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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.

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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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  • 1
    $\begingroup$ Thanks for the details. Can you explain why is $\int_{0}^{s}\int_{0}^{t}E\,[W_uW_v]\,\,du\,dv=E\left[\int_{0}^{s}W_u\,du\int_{0}^{t}W_v\,dv\right]$? $\endgroup$ – user1559897 Sep 5 '16 at 2:39
  • $\begingroup$ Fubini's theorem $\endgroup$ – user16651 Sep 5 '16 at 15:27

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