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Let $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space, $(\mathcal F_t)_{t\ge0}$ be a complete and right-continuous filtration on $(\Omega,\mathcal A,\operatorname P)$, $\left(W_t^{(i)}\right)_{t\ge0}$ be an $\mathcal F$-Brownian motion on $(\Omega,\mathcal A,\operatorname P)$ and $\left(X_t^{(i)}\right)_{t\ge0},\left(Y_t^{(i)}\right)_{t\ge0},\left(Z_t^{(i)}\right)_{t\ge0}$ with $${\rm d}X^{(i)}_t=Y^{(i)}_t{\rm d}t+Z^{(i)}_t{\rm d}W^{(i)}_t\tag1.$$ Now let $X:=\left(X^{(1)},X^{(2)}\right)$, $Y:=\left(Y^{(1)},Y^{(2)}\right)$, $Z:=\left(X^{(1)},Z^{(2)}\right)$, $W:=\left(W^{(1)},W^{(2)}\right)$ and $$f:\mathbb R^2\to\mathbb R\;,\;\;\;x\mapsto x_1x_2.$$ By the Itō formula$^1$, \begin{equation}\begin{split}f(X_t)-f(X_0)&=\frac12\int_0^t{\rm D}^2f(X_s)\:{\rm d}\:[\![X]\!]_s+\int_0^t{\rm D}f(X_s)\:{\rm d}X_s\\&=\left[X^{(1)},X^{(2)}\right]_t+\int_0^tX^{(2)}_s\:{\rm d}X^{(1)}_s+\int_0^tX^{(1)}_s\:{\rm d}X^{(2)}_s.\end{split}\tag2\end{equation}

Now my question is: In order to obtain $$\left[X^{(1)},X^{(2)}\right]_t=\int_0^tZ^{(1)}_sZ^{(2)}_s\:{\rm d}s\tag3,$$ do we need $W^{(1)}$ and $W^{(2)}$ to be independent? Or could we even take $W^{(1)}=W^{(2)}$?

$^1$ $[\![M]\!]$ denotes the tensor-quadratic variation of $M$.

0xbadf00d
  • 14,208

1 Answers1

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Actually, $[X^{(1)},X^{(2)}]_t = \int_0^t Z_s^{(1)}Z_s^{(2)}ds$ holds iff $W^{(1)}$ and $W^{(2)}$ are perfectly correlated. If they are independent, $[X^{(1)},X^{(2)}]_t = 0$. In general, if $\text{corr}(W^{(1)}_t,W^{(2)}_t) = \rho t$ then $[X^{(1)},X^{(2)}]_t = \int_0^t \rho Z_s^{(1)}Z_s^{(2)}ds$.

user6247850
  • 14,184
  • Thank you for your answer. You are right. But if $W^{(1)}=W^{(2)}$, is $W$ still a ($\mathbb R^2$-valued) martingale? I guess the answer should be (almost trivially) yes, or am I missing something? – 0xbadf00d Jun 19 '20 at 04:23
  • You're not missing anything, $W$ is still a martingale in $\mathbb{R}^2$. – user6247850 Jun 19 '20 at 04:56