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I have come across the following question in my lecture notes. I don't feel that comfortable with filtrations and showing measureability.

We consider the integral $X_t=\int_0^tB_sds$, where $Bt:t\geq 0$ is standard brownian motion.

We want to show that $X_t$ is not a markov process w.r.t to it's own filtration, but that the pair $\{(B_t,X_t),t\geq 0\}$ is a markov process w.r.t it's own filtration. Now I have seen how to show the markov property for brownian motion on it's own, by using the moment generating function, but is there some general way of going about this kind of problem?

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For $0<s<t$ you can write $X_t$ as $X_s+(t-s)B_s+\int_s^t(B_u-B_s)\,du$. The integral on the right is independent of $\mathcal F_s:=\sigma\{B_u:0\le u\le s\}$ while the first two terms on the right are $\mathcal F_s$-measurable. This should convince you that the conditional distribution of $X_t$, given $\mathcal F_s$ depends only on $X_s$ and $B_s$.

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