consider the following SDE for the 2-dimensional stochastic processes $ X_t=( ( X_1(t) ),( X_2(t) ) ) $ driven by a 2-dimensional continuous standard Brownian motion $ B_t=( ( B_1(t) ),( B_2(t) ) ) $
$ dX_1(t)=X_2(t)dt+adB_1(t) $
$ dX_2(t)=-X_1(t)dt+bdB_2(t)$
where a and b control the intensity of the noise.
find the quadratic covariation of $ X_1(t), X_2(t) $.
my result is $0$ but the professor told us $abt$.
which one is correct?
here my computations:
$<X_1(t),X_2(t)>=1/2[<X_1(t)+X_2(t)>-<X_1(t)>-<X_2(t)>]$
where
$ <X_1(t)>=a^2t$
and
$ <X_2(t)>=b^2t$
now I want to find $<X_1(t)+X_2(t)>$
$dX_1(t)+dX_2(t)=X_2(t)dt+adB_1(t)-X_1(t)dt+bdB_2(t)$
so $<X_1(t)+X_2(t)>=\int_0^ta^2ds+\int_0^tb^2ds=a^2t+b^2t$
and then $<X_1(t),X_2(t)>=1/2[a^2t+b^2t-a^2t-b^2t]=0$
I think that $<X_1(t)+X_2(t)>=\int_0^t(a+b)^2ds$ is not correct because $dB_1(t)$ and $dB_2(t)$ are different so their product is $0$ and not $dt$
Indeed we just use the polarization identity:
$$[X,Y]_t=\tfrac{1}{2}([X+Y]_t-[X]_t-[Y]_t).$$
We compute $[X_{t}^1]_t=a^{2}t$,$[X_{t}^2]_t=b^{2}t$ since Brownian motion has zero cross-variation with any increasing process. For the sum, we use that Computing cross variation of independent brownian motions cross variation of independents BMs is zero
$$E(\langle X,Y\rangle_t)^{2} = \lim_{||\Delta||\to 0} E(\sum_i(X_{t_{i+1}}-X_{t_i})(Y_{t_{i+1}}-Y_{t_i}))^{2}\leq \lim_{||\Delta||\to 0} ||\Delta|| t=0$$
to indeed get the sum of their quadratic variations $$[X+Y]_t=[X]_t+[Y]_t.$$ and thus zero overall.