Integration of Brownian

[JohnnySinz]

Given Xt = Integral of {0 -> t} of WsQ ds, where WtQ is a Q-standard Brownian, how do we find the expected value and variance under Q of Xt?

I understand the initial steps of using double integrals to acquire Xt = Integral of {0 -> t} (t - u) dWuQ where u is some dummy variable.

Any help would greatly be appreciated !

 

[Steve Hales]

Hi. Great question So that we're all clear on the notation, let's define: \[X(T)=\int\limits_{0}^{T}{W(s)ds}\] where \(W(s)\) is a standard Brownian motion under \(Q\). The job is to then find the distribution of \(X(T)\). We can discretise the integral by splitting the interval \([0,T]\) into \(n\) equal sub-intervals each of width \(\Delta s\). Therefore we have \(T=n\times \Delta s\). Then the integral can be written as a limit:
\[X(T)=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{W(k\times \Delta s)}\Delta s\] Now take the expectation (which is relatively straightforward) and find the variance (which is a bit trickier) of this expression instead. It might be easier than using double integrals? Let me know how you get on.