Covariance of Brownian Stochastic Integrals at Different Times

Summary

The document derives the covariance between two stochastic integrals driven by the same Brownian motion but ending at different times. It splits the longer integral into the part shared with the shorter integral and the later Brownian increments. By independence of disjoint Brownian increments, the later part has zero covariance with the shared part. The covariance therefore reduces to the variance of the integral over the common time interval.

It then applies Itô’s isometry to express that variance as the integral of the squared deterministic integrand up to the earlier endpoint. The result relies on the integrals being well-defined and square-integrable, and on the shared Brownian driver and integrand specified in the question. The note gives a derivation rather than an empirical application; it does not address stochastic integrands or multiple correlated Brownian drivers, where additional conditions or covariance terms may be required.

Key ideas

• The covariance is bilinear, so the longer integral can be split into shared and later increments.
• Disjoint increments of the same Brownian motion are independent, making the later segment uncorrelated with the shared segment.
• The covariance equals the variance of the stochastic integral over the common interval.
• Itô’s isometry turns that variance into the integral of the squared integrand up to the earlier endpoint.

Tags

# How to calculate the covariance between two stochastic integrals?


# How to calculate the covariance between two stochastic integrals?












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?

## Answer by Daneel Olivaw (score 5, accepted)

https://quant.stackexchange.com/a/42484

By:

- bilinearity of covariance,

- independence of Brownian increments, and

- Itô's isometry,

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

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.