Stochastic process, why this is that?

Question

Let $W_t$ be a process such that:

$dW_t = (r+\pi* (\mu-r)) ) * W_t * dt + \pi * \sigma * W_t * d_{z_t}$

where $r, \pi, \mu, \sigma $ are constant, $Z_t$ is the Geometric Brownian motion.

---

Applying Ito's Lemma on $logW_t$, then we have:

$log W_t = \int_{0}^t (r+\pi * (\mu-r)-\frac{\pi^2 \sigma^2}{2})du + \int_{0}^{t}\pi \sigma dz_{u}$

I dont know how come after applying Ito's lemma on the first equation will get to the second equation. Please help

Answer 1

Your source of confusion might be that the first equation is written in a differential form while the second one is written in an integral form.

---

Here is how to get the 2nd equation. Applying Ito Lemma to $logW_t$ we obtain \begin{align} d (\log W_t) &= \frac{1}{W_t}dW_t -\frac{1}{2 W_t^2}(dW_t)^2 \\ & = \frac{1}{W_t}( (r+\pi(\mu-r))W_t dt + \pi \sigma W_t d Z_t)\\ & \quad -\frac{1}{2 W_t^2}(\pi^2 \sigma^2 W_t^2 dt) \\ & = \left(r + \pi(\mu-r) - \frac{1}{2}\pi^2\sigma^2\right)dt + \pi \sigma d Z_t \end{align}

Written in an integral form this becomes $$ \log W_t = \int_0^t \left(r + \pi(\mu-r) - \frac{1}{2}\pi^2\sigma^2\right) du + \int_0^t \pi \sigma d Z_u $$