    
@gnometorule In usuall definitions integrand only depends on time via integration variable, i.e. $s$ but not $t$. So I wonder does this dependence breaks something or not? –  learningmath Feb 8 '13 at 12:47
    
@gnometorule I suspect that this dependence on $t$ can ruin the adaptedness of the integrand to $W(s)$ –  learningmath Feb 8 '13 at 12:49
    
I'm really rusty, but learned Ito calculus way back from both K&S, and Oksendahl. From memory, Oksendahl typically uses $\alpha$ exactly as you do, in which case you need a set of 3 or so conditions on $\alpha$ to ensure the integral exists (these are not the only possible ones obviously; something about square-integrability and adaptedness of $\alpha$). You wouldn't have access to this book maybe? I can warmly recommend it anyway, in particular for applications in finance and filtering: amazon.com/gp/aw/d/3540047581/ref=redir_mdp_mobile –  gnometorule Feb 8 '13 at 13:59
    
Well, if for example the function $(s,t)\mapsto a(s,t)$ is deterministic, then the process $r$ is very much adapted to the filtration generated by $W$. "Not adapted" does not mean "which depends on $t$", but "not measurable with respect to some sigma-algebra". And deterministic objects are measurable with respect to EVERY sigma-algebra. –  Did Feb 10 '13 at 21:10
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The question is: does dependence $a(s,t)$ on $t$ (which is not the variable for inte[r]gration) interferes somehow with the usual[l] requirement for Ito integral that $a(s,t)$ must be adapted to the filtration generated by $W$?

The answer is: it does not whatsoever.

As explained in a comment, if the function $(s,t)↦a(s,t)$ is deterministic, then the process $r$ is very much adapted to the filtration generated by $W$. "Not adapted" does not mean "which depends on $t$", but "not measurable with respect to some sigma-algebra". And deterministic objects are measurable with respect to EVERY sigma-algebra.

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