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Stochastic calculus: [itex] \int_0^td(e^{us}X(s))=\sigma\int_0^t e^{us}dB(s)[/itex] 
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#1
Nov912, 08:55 AM

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Why does:
[itex] \int_0^t d(e^{us} X(s)) = \sigma \int_0^t e^{us} dB(s)[/itex] for stochastic process [itex]X(t)[/itex] and Wiener process [itex]B(t)[/itex]? Also, why is the following true: [itex] \int_0^t d(e^{us} X(s)) = e^{ut}X(t)  X(0)[/itex] 


#2
Nov912, 09:36 AM

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I assume the limits of integration are for the variable, s, and not for the differential quantity, [itex]\displaystyle d(e^{us} X(s))\ .[/itex] We can write that differential as [itex]\displaystyle \ \ d(e^{us} X(s)) = \left(\frac{d}{ds}\left(e^{us} X(s)\right)\right)ds\ .[/itex] So that [itex]\displaystyle \ \ \int_{s=0}^{s=t}d(e^{us} X(s)) = \int_{0}^{t}\left(\frac{d}{ds}\left(e^{us} X(s)\right)\right)ds\ .[/itex] 


#3
Nov912, 09:43 AM

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[itex]\int_{0}^{t}\left(\frac{d}{ds}\left(e^{us} X(s)\right)\right)ds = \int_{0}^{t} \left( ue^{us}X(s) + e^{us} \right)ds[/itex] [itex]= u \int_0^t e^{us}X(s)ds + \int_0^t e^{us}ds[/itex] So I'm still not sure how I can get to the identity based on what you've provided? 


#4
Nov912, 09:48 AM

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Stochastic calculus: [itex] \int_0^td(e^{us}X(s))=\sigma\int_0^t e^{us}dB(s)[/itex]
What is the antiderivative of [itex]\displaystyle \ \ \frac{d}{ds}\left(e^{us} X(s)\right)\ ?[/itex] 


#5
Nov912, 09:55 AM

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$$dX_t = uX_t dt + \sigma dB_t.$$ (Ito interpretation) (I believe that is not actually the correct stochastic differential equation for that integral; it was just a first guess. Look up OrnsteinUhlenbeck process for more information and the actually stochastic differential equation. ) The important point is that a different stochastic differential equation would yield a different integralform solution. 


#6
Nov912, 10:05 AM

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[itex]\displaystyle \ \ = d(e^{ut}X(t))[/itex] It seems I've gone in a circle (obviously I didn't do what you were asking for). 


#7
Nov912, 10:06 AM

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#8
Nov912, 10:10 AM

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#9
Nov912, 10:12 AM

P: 110

Since you know about the OU process, can I ask if this is an accurate representation of a meanreverting OU? [itex] dX(t) = (mX(t))dt + \sigma X(t) dB(t)[/itex] where m is the meanreversion term, B(t) is standard Brownian Motion. I ask because (i) This is what's in my tutorial question list, (ii) Wikipedia and all other external sources I've seen state this process without X(t) in the latter part of the expression (e.g. Wikipedia et al.). 


#10
Nov912, 10:41 AM

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RGV 


#11
Nov912, 11:00 AM

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is [itex]\displaystyle \ \ e^{us} X(s)\ .[/itex] So that [itex]\displaystyle \ \ \int_{0}^{t}\left(\frac{d}{ds}\left(e^{us} X(s)\right)\right)ds=\left(e^{ut} X(t)\right)\left(e^{u0} X(0)\right)\ .[/itex] 


#12
Nov912, 11:10 AM

P: 110

nice, I understand, well done Sammy!



#13
Nov912, 12:14 PM

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[tex] f_x(x,t) = e^{ut},\: f_{xx}(x,t) = 0, \; f_t(x,t) = u e^{ut} x,[/tex] so Ito's Lemma gives [tex] d[f(X(t),t)] = \left( 0 f_x + \frac{1}{2} \sigma^2 f_{xx} + f_t \right) dt + \sigma f_x dB = u e^{ut} X(t) dt + \sigma e^{ut} dB,[/tex] so [tex] \int_0^t d\left(e^{us} X(s) \right) = \sigma \int_0^t e^{us} dB(s)  \int_0^t u e^{us} X(s) ds.[/tex] RGV 


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