Exploiting the Markov property I've encountered the following problem when dealing with short-rate models in finance and applying the Feynman-Kac theorem to relate conditional expectations to PDEs.
Let $(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge 0},\mathbb{P})$ be a filtered probability space where the filtration is the (augmented) Brownian filtration generated by a one-dimensional $\mathbb{P}$-Brownian motion $W$. Assume that $r$ is a Markov process satisfying
$$dr(t)=\mu(t,r(t))\mathrm{d}t+\sigma(t,r(t))\mathrm{d}W(t),$$
where $\mu$ and $\sigma$ are deterministic functions. Let further $\Phi:\mathbb{R}\rightarrow\mathbb{R}$ be a deterministic, bounded function.
Is there a way to prove that the following equation holds?
$$\mathbb{E}\left[\left.\exp\left\{-\int_t^T r(s)\mathrm{d}s\right\}\cdot\Phi(r(T))\right|\mathcal{F}_t\right]=\mathbb{E}\left[\left.\exp\left\{-\int_t^T r(s)\mathrm{d}s\right\}\cdot\Phi(r(T))\right|r(t)\right]$$
So far, I attempted to exploit that $r$ is a Markov process and tried to employ some Monotone-Class arguments. Unfortunately, all my attempts failed. Does somebody have an idea? Thanks a lot in advance!
 A: 
Fact 1 If $Y$ is $\mathcal G$-measurable, and $X$ is independent of $\mathcal G$, then $\mathbb E[f(X,Y)\mid \mathcal G] = \mathbb E[f(X,y)]|_{y=Y}$ for any bounded measurable $f$.

Denote by $r_{t,x}(s), s\in [t,T]$ the solution to $$dr(s) = \mu(s,r(s))ds + \sigma(s,r(s))dW(s),\tag{1}$$ on $[t,T]$ with $r(t) = x$. Clearly, $r_{t,x} = F_t(x,\{W_u-W_t,u\in [t,T]\})$ with some function $F_t\colon \mathbb{R}\times C[t,T]\to C[t,T]$ (its joint measurability is not immediately obvious, but I omit this for the moment). The Markov property, which in essence follows from the strong uniqueness, says that the solution to $(1)$ on $[0,T]$ satisfies $$r(s) = r_{t,r(t)}(s)\tag{2}$$ for $s\ge t\ge 0$. 
Let $\Psi\colon C[t,T]\to \mathbb R$ be some bounded measurable function. Then, in view of $(2)$,
$$
\mathbb E\big[\Psi(r|_{[t,T]})\mid \mathcal F_t\big] = \mathbb E\big[\Psi(r_{t,r(t)})\mid \mathcal F_t\big] = \mathbb E\big[\Psi\big(F_t(r(t), \{W_u-W_t,u\in [t,T]\})\big)\mid \mathcal F_t\big]\\
= \mathbb E\big[\Psi\big(F_t(x, \{W_u-W_t,u\in [t,T]\})\big)\big]\big|_{x=r(t)},
$$
where we have used Fact 1 with $X = \{W_u-W_t,u\in [t,T]\}$, $Y = r(t)$, $\mathcal G = \mathcal F_t$. Similarly, using it with the same $X$, $Y$ and $\mathcal G = \sigma(r(t))$ gives
$$
\mathbb E\big[\Psi(r|_{[t,T]})\mid \mathcal r(t)\big] = \mathbb E\big[\Psi\big(F_t(x, \{W_u-W_t,u\in [t,T]\})\big)\big]\big|_{x=r(t)},
$$
whence 
$$
\mathbb E\big[\Psi(r|_{[t,T]})\mid \mathcal F_t\big] = \mathbb E\big[\Psi(r|_{[t,T]})\mid \mathcal r(t)\big],
$$
as required.
