Given the stochastic differential equation:
$$dZ_t = -Z_t \theta_t dB_t, \quad Z_0 = 1.$$
for an adapted process $\theta_t$ and Brownian motion $B_t$, how exactly do I apply Itô's Lemma to obtain:
$$ Z_t = \exp\left(- \int_{0}^{t}\theta_u \;dB_u - \frac{1}{2}\int_{0}^{t}\theta_u^2 \;du\right)? $$
This is the SDE for a geometric Brownian motion with time dependent volatility $\theta_t$. It can be easily solved with the substitution $$ X_t = \log Z_t =: f(Z_t). $$
According to Ito's Lemma we have that \begin{align} dX_t &= d f(Z_t) = \frac{\partial f}{\partial z}(Z_t) \;d Z_t + \frac 12 \frac{\partial^2 f}{\partial z^2}(Z_t) \;\bigl(dZ_t \bigr)^2 = \\[2mm] &= -\frac{1}{Z_t} Z_t \theta_t \; dB_t - \frac{1}{2} \frac{1}{\bigl(Z_t\bigr)^2} \bigl(Z_t \theta_t\bigr)^2 \; dt = \\[2mm] &=-\theta_t \; dB_t - \frac 12 \theta_t^2 \; dt. \end{align}
Therefore, $$ X_t = X_0 - \int_0^t \theta_u \; dB_u - \frac 12 \int_0^t \theta_u^2 \; du, $$ and the result follows by simply applying $\exp$ to both sides.
