# $X_t=\frac{tW_t}{e^{W_t}}$ calculate $dX_t$ and $⟨X⟩_t$

Let $W_t$ be a Brownian motion and $X_t=\frac{tW_t}{e^{W_t}}$. Applying Itô's Lemma gives

$X_t=\int_0^ts\frac{1-W_s}{e^{W_s}}dW_s+\int_0^t\frac{W_s}{e^{W_s}}ds+\frac{1}{2}\int_0^ts\frac{W_s-2}{e^{W_s}}ds$, but that doesn't seem too helpful to me.

• Why isn't is useful? It seems like you can calculate everything. Just forget about the integrals to get $dX_t$ and square the integrand in the stochastic integral to get the quadratic variation. – Kore-N Dec 11 '17 at 10:16
• Just seemed like a ugly solution to me ..so basicly $$dX_t=t\frac{1-W_t}{e^{W_t}}dW_t+\frac{W_}{e^{W_t}}dt+\frac{1}{2}t\frac{W_t-2}{e^{W_t}}dt$$ and i am done. The quadratic variation part is clear to me. – Andrew Dec 11 '17 at 10:29