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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?

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By:

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

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

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  • $\begingroup$ 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
    Commented Aug 8, 2020 at 17:32
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    $\begingroup$ $x\wedge y=\min(x,y)$, $x\vee y=\max(x,y)$. The expectation is obtained by noting that the covariance is null and using Itô’s Isometry. $\endgroup$
    – Daneel Olivaw
    Commented Aug 8, 2020 at 20:15

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