Let $w = e^{sz-\frac{1}{2}t^2}$ where z is a standard wiener process. Find the equation governing w.

Question

First I applied logarithms to simplify this. $$ lnw = sz-\frac{1}{2}t^2 $$

Then I found the partial derivatives with respect to $t$ and then $z$. $$ \frac{\partial lnw}{dt} = -t $$ $$ \frac{\partial lnw}{dz} = s $$

Then I solved both equations for $dlnw$: $$ dlnw = sdz \hspace{1cm} dlnw = -tdz$$

Putting these together I get what is my solution. $$dlnw = \frac{1}{2}(sdz - tdz)$$ But this is just plain wrong because we should get an answer the form should be $dw = adt + bdz$. What am I missing?

The context of the problem comes from a financial mathematics text where we are working with Ito calculus and Wiener processes.

Answer 1

Apply Ito’s calculus $$dw = \frac{\partial w}{\partial z}dz + \frac12 \frac{\partial^2 w}{\partial z^2}dz^2+ \frac{\partial w}{\partial t} dt = w[sdz +(s^2-t)dt] $$ where $ dz^2=dt$.