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?
2026-03-29 16:39:56.1774802396
Ito's formula for the process $M_t=f(t)B_t-\int_{0}^{t}f'(s)B_sds$
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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 $M_{t}=\int_{0}^{t}f(s)dB_{s}$.