Ito differential equation

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Define

$$X_t := \left( \begin{matrix} \cos W_t \\ \sin W_t \end{matrix} \right).$$

where $W = \left( W_t,\mathcal F_t \right) _{t\ge0}$ is a standard Wiener process. Find the Ito differential of X and show that

$$\mathbb EX_t = \left( \begin{matrix} \exp \left( -\frac t2 \right) \\ 0 \end{matrix} \right).$$

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By applying Itô's Formula $$f(W_t)-f(W_0) = \int_0^t f'(W_s) \, dW_s + \frac{1}{2} \int_0^t f''(W_s) \, ds$$

to $f(x) := \cos x$ we obtain

$$\cos(W_t)-1 = -\int_0^t \sin(W_s) \, dW_s - \frac{1}{2} \int_0^t \cos(W_s) \, ds$$

The mapping $$(t,w) \mapsto M(t,w) := \int_0^t \sin(W_s) \, dW_s$$ is a martingale since it's a stochastic integral with respect to a Brownian motion. This implies $\mathbb{E}M_t=0$. Thus,

$$\mathbb{E}(\cos(W_t))-1 = - \frac{1}{2} \int_0^t \mathbb{E}(\cos(W_s)) \, ds$$

Define $\varphi(s) := \mathbb{E}(\cos(W_s))$. The last equation is equivalent to

$$\varphi(t)-1 = - \frac{1}{2} \int_0^t \varphi(s) \, ds$$

Solve this differential equation and you are done with the first component of $X_t$. Similar argumentation works for $\sin(W_t)$.