Expectation of time integral of Wiener process - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T11:44:54Z http://mathoverflow.net/feeds/question/42104 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42104/expectation-of-time-integral-of-wiener-process Expectation of time integral of Wiener process Cosmonut 2010-10-14T03:11:28Z 2013-03-30T05:42:33Z <p>I am trying to calculate $E(\int_0^T {W_s ds})$, where $W_s$ is a standard Brownian motion.</p> <p>Now two approaches I can think of:</p> <p>1) Take a partition of $[0,T]$. Calculate $E(\sum {W_{t_i}(t_{i+1} - t_i)})$ and take the limit as you shrink the size of the partition.</p> <p>2) Calculate $\int_0^T { E(W_s)ds }$.</p> <p>However, for approach 1), its not clear what function would dominate the absolute value of the terms inside the E() for all possible partitions, and that it would have a finite expectation. So, interchanging limit and expectation is dicey.</p> <p>For approach 2), Fubini's theorem would require me to know a-priori that $E(\int_0^T {|W_s|ds})$ is finite, and I don't see how I could show that.</p> <p>How can any of these approaches be fixed, if at all? Or is there another way to solve the problem? </p> http://mathoverflow.net/questions/42104/expectation-of-time-integral-of-wiener-process/42152#42152 Answer by MarkV for Expectation of time integral of Wiener process MarkV 2010-10-14T14:16:31Z 2010-10-14T17:23:01Z <p>It is possible to integrate by parts in $\int_0^T B(t) dt$ and obtain</p> <p>$-B(t) (T-t)|_{t=0}^{t=T} + \int_0^T (T-t) dB(t) \overset{d}{=} \int_0^T (T-t) dB(t)$</p> <p>The Wiener integral on the right has a normal distribution with mean $0$ and variance $\int_0^T (T-t)^2 dt = T^3/3$.</p> <p>Edit: Sorry, I used $B$ instead of $W$ to denote Brownian motion.</p> http://mathoverflow.net/questions/42104/expectation-of-time-integral-of-wiener-process/42156#42156 Answer by Nate Eldredge for Expectation of time integral of Wiener process Nate Eldredge 2010-10-14T14:44:35Z 2010-10-14T14:44:35Z <p>For approach 2, Fubini's theorem works just as well if you show $$\int_0^T E|W_s|ds &lt; \infty$$ which is easy. Indeed, perhaps even easier is to note $$\int_0^T E(|W_s|^2)ds = \int_0^T s ds = \frac{1}{2}T^2 &lt; \infty$$ and use Jensen/H&ouml;lder/Cauchy-Schwarz.</p> http://mathoverflow.net/questions/42104/expectation-of-time-integral-of-wiener-process/125917#125917 Answer by unknown (google) for Expectation of time integral of Wiener process unknown (google) 2013-03-29T14:46:52Z 2013-03-29T14:52:12Z <p>Sorry if I ask a question in the answer section because my problem is related to the variance of the time integral of wiener process. I use the first approach of @cosmonut and come up with the variance equal limit of <code>\sum {t_{t_i} (t_{i+1} - t_i)^2}+\sum\sum {min(t_t{t_i},t_t{t_j}) (t_{i+1} - t_i)(t_{j+1} - t_j)}</code> and don't know how to reduce the sequence and get the result of <code>T^3/3</code>. Can anyone help me? Thanks.</p> http://mathoverflow.net/questions/42104/expectation-of-time-integral-of-wiener-process/125980#125980 Answer by unknown (google) for Expectation of time integral of Wiener process unknown (google) 2013-03-30T05:42:33Z 2013-03-30T05:42:33Z <p>I think it's related to the expectation of the time integral of wiener process. I'm curious to use the first approach to find its variance. I'm the beginner in stochastic calculus.</p>