partial derivative of geometric brownian motion wrt time? $S_t = S_0 e^{\mu t - \frac{1}{2}\sigma^2t + \sigma W_t}$

Question

If geometric brownian motion is given by:

$$S_t = S_0 e^{\mu t - \frac{1}{2}\sigma^2t + \sigma W_t}$$

Then what would $\frac{\partial S_t}{\partial t}$ be?

Answer 1

With respect to Itô's differentiation formula, the first thing is to take $W_t$ as $x_t$ or because it's not differentiable, so $$g(t,x_t)= e^{\mu t - \frac{1}{2}\sigma^2t + \sigma x}$$then using Itô's differentiation formula as below $$dg(t,x)=\frac{\partial g}{\partial t}dt +\frac{\partial g}{\partial x}dx+\frac 12 \frac{\partial^2 g}{\partial^2 x}(dx)^2\\ dg(t,x)=e^{\mu t - \frac{1}{2}\sigma^2t + \sigma x}(\mu t - \frac{1}{2}\sigma^2)dt +e^{\mu t - \frac{1}{2}\sigma^2t + \sigma x}(1\sigma)dx+\frac 12 e^{\mu t - \frac{1}{2}\sigma^2t + \sigma x}(1\sigma)^2(dx)^2\\$$now you can simplify by factoring $e^{\mu t - \frac{1}{2}\sigma^2t + \sigma x}$ and this fact $$\boxed {(dx)^2=(dW_t)^2=dt}$$ so

$$dg(t,x)=e^{\mu t - \frac{1}{2}\sigma^2t + \sigma x}((\mu t - \frac{1}{2}\sigma^2)dt+(1\sigma)dx+\frac 12(1\sigma)^2(dx)^2) |_{x=W_t}\\\to\\ dg(t,x)=e^{\mu t - \frac{1}{2}\sigma^2t + \sigma W_t}((\mu t - \frac{1}{2}\sigma^2)dt+(1\sigma)dW_t+\frac 12(1\sigma)^2\underbrace{(dt)}_{(dx)^2=(dW_t)^2=dt}))\\$$ finally $$dg(t,x)|_{x=W_t}=e^{\mu t - \frac{1}{2}\sigma^2t + \sigma W_t}((\mu t - \frac{1}{2}\sigma^2+\frac 12(1\sigma)^2)dt+(1\sigma)dW_t)\\dg(t,x)|_{x=W_t}=e^{\mu t - \frac{1}{2}\sigma^2t + \sigma W_t}((\mu t )dt+(1\sigma)dW_t)$$