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 $dX_t = t\,dS_t$.


1 Answer 1


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: $$\begin{align} E(X_t)&=E\left(\int dX_u\right) \\ &=\int uE(dS_u) \\ &=\int u\mu E(S_u)du \\ &=S_0\int u\mu e^{\mu u}du \\ &=S_0\left(te^{\mu t}-\int e^{\mu u}du\right) \\ &=S_0\left(te^{\mu t}-\frac{1}{\mu}(e^{\mu t}-1)\right) \end{align} $$

[Note: my previous variance calculation was wrong, will fix it when available.]

  • $\begingroup$ Thanks. Quite to the point. Solving the equation for Xt, can I use ito lemma with f(x,t)=d((log s)/t)? $\endgroup$ Commented Oct 17, 2019 at 8:26
  • $\begingroup$ @CharlesSmith I have updated my answer. $\endgroup$ Commented Oct 17, 2019 at 10:42
  • $\begingroup$ I am not sure whether $f(S_t,t)=\log (S_t)/t$ is the best option. $\endgroup$ Commented Oct 17, 2019 at 10:46
  • $\begingroup$ @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. $\endgroup$
    – Slade
    Commented Oct 17, 2019 at 13:20
  • $\begingroup$ @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. $\endgroup$ Commented Oct 17, 2019 at 15:00

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