I am currently learning about stochastic differential equations and Ito's lemma and this problem came up.
$$dZ(t) = \frac{1}{2} \, (W(t))^2 \cdot Z(t) \cdot dt + W(t) \cdot Z(t) \cdot dWt; \hspace{5mm} Z(0)=1. $$
I know that I need to use Ito's lemma, and I believe the easiest way to solve this problem is to somehow analyze the result for $\ln(Z(t))$, but I am not sure exactly how to go about doing this. Any help or tips would be greatly appreciated!
Thanks!
Suppose that $(Z_t)_{t \geq 0}$ solves the SDE
$$dZ_t = \frac{1}{2} W_t^2 Z_t \, dt + W_t Z_t \, dW_t, \qquad Z_0=1 \tag{1}$$
Applying (formally) Itô's lemma for $f(x) := \log x$, we find
$$\begin{align*} \log Z_t - \underbrace{\log Z_0}_{0} &= \int_0^t f'(Z_s) \, dZ_s + \frac{1}{2} \int_0^t f''(Z_s) \, d\langle Z \rangle_s \\ &= \int_0^t \frac{1}{Z_s} \, dZ_s - \frac{1}{2} \int_0^t \frac{1}{Z_s^2} \, d\langle Z \rangle_s. \tag{2} \end{align*}$$
where, by $(1)$, $$d\langle Z \rangle_s = W_s^2 Z_s^2 \, ds.$$ Plugging this into $(2)$ and using $(1)$, we get
$$\log Z_t = \int_0^t W_s \, dW_s + \frac{1}{2} \int_0^t W_s^2 \, ds - \frac{1}{2} \int_0^t W_s^2 \, ds = \int_0^t W_s \, dW_s.$$
The latter integral can be calculated explicitly using Itô's formula, and therefore we obtain
$$\log Z_t = \frac{1}{2} (W_t^2-t).$$
Hence,
$$Z_t = \exp \left( \frac{1}{2} (W_t^2-t) \right).$$
Since the above calculations were just formal calculations, we have to check that this process is indeed the solution to $(1)$; this follows from another application of Itô's formula.