I want Find $d(\frac{X(t)}{Y(t)})$ where $B(t)$ is a brownian motion and $X(t)=tB (t)$ and $Y(t)=e^{B(t)}$.

Question

Let $B(t)$ is a brownian motion and $X(t)=tB (t)$ and $Y(t)=e^{B(t)}$.
Find $d(\frac{X(t)}{Y(t)})$

Thanks for help.

Answer 1

Name $Z=X/Y$ and use the Ito transformation formula. Or just expand for the infinitesimal time step:

\begin{align*} Z_{t+dt}&=(t+dt)(B_t+dB_t)e^{−B_t-dB_t}\\ &=(t+dt)(B_t+dB_t)e^{B_t}(1-dB_t+\tfrac12dB_t^2+...)\\ &=tB_te^{−B_t}+te^{−B_t}(1-B_t)dB_t+e^{−B_t}(B_t+\tfrac12tB_t-t)dt+O(\sqrt{dt}^3) \end{align*}

using the formal rule $(dB_t)^2=dt$. Then $dZ=e^{−B_t}(B_t+\tfrac12tB_t-t)dt+te^{−B_t}(1-B_t)dB_t$

Answer 2

You can just differentiate the given function by using the product rule twice. $$ \frac d{dt}tB(t)e^{-B(t)}$$