deriving the joint distribution of $(B_s,B_t,B_u)$ of brownian motions using conditional expectations

Question

I am reading Rene Schilling's Brownian Motion and there is a part I have difficulty figuring out.

The goal is to find the joint distribution of three Brownian motions at different times ($0\le s<t<u$) using the conditional expectation rules, where $\phi,\psi,\chi:\mathbb{R}^d \to \mathbb{R}$ are bounded measurable. We want to show that $$E[\phi(B_s)\psi(B_t)\chi(B_u)]=E[\phi(B_s)E^{B_s}(\psi(B_{t-s})E^{B_{t-s}}\chi(B_{u-t}))],$$ and write the expectations as integrals.

Initially, we condition on $\mathscr{F}_t$ then we have $$E[\phi(B_s)\psi(B_t)\chi(B_u)]=E[\phi(B_s)\psi(B_t)E^{B_t}\chi(B_{u-t})]$$ from the Markov property.

But I don't know how we get the next step which is $E[\phi(B_s)\psi(B_t)E^{B_t}\chi(B_{u-t})]=E[\phi(B_s)E^{B_s}(\psi(B_{t-s})E^{B_{t-s}}\chi(B_{u-t}))]?$

Finally why do we get $E^{B_s}(\psi(B_{t-s})E^{B_{t-s}}\chi(B_{u-t}))=\int \psi(a)\int \chi(b)P^a(B_{u-t}\in db)P^{B_s}(B_{t-s}\in da)$ when writing the expectations as integrals?

I would greatly appreciate some help as I have been struggling on getting these for a while.

Answer 1

\begin{align*} \mathbb{E}[\phi(B_s) \psi(B_t) \chi(B_u)] & = \mathbb{E}[\mathbb{E}[\phi(B_s) \psi(B_t) \chi(B_u)| \mathcal{F}_s]] \\ & = \mathbb{E}[\phi(B_s) \mathbb{E}[\psi(B_t) \chi(B_u)| \mathcal{F}_s]] \\ & = \mathbb{E}[\phi(B_s) \mathbb{E}[\psi(B_{t-s}) \chi(B_{u-s})]] \\ & = \mathbb{E}[\phi(B_s) \mathbb{E}[\mathbb{E}[\psi(B_{t-s}) \chi(B_{u-s})|\mathcal{F}_{t-s}]]] \\ & = \mathbb{E}[\phi(B_s) \mathbb{E}[\psi(B_{t-s})\mathbb{E}[ \chi(B_{u-s})|\mathcal{F}_{t-s}]]] \\ & = \mathbb{E}[\phi(B_s) \mathbb{E}[\psi(B_{t-s})\mathbb{E}[ \chi(B_{u-t})]]] \\ \end{align*}