Quadratic covariation of Itô processes I haven't found any similar question in the forum, so I trust some of you will find this thought-provoking (at the very least). Perhaps you can help me.
Let's consider first the two following stochastic differential equations:
$$dX_t = (6+3X_t)dt + (2+X_t)dB_t, X_0 = 1$$
$$dY_t = 3Y_tdt + Y_tdB_t, Y_0 = 1$$
Of course $B_t$ is a standard Brownian Motion started at zero.
Now the question is: how can I compute the quadratic covariation of these two processes, and then obtain its expectation? That is, how can I compute $E\left(\left<X,Y\right>_t\right)$.
Regards and thanks in advance.
 A: Define the quadratic variation of a semimartingale $(X_t)_{t \geq 0}$ by
$$[X,X]_t := \mathbb{P}-\lim_{n \to \infty} \sum_{j=0}^n (X_{t_j}-X_{t_{j-1}})^2$$
where $\Pi_n := \{t_0<\ldots<t_n<t\}$ is a sequence of partitions such that $|\Pi_n| \to 0$. Moreover, we set $$[X,Y]_t := \frac{1}{4} ([X+Y]_t-[X-Y]_t).$$
Using this definition it is not difficult to show the following result.

Lemma 1 Let $(M_t)_{t \geq 0}$ a continuous square-integrable martingale and $(A_t)_{t \geq 0}$ a continuous process of bounded variation. Then,
  
  
*
  
*$[M]_t$ is the unique previsible process such that $M_t^2-[M]_t$ is a martingale.
  
*$[M,A]_t=[A,A]_t=0$
  

Now we are ready to calculate the quadratic covariation. By definition,
$$X_t\pm Y_t= \underbrace{\int_0^t (2+X_s \pm Y_s) \, dB_s}_{=:M_t} + \underbrace{c_{\pm}+ \int_0^t (6+3X_s \pm 3Y_s) \, ds}_{=:A_t}.$$
where $c_+=2$, $c_-=0$. Obviously, $(M_t)_{t \geq 0}$ is a (continuous) martingale and $(A_t)_{t \geq 0}$ of bounded variation. Consequently, we obtain by applying Lemma 1
$$[X \pm Y,X \pm Y]_t = [M,M]_t+2[M,A]_t+[A,A]_t = \int_0^t (2+X_s \pm Y_s)^2 \, ds.$$
Hence,
$$[X,Y]_t = \frac{1}{4} ([X+Y]_t-[X-Y]_t) = \int_0^t (2+X_s) \cdot Y_s \, ds \tag{1}$$
In order to compute $\mathbb{E}[X,Y]_t$, we have to find $\mathbb{E}Y_t$ and $\mathbb{E}(X_t \cdot Y_t)$. Since stochastic integrals with respect to a Brownian motion are martingales, we find
$$f(t) :=\mathbb{E}(Y_t)=1+\underbrace{\mathbb{E} \left( \int_0^t Y_s \, dB_s \right)}_{0} + 3 \int_0^t \mathbb{E}(Y_s) \, ds$$
i.e. $f$ satisfies the ODE
$$f'(t) = 3f(t) \qquad f(0)=1$$
Obviously, the unique solution is given by $$\mathbb{E}Y_t = f(t)= e^{3t} \tag{2}$$ Similarly, we obtain from Itô's formula that
$$g(t) := \mathbb{E}(X_t \cdot Y_t) =1+\mathbb{E} \left( \int_0^t(7X_s Y_s+8Y_s) \, ds \right) = 8 \int_0^t f(s) \, ds + 7 \int_0^t g(s) \, ds$$
i.e. $$g'(t) = 8f(t)+7g(t) = 8e^{3t}+7g(t) \tag{3}$$ 
This ODE can be solved explicitely; I leave it to you. Combining $(1)$, $(2)$ and $(3)$ allows us to compute $\mathbb{E}[X,Y]_t$.
