2
$\begingroup$

How to calculate the covariance between the integral of a Brownian motion at different times: $$\text{Cov}\left(\int^{t_1}_0\sigma(t)dW_t,\int^{t_2}_0\sigma(t)dW_t\right)\ ?$$ I know the answer is: $$\int^{t_1\wedge t_2}_0\sigma^2(t)dt.$$

If $\int^{s}_0\sigma(t)dW_t$ was a Brownian motion, then the above answer would be obvious, but unfortunately it's not. So how to calculate such covariance?

$\endgroup$
5
$\begingroup$

By:

  1. bilinearity of covariance,
  2. independence of Brownian increments, and
  3. Itô's isometry,

we obtain: $$\begin{align} & \text{Cov}\left(\int^{t_1}_0\sigma(t)dW_t,\int^{t_2}_0\sigma(t)dW_t\right) \\[6pt] & \qquad = \text{Cov}\left(\int^{t_1\wedge t_2}_0\sigma(t)dW_t,\int^{t_1\wedge t_2}_0\sigma(t)dW_t +\int^{t_1\vee t_2}_{t_1\wedge t_2}\sigma(t)dW_t\right) \\[6pt] & \qquad \overset{1}{=} V\left(\int^{t_1\wedge t_2}_0\sigma(t)dW_t\right)+\text{Cov}\left(\int^{t_1\wedge t_2}_0\sigma(t)dW_t,\int^{t_1\vee t_2}_{t_1\wedge t_2}\sigma(t)dW_t\right) \\[6pt] & \qquad \overset{2}{=} E\left(\left(\int^{t_1\wedge t_2}_0\sigma(t)dW_t\right)^2\right) \\[6pt] & \qquad \overset{3}{=} \int^{t_1\wedge t_2}_0\sigma^2(t)dt \end{align}$$

| improve this answer | |
$\endgroup$
  • $\begingroup$ What do the notations $t_1\wedge t_2 $ and $t_1 \vee t_2 $ mean? How can we get the expectation in 2nd-last line? $\endgroup$ – Bogaso Aug 8 at 17:32
  • $\begingroup$ $x\wedge y=\min(x,y)$, $x\vee y=\max(x,y)$. The expectation is obtained by noting that the covariance is null and using Itô’s Isometry. $\endgroup$ – Daneel Olivaw Aug 8 at 20:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.