How to find the Covariance of $Z(p)$ and $Z(q)$, where $$Z(p)=\int_0^p B(u)du+\int_p^1 \frac{B(u)}{1-u}du$$ and $B(u)$ is the standard Brownian Bridge on $[0,1]$. My approach is that $Z(p)=\int_0^p B(u)du+\int_p^1 \frac{B(u)}{1-u}du$ is a Gaussian process with mean $0$. To find the covariance, I proceed as \begin{eqnarray*} Cov(Z(p),Z(q))&=&\int_0^p\int_0^q E[B(u)B(v)]dv du+\int_0^p\int_q^1\frac{E[B(u)B(v)]}{1-u} dv du\\ &+&\int_p^1\int_0^q\frac{E[B(u)B(v)]}{1-u} dv du +\int_p^1\int_q^1\frac{E[B(u)B(v)]}{(1-u)^2} dv du\\ &=& \int_0^p\int_0^q [min\{u,v\}-uv]dv du+ \int_0^p\int_q^1\frac{[min\{u,v\}-uv]}{1-u} dv du\\&+& \int_p^1\int_0^q\frac{[min\{u,v\}-uv]}{1-u} dv du +\int_p^1\int_q^1\frac{[min\{u,v\}-uv]}{(1-u)^2} dv du. \end{eqnarray*}
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its covariance with what? – Mark Joshi Sep 12 '17 at 06:15
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if we consider the functional as Z(p), then i am trying to to find Cov(Z(p),Z(q)). thank you...@MarkJoshi – Dhrubasish Bhattacharyya Sep 12 '17 at 12:33
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Looks good, go ahead. – zhoraster Sep 12 '17 at 18:38
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I am looking forward if I am going in right way or [email protected] you – Dhrubasish Bhattacharyya Sep 13 '17 at 13:27