Let $X_t:=e^{W_t}$ where $W_t$ follows the Wiener process. Calculate the drift.

The answer is given as $X_t/2$. My attempt at a solution (which I'm afraid is poor from a mathematical standpoint):

I applied Ito's lemma as $$dX_t=\frac{\partial X_t}{\partial W_t}dW_t+\frac{1}{2}\frac{\partial^2 X_t}{\partial W_t^2}(dW_t)^2$$ and using the fact that $(dW_t)^2=dt$, we get: $$dX_t=\frac{e^{W_t}}{2}dt+e^{W_t}dW_t$$ Therefore the drift is indeed $X_t/2$.

Is my derivation correct? I would appreciate any input on that.


Your solution is correct. Generally speaking, for any $\alpha,\beta\in\mathbb{R}$, the drift $\mu_{X^{\alpha\beta}}$ of the process: $$X_t^{\alpha\beta}:=e^{\alpha t+\beta W_t}$$ will be equal to: $$\mu_{X^{\alpha\beta}}=\left(\alpha+\frac{\beta^2}{2}\right)X_t^{\alpha\beta}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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