Given the information about the correlation of two Brownian motions as $E[dW_1 dW_2] = \rho dt$ and knowing that $E[W_1(t)W_1(t')] = \min(t,t')$, I want to compute
$E[W_1(t)W_2(t')]$
I interpret $dW_1 as \int^{t + \Delta t}_t dW_1(s)$ and I also know that $E[(dW_1)^2] = dt$
Can I follow from the correlation property that $E[W_1(t)W_2(t')] = E[W_1(t) \rho W_1(t')] = \rho \min(t,t')$ ?
Thank you and best regards !
Best Answer
You would say: $$\begin{eqnarray} \mathsf{E}\left(W_1(t_1) W_2(t_2)\right) &=& \mathsf{E}\left(\int_0^{t_1} \mathrm{d}W_1(s_1) \int_0^{t_2}\mathrm{d}W_2(s_2)\right) \stackrel{\text{Ito isometry}}{=} \mathsf{E}\left(\int_0^{\min(t_1,t_2)} \rho\, \mathrm{d}t \right) \\ &=& \int_0^{\min(t_1,t_2)} \rho\, \mathrm{d}t = \rho \min\left(t_1, t_2\right) \end{eqnarray} $$