Which function makes $f(t)\sin(W(t))$ a martingale?

Question

Let $W(t)$ be a Wiener process. What function $f$ makes $f(t)\sin(W(t))$ a martingale?

I feel a bit confused by the question.

I know that $\sin(W(t))$ has the differential

$d\sin(W(t)) = -\frac{1}{2}\sin(W(t))dt +\cos(W(t))dW(t).$

I also know that I must pick $f$ such that the $dt$ term disappears, but I am unsure how to do this. Could I just pick $f(t) = 0$?

Answer 1

Let $F(t) = f(t)\sin W(t)$. By Ito's lemma,

$$dF(t) = f^\prime(t) \sin W(t) d t-\frac{1}{2} f(t) \sin W(t)dt + f(t) \cos W(t) dW(t)$$

Martingales have zero drift, so choosing any function with derivative $f^\prime(t)=\frac{1}{2}f(t)$ works. As you observe, setting $f(t)=0$ satisfies this.

More generally, any function of the form $$f(t)=Ce^{\frac{1}{2}t}$$ where $C$ is an arbitrary constant will do.