I have been given the following question:

Given that $S_t$ follows Geometric Brownian Motion, write down the dynamic of $S_t$ and then compute the dynamic of $f(t,S_t) = e^{tS^{2}}$

For the first part of the question, I have got this answer: $$dS_t = \mu S_tdt + \sigma S_t dWt$$

Is it correct?

And for the second part, I know that the price $f(t,S_t)$ follows the process $$df = (\frac{\partial f}{\partial t}+\mu S_t \frac{\partial f}{\partial S_t}+\frac{1}{2} \sigma ^2S_t\frac{\partial^2f}{\partial S_t^2})dt +\sigma S_t dWt$$

I am having trouble finding the answer using this process and given the information.

Any help is appreciated.


1 Answer 1


The above equation should correctly read as follows:

$df=\big(\frac{\partial f}{\partial t}+\mu S_t \frac{\partial f}{\partial S_t}+\frac{1}{2}\sigma^2 S_t^2\frac{\partial^2 f}{\partial S_t^2}\big)+\sigma S_t \frac{\partial f}{\partial S_t}dW$


(a) $\frac{\partial f}{\partial t}=S_t^2f$

(b) $\frac{\partial f}{\partial S_t}=2S_ttf$

(c) $\frac{\partial^2 f}{\partial S_t^2}=2tf+4S_t^2t^2f$

The Stochastic Differential Equation (SDF) governing the dynamics of $f$ becomes:

$\frac{df}{f}=dt \big(S_t^2+2 \mu S_t^2t+\sigma^2S_t^2t+2\sigma^2S_t^4t^2 \big)+2S_t^2t\sigma dW$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.