$d(f(t)W_t) = f'(t)W_t dt + f(t)dW_t $

Question

I keep reading about one proof on internet and I found there a specific implication which I don't understand. Let $W_t$ be a Wiener process. Could you please explain to me why from

$$d(f(t)W_t) = f'(t)W_t dt + f(t)dW_t$$

We get

$$\int_0^t f(s)dW_s = f(t)W_t - \int_0^tf'(s)W_s ds$$

Answer 1

Answer assuming that Ito's formula holds.

You can consider the process $X$ as $d X_t = A_t dt+B_td W_t$. In this setting, your process $X$ is the brownian motion $W$, i.e. $A_t=0$ and $B_t=1$ for all $t$. In order to apply Ito's formula we should compute the derivatives of the function $g(t,x):= f(t)x$. In this case we have $g_t(t,x)=f'(t)x$, $g_x(t,x)=f(t)$ and $g_{xx}(t,x)=0$. Applying Ito's formula to the function $g$ we get $$g(t,W_t)=g(0,W_0)+\int_0^t(g_t(s,W_s)+A_sg_x(s,W_s)+\tfrac{1}{2}B_s^2g_{xx}(s,W_s))ds+\int_0^t B_sg_x(s,W_s)dW_s.$$ Using the previous information and thanks to the fact that $W_0=0$ you have $$f(t)W_t=\int_0^tf'(s)W_s ds+\int_0^t f(s)dW_s.$$

Answer 2

$$d(f(t)W_t)=f'(W_t)dt+f(t)dW_t,$$ is a short notation for $$f(t)W_t-f(0)W_0=\int_0^t f'(W_s)ds+\int_0^t f(s)dW_s.$$