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 $\sin^2B_t+\cos^2B_t$
Best Answer
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: $\int_0^tF'(B_s)^2d[B]_s = \int_0^t e^{2B_s}ds$