In the following exercise, I can't get started on question 2) as I am not sure what to do when there is an integral inside:


Could you help me out?

  • 2
    $\begingroup$ Could you state what you tried and at which step exactly you are in trouble? As a hint: exercise 1 in combination with itos lemma should help $\endgroup$
    – muffin1974
    Jun 14 '15 at 22:39
  • $\begingroup$ @muffin1974 as he's stuck with the integral, I guess he doesn't know where to start so I can understand why he wrote the question this way. But providing the hint is really good. $\endgroup$
    – SRKX
    Jun 16 '15 at 8:01

Let \begin{align*} X_t = W(t)W_*(t) - \frac{1}{2}\int_0^t\big(W_*(u)^2+ W(u)^2\big)du. \end{align*} Then, \begin{align*} dX_t &= W(t) dW_*(t) + W_*(t) dW(t) -\frac{1}{2}\left(W_*(t)^2+ W(t)^2\right)dt, \end{align*} as $W$ and $W_*$ are independent. Consequently, \begin{align*} X_t = \int_0^t \big[W(s) dW_*(s) + W_*(s) dW(s)\big] -\frac{1}{2}\int_0^t\left(W_*(s)^2+ W(s)^2\right)ds. \end{align*} Moreover, \begin{align*} \langle X, X\rangle_t = \int_0^t\left(W_*(s)^2+ W(s)^2\right)ds. \end{align*} That is, \begin{align*} d\langle X, X\rangle_t &= \left(W_*(t)^2+ W(t)^2\right)dt. \end{align*} Since $R_2(t) = e^{X_t}$, \begin{align*} dR_2(t) &= e^{X_t} dX_t + \frac{1}{2}e^{X_t}d\langle X, X\rangle_t\\ &= R_2(t)\Big( W(t) dW_*(t) + W_*(t) dW(t) \Big). \end{align*}

  • $\begingroup$ You'll need to comment what you do at each step if you want to help him understand the solution. $\endgroup$
    – SRKX
    Jun 16 '15 at 8:06
  • $\begingroup$ @Gordon can you just explain how did you get <dXt,dXt> $\endgroup$
    – Lior
    Jun 16 '15 at 8:52
  • $\begingroup$ @Lior, I added more details. $\endgroup$
    – Gordon
    Jun 16 '15 at 12:31
  • $\begingroup$ @Lior, a more heuristic way to compute $\langle dX_t,dX_t\rangle$ is to note that $dW(t)dW(t)=dt$, $dW(t)dt=0$, and $dW(t)dW_∗(t)=0$ (because of independence) etc.. $\endgroup$
    – Gordon
    Jun 16 '15 at 13:58

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