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Covariance of Integrated Brownian Motion Across Two Time Horizons

Article Quant Q&A · Author: user16692
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Summary

The document derives the covariance between two time integrals of standard Brownian motion, taken over horizons s and t. Because each integral has zero expectation, their covariance is the expected value of their product. Interchanging expectation and integration reduces the problem to a double integral of the Brownian covariance kernel, which equals the smaller of the two time arguments.

For unequal horizons, the integration region is split at the shorter horizon: the square up to that point and the remaining rectangular strip. Evaluating those pieces gives a symmetric covariance expression in terms of the smaller horizon and the difference between horizons. This is a concise derivation based on standard Brownian motion assumptions; it does not discuss applications, alternative stochastic processes, or empirical evidence. The printed case for s greater than t appears to contain a typographical inconsistency, so its formula should be checked against the symmetric general result.

Key ideas

  • The covariance of the two integrals is a double integral of the Brownian covariance kernel.
  • For standard Brownian motion, the covariance between values at times u and v is the smaller of u and v.
  • Splitting the integration region at the shorter horizon makes the double integral manageable.
  • The resulting covariance should be symmetric in s and t; one displayed case appears to have a typo.

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# How can I calculate $Cov\left(\int_{0}^{s}W_u\,du\,\,\,,\,\int_{0}^{t}W_v\,dv\right)$


# How can I calculate $Cov\left(\int_{0}^{s}W_u\,du\,\,\,,\,\int_{0}^{t}W_v\,dv\right)$












How can I calculate? \begin{align} Cov\left(\int_{0}^{s}W_u\,du\,\,\,,\,\int_{0}^{t}W_v\,dv\right) \end{align}

Thank you for your attention.

## Answer by user16651 (score 8, accepted)

https://quant.stackexchange.com/a/18541

You know that $E\left[\int_{0}^{s}W_udu\right]=E\left[\int_{0}^{t}W_vdv\right]=0$. By definition \begin{align} & Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=E\left[\int_{0}^{s}W_u\,du\int_{0}^{t}W_v\,dv\right]-0 \end{align} then \begin{align} & Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=\int_{0}^{s}\int_{0}^{t}E\,[W_uW_v]\,\,du\,dv \end{align} Since $E\,[W_uW_v]=min \{\,u\,,v \}$ therefor \begin{align} & Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=\int_{0}^{s}\int_{0}^{t}min \{\,u\,,v \}\,\,du\,dv \end{align} For the case $s<t$

\begin{align} \int_{0}^{s}\int_{0}^{t}min \{\,u\,,v \}\,\,du\,dv=\int_{0}^{s}\int_{0}^{s}min \{\,u\,,v \}\,\,du\,dv+\int_{0}^{s}\int_{s}^{t}min \{\,u\,,v \}\,\,du\,dv \end{align} we immediately have \begin{align} \int_{0}^{s}\int_{0}^{t}min \{\,u\,,v \}\,\,du\,dv=\frac{1}{3}s^3+\frac{1}{2}(t-s)s^2 \end{align} Following the same steps as described above, for the case $s > t$ we can also show \begin{align} \int_{0}^{s}\int_{0}^{t}min \{\,u\,,v \}\,\,du\,dv=\frac{1}{3}t^3+\frac{1}{2}(s-t)s^2 \end{align} Thus, \begin{align} Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=\frac{1}{3}min\{s^3\,\,,t^3\}+\frac{1}{2}|t-s|min\{s^2\,\,,t^2\} \end{align}

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