G'day, I am trying to solve for the Geometric Brownian motion
$()=_t_t+_tS_t_t$
using Ito's formula. Starting:
$d ln S_t = \frac{1}{S_t}dS_t + \frac{1}{2}(-\frac{1}{S_t^{2}})(dS_t)^2$
Computing the following:
a) $\frac{1}{2}(-\frac{1}{S_t^{2}}) = -\frac{1}{2S_t^{2}}$
b) $(_t_t+_tS_t_t)^2 = u^2S^2dt^2 + ^2S^2dW^2 - 2_t_t_tS_t_t$
Continuing with the term $(u^2S^2dt^2 + ^2S^2dW^2 - 2_t_t_tS_t_t)dt$
$dt^2=0 ,u^2S^2dt^2 $ disappears;
$dW^2 = dt, (^2S^2dW^2)dt$ leads to $dt^2$ and hence this term is 0 as well;
Last term $(2_t_t_tS_t_t)dt$ leads to zero as well if we would follow my reasoning above however this doesn't seem correct and the solution given is as follows:
$= \frac{1}{S_t}[_t_t+_tS_t_t] -\frac{1}{2S_t^{2}}(_tS_t)^2dt$
$= (_t-\frac{1}{2}_t^{2})dt + _t_t$
Can someone explain to me
1) Where do I go wrong from step b where I try to eliminate terms and
2) How the last 2 steps are being calculated?
The way you can apply Ito is to use $$ f(S_t) = \ln S_t $$ then we have $$ df = \frac{\partial f}{\partial S_t} dS_t + \frac{1}{2}\frac{\partial^2 f}{\partial S_t^2} (dS_t)^2 =\frac{1}{S_t}dS_t + \frac{1}{2}\frac{1}{S_t^2}dS_t^2 $$
We now need to compute $dS_t^2$
$$ dS_t^2 = \left(\mu_tS_tdt + \sigma_tS_tdW_t \right)^2 = \sigma_t^2S_t^2dW_t^2 + 2\mu_t\sigma_tS_t^2dW_tdt + \mu_t^2S_t^2dt^2 $$ I have ordered the terms in order of $dt$. The first term is order $dt$ the other terms are all higher orders of $dt$ namely, $dW_tdt\sim dt^{3/2}$. We ignore terms higher than $dt$ as they vanish to zero faster than $dt$.
So we end up with $$ df = \left(\mu_tS_tdt + \sigma_tS_tdW_t\right)\cdot \frac{1}{S_t} + \sigma_t^2S_t^2dt \cdot \frac{1}{2}\frac{1}{S_t^2} $$ or $$ df = \mu_t dt + \sigma_t dW_t + \frac{\sigma_t^2}{2}dt = \left(\mu_t + \frac{\sigma_t^2}{2}\right)dt + \sigma_tdW_t $$