Let $x$ be an $\mathbb{R}$-valued random variable, then for any bounded and continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$ one may write
$$
\mathbb{E}[f(X)] = \int\_{x \in \mathbb{R}} f(x)\pi(x)dx,
$$
where $\pi$ is the density of the law of $X$ *(granted that it exists)*. 


In general, if $f:C([0,T];\mathbb{R})$ is a continuous map to $\mathbb{R}$; then given an $\mathbb{R}$-valued process $X\_t$ one can view $f(X\_{\cdot})(\omega)$ as a random-variable (defined path-by-path). Is there way to write
$$
E[f(X\_{\cdot})] = \int\_{t=0}^T \int\_{x \in \mathbb{R}} f(\cdot) ...\pi\_t(x)dxdt?
$$
Or something similar (without integrating over the path-space directly)?



Writing path-dependent conditional expectation in terms of distribution
The answer is no. In general, to find $Ef(X\_\cdot)$, you need to know the distribution of the entire path $X\_\cdot\,$. So, as a minimum, you need to know the so-called finite-dimensional distributions of the process $X\_\cdot$, that is, the joint distributions of the random variables $X\_{t\_1},\dots,X\_{t\_k}$ for any $t\_1,\dots,t\_k$ in $[0,T]$ -- whereas the densities $\pi\_t$ determine only the individual distributions of the $X\_t$'s for $t\in[0,T]$. 


