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.


1 Answer 1


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: $$\mu_{X^{\alpha\beta}}=\left(\alpha+\frac{\beta^2}{2}\right)X_t^{\alpha\beta}$$


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