Let $Y_t=t\sin W_t$ where $W_t$ - Wiener process. Calculate $\left< Y \right>_t$.

Question

Let $Y_t=t\sin W_t$ where $W_t$ - Wiener process. Calculate $\left< Y \right>_t$.

The correct answer is: $\left< Y \right>_t=\int_0^t(2s\sin W_s+s^2 \cos (2W_s))ds$

I received something different and I don't know where I made a mistake: $$\begin{align} Y_t &= t\sin W_t \\ dY_t &= td(\sin W_t)+\sin(W_t)dt \\ \sin W_t &=0+\int_0^t\cos W_s dW_s - \frac 12 \int_0^t \sin W_s ds \\ d\sin W_t&= \cos W_t dW_s - \frac 12 \sin W_t dt \\ dY_t&= t\cos W_t dW_t+\sin W_t (1-\frac 12 t)dt \\ \left<Y\right>_t&=\int_0^t(s\cos W_s)^2 ds\end{align}$$ Can someone explain to me what I'm doing wrong?

Answer 1

The one dimensionale Ito's rule for the brownian motion states that for every $F\in C^{2}(\mathbb{R})$ it holds the following $$dF(W_t)=F'(W_t)dW_t+\frac{1}{2}F''(W_t)dt$$

Applying this formula with $F(x)=\sin(x)$, we have that $$d(\sin(W_t))=\cos(W_t)dW_t-\frac{1}{2}\sin(W_t)dt$$ Now, using the stochastic integration by part formula

If $$dX_t=\varphi_t^Xdt+\psi_t^XdW_t,\quad dY_t=\varphi_t^Ydt+\psi_t^YdW_t$$ then $$d(X_tY_t)=X_tdY+Y_tdX_t+\psi_t^X\psi_t^Ydt$$

with $\varphi_t^X=1, \psi_t^X=0$ and $\varphi_t^Y=-\frac{1}{2}\sin(W_t), \psi_t^Y=\cos(W_t)$ we have that \begin{align} dY_t &= \sin(W_t)dt+t\,d(\sin(W_t))\\ &= \left(\sin(W_t) - \frac{1}{2}t\sin(W_t)\right)dt+t\cos(W_t)dW_t \end{align} And so we get that $$\left<Y_t\right>=\int_0^t\left(s\cos(W_s)\right)^2\,ds$$ Which is exactly what you get, maybe i've done something wrong too. If someone finds the mistake, please let me know so i can correct the answer.