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Let $X_t=e^{-t}B_{e^{2t}}$ be an Ornstein Uhlenbeck process with $\mathbb{E}[X_tX_s]=e^{-|t-s|}$. Is $Y_t = X_t - \frac{1}{2} \int_o^{t}X_u du$ a Brownian motion? I managed to prove that $Y_t$ is a centered Gaussian process. The only thing I need to show is that it has $\mathbb{E}[Y_sY_t]=min(s,t)$, but I could not figure this out. Any helps would be appreciated.

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Hint: $EX_tX_s=e^{-t-s} (e^{2t}\wedge e^{2s})$. $EX_t\int_0^{s}X_u du=\int_0^{s} E(X_tX_u)du$ and $E\int_0^{t} X_u du \int_0^{s}X_vdv=\int_0^{t}\int_0^{s} E(X_uX_v)dudv$. All these are easy to compute.

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  • $\begingroup$ It should be $E\int_0^{t} X_u du \int_0^{s}X_vdv=\int_0^{t}\int_0^{s} E(X_uX_v)dudv$. This is still easy to compute but it does not end up with what I expect. $\endgroup$ – pac Sep 24 at 0:38

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