# Ito's formula for the process $M_t=f(t)B_t-\int_{0}^{t}f'(s)B_sds$

I have this problem: let $B$ be a Brownian motion, $f:\mathbb{R}\rightarrow\mathbb{R}$ a function of class $C^{1}$ and $M_t=f(t)B_t-\int_{0}^{t}f'(s)B_sds$. I want to compute $dM_{t}$ by Ito's formula. Usually, in this situation, I define a function $g:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ such thath $M_{t}=g(t,B_t)$ and then compute $dM_t$ by the multidimensional Ito's formula for the Brownian motion. The problem is that I don't know how to handle the $\int_{0}^{t}f'(s)B_sds$ term, due to the presence of $B_s$ inside the integral. The question is: can someone give me a hint on how to proceed?

• Hint: Apply Itô to $$g(t,B_t)=f(t)B_t$$ – Did Feb 11 '17 at 22:21
• This worked. If you post it as an answer I'll accept it (just not to leave the question unanswered) – Uskebasi Feb 11 '17 at 22:47
• Please post an answer yourself. – Did Feb 11 '17 at 22:47

Following the hint given by Did: $$\frac{\partial {g(t,B_t)}}{\partial{t}}=f'(t)B_{t}$$ $$\frac{\partial {g(t,B_t)}}{\partial{x}}=f(t)$$ $$\frac{\partial^{2} {g(t,B_t)}}{\partial{x^{2}}}=0$$ where by $\frac{\partial {g(t,B_t)}}{\partial{x}}$ I mean the partial derivative taken with respect to the second variable. By Ito's formula $$f(t)B_{t}=\int_{0}^{t}f'(s)B_{s}ds + \int_{0}^{t}f(s)dB_{s}.$$ Thus $M_{t}=\int_{0}^{t}f(s)dB_{s}$.
• So the differential form of $M_t$ is $dM_t = f(t) dB_t$. Great! $M_t$ is a martingale with mean zero and variance $\int_0^t f^2(s)ds$. – Pandaaaaaaa Feb 12 '17 at 21:22