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Let $B$ be a Brownian motion. Fix times $0 < r < t$. Write $\mathcal{D}$ for the space of paths traced out by continuous maps from $[0,t]$ to $\mathbb{R}$ with a suitable (e.g. Skorokhod) topology. Let $F : \mathcal{D} \times \mathbb{R} \rightarrow \mathbb{R}_+$ be a measurable map.
Now I define a random variable in two ways. First, define $f : \mathbb{R} \rightarrow \mathbb{R}_+$ by $$ f(x) := E\bigg[F \big((B_s)_{s \in [0,t]}, x\big)\bigg] . $$
Then let $X := f(B_r)$.

Next I define a random variable $Y$ in the following way (letting $\mathcal{F}_s$ stand for the sigma-algebra generated by $B_s$): $$ Y := E \bigg[ F\big((B_{s+r}-B_r)_{s \in [0,t]}, B_r\big) \bigg\vert \mathcal{F}_r \bigg] . $$

Question: do we have $X = Y$ almost surely?

Many thanks for your help.

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Yes, this is true. Usually, I would say that since $(B_{s+r}-B_r)_s$ is a Brownian motion independent of $\mathcal F_r$, one can deduce the result immediately from the Markov property. Since you were asking for more detail in your other question, I will provide more here.

To start with, define $\mathcal D_s$ to be $\mathcal D$ when $t$ is replaced by $s$. Since $X$ is $\mathcal F_r$-measurable, by definition we will have $X=Y$ almost surely if and only if for every Borel set $A\subseteq\mathcal D_r$, we have $$E\Big[X\mathbf1_{\{(B_s)_{s\in[0,r]}\in A\}}\Big]=E\bigg[F((B_{s+r}-B_r)_{s\in[0,t]},B_r)\mathbf1_{\{(B_s)_{s\in[0,r]}\in A\}}\bigg].$$

Let $\mu_s$ be the law of $(B_u)_{u\in[0,s]}$, which is a measure on $\mathcal D_s$. By definition, $$f(x)=\int_{\mathcal D_t} F(\gamma,x)\mu_t(d\gamma)$$ and so $$X=\int_{\mathcal D_t}F(\gamma,B_r)\mu_t(d\gamma).$$ By the independent increments of Brownian motion, the pair $((B_{s+r}-B_r)_{s\in[0,t]},(B_s)_{s\in[0,r]})$ has law $\mu_t\otimes \mu_r$. Thus, we deduce $$E\bigg[F((B_{s+r}-B_r)_{s\in[0,t]},B_r)\mathbf1_{\{(B_s)_{s\in[0,r]}\in A\}}\bigg]=\int_{A}\int_{\mathcal D_t}F(\gamma,\phi(r))\mu_t(d\gamma)\mu_r(d\phi) =E\Big[X\mathbf1_{\{(B_s)_{s\in[0,r]}\in A\}}\Big],$$ which concludes the proof. Note this also verifies the Markov property holds in this case, since $X$ is $\sigma(B_r)$-measurable and clearly $\sigma(B_r)\subset\mathcal F_r$.

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