$\sin(W_T)$ and Ito / Martingale Representation Theorem

Question

I've been solving some exercises which require a function to be represented as an adapted stochastic process such that $$ X = \mathbb{E}[X] + \int_0^T \Theta(s)\,dW(s) $$ For example,

1. $X = W(T)$
   The solution was $W(T) = \int_0^T \,dW(s)$.

2. $X = \int_0^T W(s)\,ds$ Using partial integration on $\int_0^T s \,dW(s)$, the solution was $$\int_0^T W(s) \, ds = \int_0^T (T-s)\,dW(s).$$

3. $X = W^2(T)$
   Using $f(x) = x^2$, the solution was $$W^2(T) = T + \int_{0}^{T} 2W(s)\,dW(s).$$

4. $X = \int_0^T W^2(s)\,ds$ Using $f= tx^2$, the solution was $$\int_0^T W^2(s)\,ds = \tfrac{1}{2}T^2 + \int_0^T 2(T-s)\,W(S)\,dW(s).$$

$\textbf{Question:}$ Find $x\in \mathbb{R}$ and a stochastic process $\Theta(t)$ such that $$ \sin(W(T)) = x + \int_0^T \Theta(s)\,dW(s) $$

I'm not sure how to proceed on this one.

All help is appreciated.

Many thanks,
John

Answer 1

Hint: Using Itô's formula, show that $X_t := \sin(W_t) e^{t/2}$ satisfies

$$dX_t = \cos(W_t) e^{t/2} \, dW_t.$$

Conclude that $$\sin(W_T) = \int_0^T e^{-(T-t)/2} \cos(W_t) \, dW_t.$$

Remark: The idea is to find a determinstic function $f$ such that $X_t := f(t) \sin(W_t)$ is a martingale. By Itô's formula, a sufficient condition is

$$\frac{1}{2} \sin''(W_t) f(t) + \sin(W_t) f'(t)=0. \tag{1}$$

This is equivalent to $$-\frac{1}{2} \sin(W_t) f(t)+ \sin(W_t) f'(t)=0 \iff f'(t) =\frac{1}{2} f(t)$$

Obviously, $f(t) := e^{t/2}$ is a solution to this ODE. Applying again Itô's formula, we get

$$X_t = \int_0^t f(s) \cos(W_s) \, dW_s + \underbrace{\int_0^t \frac{1}{2} f(s) \sin''(W_s) + f'(s) \sin(W_s) \, ds}_{\stackrel{(1)}{=}0},$$

i.e.

$$\sin(W_t) = \frac{1}{f(t)} \int_0^t f(s) \, \cos(W_s) \, dW_s.$$

Plugging in $f$ yields the desired result.