Integration – Integration by Parts with Brownian Motion and Non-Random Function

Question

Let $B$ be a standard one-dimensional Brownian motion. I want to show for a continuously differentiable non-random function $\phi$ that,
\begin{align}
\int_0^t \phi(s) dB_s = \phi(t) B_t – \int_0^t B_s \phi'(s) ds.
\end{align}
I am familiar with the integration by parts formula for two semimartingales,
\begin{align}
X_tY_t = X_0Y_0 + \int_0^t X_s dY_s + \int_0^t Y_s dX_s + \langle X, Y \rangle _t.
\end{align}
However, I am not sure how to use this identity to prove the above statement. Any ideas?

Answer 1

You need three ingredients: 1. Any continuously differentiable function is a semimartingale, since it is of locally finite total variation, 2. for a Riemann-Stieltjes / Lebesgue-Stieltjes integral you have $d\phi(s) = \phi'(s) ds$ and 3. the covariation is defined in terms of the martingale parts, hence we have $[X,\phi]_t = [X,0]_t = 0$. Hence combining it yields

$\phi(t)B_t = \phi(0)B_0 + \int_0^t \phi(s) dB_s + \int_0^t B_s d\phi(s) = \int_0^t \phi(s) dB_s + \int_0^t B_s \phi'(s) ds$