Expectation of time integral of Wiener process - MathOverflow

Expectation of time integral of Wiener process

2010-10-14T03:11:28Z

I am trying to calculate $E(\int_0^T {W_s ds})$, where $W_s$ is a standard Brownian motion.

Now two approaches I can think of:

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.

2) Calculate $\int_0^T { E(W_s)ds }$.

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.

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.

How can any of these approaches be fixed, if at all? Or is there another way to solve the problem?

Answer by MarkV for Expectation of time integral of Wiener process

2010-10-14T14:16:31Z

It is possible to integrate by parts in $\int_0^T B(t) dt$ and obtain

$-B(t) (T-t)|_{t=0}^{t=T} + \int_0^T (T-t) dB(t) \overset{d}{=} \int_0^T (T-t) dB(t)$

The Wiener integral on the right has a normal distribution with mean $0$ and variance $\int_0^T (T-t)^2 dt = T^3/3$.

Edit: Sorry, I used $B$ instead of $W$ to denote Brownian motion.

Answer by Nate Eldredge for Expectation of time integral of Wiener process

2010-10-14T14:44:35Z

For approach 2, Fubini's theorem works just as well if you show $$ \int_0^T E|W_s|ds < \infty $$ which is easy. Indeed, perhaps even easier is to note $$ \int_0^T E(|W_s|^2)ds = \int_0^T s ds = \frac{1}{2}T^2 < \infty$$ and use Jensen/Hölder/Cauchy-Schwarz.

Answer by unknown (google) for Expectation of time integral of Wiener process

2013-03-29T14:46:52Z

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 \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)} and don't know how to reduce the sequence and get the result of T^3/3. Can anyone help me? Thanks.

Answer by unknown (google) for Expectation of time integral of Wiener process

2013-03-30T05:42:33Z

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.