3
$\begingroup$

Let $V_t, W_t : [0,T] \times \Omega \to \mathbb R$ be indpendent Brownian motions defined up to time $T$ and let $X_t,Y_t : [0,T] \times \Omega \to \mathbb R$ be stochastic processes adapted with respect to the standard filtrations $\mathcal F^V_*$ and $\mathcal F^W_*$, respectively.

Can I conlude the following?

$$\mathbb E \left[\left(\int_0^T X_t d V_t \right) \left(\int_0^T Y_t d W_t\right) \right] = 0$$

I'm guessing this by in turn guessing some sort of generalisation of the Ito isometry, namely

$$\mathbb E \left[\left(\int_0^T X_t d V_t\right) \left( \int_0^T Y_t d W_t\right) \right] \overset{?}{=} \mathbb E \left[\int_0^T X_t Y_t d V_t d W_t \right] = 0.$$

$\endgroup$

1 Answer 1

5
$\begingroup$

No, you don't need a generalization of Itô's isometry; you have to use the independence of the processes.

By assumption, $(V_t)_{t \geq 0}$ and $(W_t)_{t \geq 0}$ are independent processes and therefore the correspdonding canonical filtrations are independent, i.e. $\mathcal{F}_T^V$ and $\mathcal{F}_T^W$ are independent for all $T \geq 0$. Since $\int_0^T X_t \, dV_t$ is $\mathcal{F}_T^V$-measurable and $\int_0^T Y_t \, dW_t$ is $\mathcal{F}_T^W$-measurable, we find $$\mathbb{E} \left[ \left( \int_0^T X_t \, dV_t \right) \left( \int_0^T Y_t \, dW_t \right) \right] = \mathbb{E} \left( \int_0^T X_t \, dV_t \right) \mathbb{E} \left( \int_0^T Y_t \, dW_t \right) = 0.$$

$\endgroup$
1
  • $\begingroup$ @Stefano: You do need to impose the appropriate square-integrability condition to ensure that the expectation in question is defined. $\endgroup$ Mar 5, 2017 at 17:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .