Time Integrals of Brownian Motion and Their Itô Representations

# More questions about integral of Brownian Motion w.r.t time


# More questions about integral of Brownian Motion w.r.t time












A similar question have been posted earlier but one part has remained unanswered. Let us define: $$X_t = \int_0^t W_s ds,$$

where $W_t$ is a standard Brownian Motion. Is $X_t$ an Itô process or a Riemann integral? How to write the Itô form of: $$\int_0^tW_sds\text{ ?}$$ Is the following formula correct? Why? $$d\biggl(\int_0^tW_sds\biggl) = W_tdt $$

## Answer by Quantuple (score 9, accepted)

https://quant.stackexchange.com/a/49344

As usual with those kind of integrals, another way to reach the result is to:

- Express $W_s$ in integral form as $\int_0^s dW_u$

- Use Fubini theorem to change the integration bounds of the resulting double integral

More specifically, \begin{align} \int_0^t W_s ds &= \int_0^t \int_0^s dW_u ds \\ &= \int_0^t \int_u^t ds dW_u \\ &= \int_0^t (t-u) dW_u \end{align} which is indeed an Ito integral and in this case a Gaussian r.v. with mean zero and variance given by Ito isometry.

## Answer by Magic is in the chain (score 9)

https://quant.stackexchange.com/a/49340

It is indeed Riemann integrable, so you don't need stochastic integration. For a given path, you can interpret the integral in the Riemann sense. For a given t, the paths are random, so it is a random variable.

You can also express it as an Ito’s process. To see the connection, just apply ito's lemma to $tW_t$:

$d \left(tW_t\right)=tdW_t+W_tdt$

$W_tdt=d \left(tW_t\right)-tdW_t$

Then integrate:

$X_t=\int_0^t{W_sds}=tW_t-\int_0^t{sdW_s}$

$\quad =t\int_0^t{dW_s}-\int_0^t{sdW_s}$

$\quad =\int_0^t{\left(t-s\right)dW_s}$

So it is normally distributed. Easy to check mean is zero, and variance is:

$V\left[X_t\right]=\int_0^t{\left(t-s\right)^2ds}=\frac{1}{3}t^3$

Please see more detailed discussion here: Integral of Brownian motion w.r.t. time