# Evaluating the SDE $dX_t = t\,dS_t$

The process $$S$$ is a geometric Brownian motion with an SDE: $$dS_t = S_t(\sigma\, dB_t + \mu\, dt)$$. I'm stuck evaluating $$E(X_t)$$ and $$V(X_t)$$, where $$dX_t = t\,dS_t$$.

Using Itô's Lemma, notice that: $$d(tS_t)=tdS_t+S_tdt=dX_t+S_tdt$$ Hence: $$X_t=tS_t-\int S_udu$$ Using independence of Brownian increments, $$E(S_udW_u)=E(S_u)E(dW_u)=0$$, and the chain rule for the 4th step: \begin{align} E(X_t)&=E\left(\int dX_u\right) \\ &=\int uE(dS_u) \\ &=\int u\mu E(S_u)du \\ &=S_0\int u\mu e^{\mu u}du \\ &=S_0\left(te^{\mu t}-\int e^{\mu u}du\right) \\ &=S_0\left(te^{\mu t}-\frac{1}{\mu}(e^{\mu t}-1)\right) \end{align}
• I am not sure whether $f(S_t,t)=\log (S_t)/t$ is the best option. Oct 17, 2019 at 10:46
• @DaneelOlivaw I am pretty sure the plus signs in the last two expressions in the calculation of the expectation should be minus signs. Also, (you seem much more experienced so I may be totally off base here) I think that there may be an issue in calculating the Variance's first term since in the Variance the $\mathbb E [(\int u \mu S_u du)^2]$ term should appear as a double integral using two different dummy variables. So the $dt^2$ argument doesn't seem to work. Oct 17, 2019 at 13:20
• @Slade correct on both counts, I did this too hastily. For variance, I was using independence of $S_u$ and $dW_u$ to separate the total variance into the sum of the "drift part" and the "diffusion part", but I messed it up as $S_u$ multiplies $dW_u$ so the two integrals are not independent. Oct 17, 2019 at 15:00