How to compute $\int_0^t W_sds$?

Question

I am new to this stuff. Can some one explain how I could compute the stochastic integral of the form $\int_0^t W_sds$, where $W_t$ is Brownian process?

Answer 1

What to compute the integral means is unclear in this context but one can say this: for every $t\geqslant0$, the random variable $$ X_t=\int_0^tW_s\mathrm ds $$ is centered normal vith variance $\sigma_t^2$ where $$ \sigma_t^2=\mathbb E(X_t^2)=2\int_0^t\int_0^s\mathbb E(W_sW_u)\mathrm du\mathrm ds=\int_0^t\int_0^s2u\mathrm du\mathrm ds=\frac{t^3}3. $$ The process $(X_t)_{t\geqslant0}$ is called integrated Brownian motion and is the subject of some active research, for a sample see this paper and the list of references therein.

Answer 2

As Learner pointed out, the integral $\omega \mapsto \int_{0}^t W_s(\omega) \, ds$ is not a stochastic integral, it's a pathwise Lebesgue integration.

But anyway: If we would like to obtain another expression for this integral, we can apply Itô's formula:

$$f(W_t)-f(W_0) = \int_0^t f'(W_s) \, dW_s + \frac{1}{2} \int_0^t f''(W_s) \, ds \tag{1}$$

Since we are looking for the "$ds$-part", it would be nice to have

$$f''(W_s) = 2 W_s$$

i.e. $f''(x)=2x$. We obtain this by choosing $f(x) := \frac{x^3}{3}$. By applying $(1)$:

$$\frac{W_t^3}{3} - 0 = \int_0^t W_s^2 \, dW_s + \frac{1}{2} \int_0^t 2 W_s \, ds \\ \Rightarrow \int_0^t W_s \, ds = \frac{W_t^3}{3} - \int_0^t W_s^2 \, dW_s$$