$W_{0}=x$ for a Brownian motion - what is meant by $E_{x}(.)$? That is, it it such that the expectation is taken with respect to a degenerate distribution, $W_{0}\text{~}DEGEN(x)$?
This comes up, for example, when dealing with the Feynman-Kac Theorem:
The solution to the heat equation is $E_{x}[e^{\int_{0}^{t}K(W_{s}ds}f(W_{t})], W_{t}$ is standard B.M.. Is it then the case that $E_{x}[e^{\int_{0}^{t}K(W_{s}ds}f(W_{t})]=\int[e^{\int_{0}^{t}K(W_{s}ds}f(W_{t})]d[P(W^{-1}_{0}(\omega))]$ where $W_{0}$ has a degenerate law?
That notation is just making explicit in the formula the fact that $W_0 = x$, i.e. that we are starting the Brownian motion at $x$.
If it's stated elsewhere that $W_0 = x$ then technically the subscript $x$ is unnecessary, but I still think it's a good idea to keep the dependence on $x$ explicit in the formula rather than mentioned implicitly elsewhere - especially if you're going to vary $x$.