how do I integrate W(t)dt , where W(t) is standard Brownian motions?

this isnt too hard is it?

W*dt = d(W*t) - t*dW, then integrate it.

How did you get there?and how do I integrate that?

Integration by parts.

sorry I dont understand, could you show me all the steps?

Start by considering W(t)tThe SDE for this isd(W(t) t) = tdW(t) + W(t)dt + <t,W(t)> - from ito = tdW(t) + W(t)dt (Since dtdW(t) = 0)now integrate to get W(t)t = \int tdW(t) + \int W(t)dt so \int tdW(t) = W(t)t - \int W(t)dtwhich is normally distributed with mean 0 and Var = t^2(I think... this might well be rubbish)

thanks I understand your steps but still dont see how I come up with a final answer in terms of t, and W(t), anyone else?

Hi,it is a random variable, with expectation value 0.To work out what the variance is, you need to put limits to the integration.(Basically you can swap the expectation operator E with the integral sign)amleto

QuoteOriginally posted by: Oliver38how do I integrate W(t)dt , where W(t) is standard Brownian motions?Suppose the integral is from 0 to T.Then, if my arithmetic is right, you can say that the resulting r.v. hasthe same distribution as T W(T)/Sqrt[3], which has the same distribution asT^(3/2) W(1)/Sqrt[3]. This will at least get you all the moments easily. Otherwise,I believe it's a primitive form.regards,

Thanks Alan,Can I ask you how you got this?"Then, if my arithmetic is right, you can say that the resulting r.v. hasthe same distribution as T W(T)/Sqrt[3]"Thanks

Well, it is pretty long-winded and done quickly, so maybethere is a mistake, but here it is.Since W(0) = 0, let me introduce anotherBrownian motion process B(t), where B(t) is also a standardBM but it starts at B(0) = x. Then, I consideredthe moment generating function f(x,t) = E [ e^(-c int_{0,t} B(s) ds ) ].I knew this satisfied (by Feynman-Kac), the PDE problemf_t = 1/2 f_xx - c x f for f(x,t) with f(x,0)=1. I guessed a PDE solnf(x,t) = e^(a(t) + b(t) x) and found f(x,t) = e^((1/6) c^2 t^3 - c x t).Then, the m.g.f. for your problem, where B(0) = 0, is f(0,t) = e^((1/6) c^2 t^3).I recognized the latter as the m.g.f. or the Laplace transform w.r.t. c ofthe distribution of t W(t)/Sqrt[3] or the distribution of t^(3/2) W(1)/Sqrt[3].regards,

Hey Guys interesting thread, does anyone have any good online resources about integrals involving Brownian motion??? Would be a big help... Cheers

Let try, by Itod(W(t))^3=3*(W(t))^2*dW(t)+1/2*3*2*W(t)*(dW(t))^2=3*(W(t))^2*dW(t)+3*W(t)*dt1/3*d(W(t))^3-(W(t))^2*dW(t)=W(t)*dtthen integrate?when you integrate, the second term on left hand side is an Ito integral...

thanks anchorite, that makes sense. how do I compute the ito integral? lets assume we are integrating of 0 and t for simplicity.ThanksOliver