Let B be Brownian motion. Use Itos formula to compute the quadratic variation of $\left[X_t^i\right]$ for $\left[X_t^1\right]=e^{B_t}$, $\left[X_t^2\right]=\ln(B_t^2+1)$ and $\left[X_t^1\right]=\sin^2B_t+\cos^2B_t.$

I wanted to use this formula: quadratic variation=$\int_{0}^{t} (F'(X_s))^2d\left[X\right]s$, but I dont know how to get the derivative $F'(X_s)$, as there is no $X$ in $e^{B_t}, \ln(B_t^2+1)$ nor $\sin^2B_t+\cos^2B_t$


I think the wording in the problem is wrong. We are looking for the "quadratic variation"(co-variation) of the following processes : $\forall t\in \mathbb{R^+}$, $X_t^1=e^{B_t}, X_t^2 = \ln(1+B_t^2), X_t^3=\cos^2(B_t)+\sin^2(B_t)$.

Let's do it for $X_t^1$ : First, we express $X_t^1$ as an Itô process by applying the Itô's lemma to the function $F(x) = e^x$. In the differential form, we will have : \begin{equation} dX_t^1 = d(F(B_t)) = e^{B_t}dB_t + \frac12e^{B_t}dt \end{equation} Then we compute the quadratic variation using the informal way: \begin{equation} d[X_.^1,X_.^1]_t "=" dX_t^1.dX_t^1 = e^{2B_t}dt \end{equation} Therfore, we have $[X^1]_t = \int_0^t e^{2B_s}ds$. Note that using your definition of the quadratic variation, we have: $\int_0^tF'(B_s)^2d[B]_s = \int_0^t e^{2B_s}ds$

  • $\begingroup$ Thanks. Did you somehow combine differential and informal way? The differential seems redundant... Lastly, following your steps, the quadratic variation of $ln(B_t^2+1)$ would be $\int _{0}^{t} (\frac {2B_s} {B_s^2+1})^2d \left[B_s \right]$? $\endgroup$ – Maria May 24 at 6:28
  • $\begingroup$ So there are some rules of thumb to compute the quadratic variations of a process. You can look at this : section 3.1 of columbia.edu/~mh2078/FoundationsFE/IntroStochCalc.pdf. Regarding your last question, it is correct but we know that $[B]_s = s$. $\endgroup$ – Sesame May 24 at 9:46

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