How to compute integrals of functions of Brownian Motion with respect to time and Brownian motion?

Question

How are functions of Brownian Motion integrated with respect to time? For example, how to calculate the integral: $$I_1 = \int_0^t \lvert \cos(B(s)) \rvert^2 \; ds $$

Secondly, how would an integral such as $I_2$ be calculated: $$I_2 = \int_0^t \lvert \cos(B(s)) \rvert^2 \; dB(s) $$

Finally, in general what is the approach to calculating integrals of the type: $$I_3 = \int_0^t f(B(s))\; ds $$ and $$I_4 = \int_0^t f(B(s)) \; dB(s) $$

Thank you very much for your help.

Answer 1

In general, you cannot "compute" integrals involving Brownian motion in a traditional sense as these integrals are not numbers, but rather random variables. So, typically, the best you can hope for is to find their distributions.

• The easiest case is when you consider non-random function $f(t)$ and you integrate it w.r.t. a Brownian motion, $B_t$. In that case, it is easy to show from the definition that \begin{equation} \int_0^t f(s)dB_s \sim N\left(0,\int_0^t|f(s)|^2ds\right) \end{equation}
• An integral of the form $\int_0^tf(B_s)ds$ can be "computed" just like a standard integral using Riemann sums.
• For more general integrals we can try to use Ito's Lemma to transform the original integral into something more manageable. Ito's Lemma, in it simplest form, tells us that \begin{equation} f(B_t) = f(B_0) + \int_0^t f'(B_s)dB_s + \frac{1}{2}\int_0^tf''(B_s)ds \end{equation} under suitable conditions on $f$. For example, using the above formula we can easily see that $\int_0^tB_sdB_s = 1/2B_t^2 - 1/2t$.