Quadratic covariation of two Itô processes

Question

If $dX(t)=\Delta_x (t) dW(t) + \ominus_x(t) dt$ and $dY(t)=\Delta_Y(t) dW(t)+ \ominus_Y(t) dt$, where $X(t), Y(t)$ are two Ito processes.

I need show that $d[X,Y](t)=\Delta_x(t)\Delta_Y(t)dt$, where $\ominus$ stochastic process adapted to Brownian motion.

I can't provide any first steps, that is why I would really appreciate any hints.

Answer 1

You already know that $[Z,Z]_t = \int_0^t \sigma^2(s) \,ds$ for any Itô process $$dZ_t = b(t) \, dt + \sigma(t) \, dW_t.$$ Moreover, the quadratic covariation is defined via the polariation formula, i.e.

$$[X,Y]_t = \frac{1}{2} [X+Y,X+Y]_t - [X,X]_t-[Y,Y]_t. \tag{1}$$

Since both $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are Itô processes, we have

$$[X,X]_t = \int_0^t \Delta_X(s)^2 \, ds$$

and

$$[Y,Y]_t = \int_0^t \Delta_Y(s)^2 \, ds.$$

On the other hand, the sum $(X(t)+Y(t))_{t \geq 0}$ is also an Itô process since

$$d(X+Y)_t = (\Delta_X+\Delta_Y)(t) \, dW_t + (\Theta_X+\Theta_Y)(t) \, dt.$$

Hence,

$$[X+Y,X+Y]_t = \int_0^t (\Delta_X+\Delta_Y)^2(s) \, ds.$$

Plugging this into $(1)$, we get

$$\begin{align*} [X,Y]_t &= \frac{1}{2} \int_0^t (\Delta_X+\Delta_Y)^2(s) \, ds - \int_0^t \Delta_X(s)^2 \, ds - \int_0^t \Delta_Y(s)^2 \, ds \\ &= \int_0^t \Delta_X(s) \Delta_Y(s) \, ds. \end{align*}$$

Answer 2

It looks like there is a missing Brownian term in your expression for $X(t)$

If so then the only term that will appear in the quadratic variation is the product of the two random terms as these are effectively order $dt$, with all order terms in the product now higher order Hence the resulting quadratic variation will be $\Delta_x \Delta_y \cdot dt $

For a more theoretical argument see Shreve Stochastic calculus Vol 2 page 143