$\require{begingroup}\begingroup\newcommand{\dd}[1]{\,\mathrm{d}#1}$When studying diffusion processes, I often see the notion of expectation conditioned on an initial value. By this I mean the following:
Consider an Iito-diffusion, that is a Stochastic process solving the SDE: $$ \dd{ X(t) } = b( X(t) ) \dd{t} + \sigma(x(t)) \dd{ W(t) }\,,~ t>0 \,, \quad X(0) = x ~~\text{a.s.} $$ where $b,\sigma$ have certain regularity properties which are not relevant (to the question I have). It is later stated that Ito diffusions satisfy the Markov Property, that is:
$$ \mathbb{E}^{x} \left[ f(X(t+h)) ~\middle|~ \mathscr{F}_T^W \right] = \mathbb{E}^{X(t)} \left [ f(X(h))\right] $$ My biggest question is, how am I to interpret the term $\Bbb{E}^x$? I am told it is to be interpreted as $\mathbb{E}^x[ f(X(t))] = \mathbb{E}[f(X(t)) \mid X(0) = x]$. But I thought that it was fairly meaningless to condition on a probability $1$ event. Am I wrong about this? $\endgroup$
If $b$ and $\sigma$ are regular enough, then for all $x$, there exists a unique solution $X$ solving the SDE $dX(t)=b(X(t))\,dt+\sigma(X(t))\,dW(t)$ and $X(0)=x$.
Of course this solution depends on the initial condition $x$, so I would write for instance $X(x,t)$ instead of $X(t)$ the unique solution of the SDE which starts at $x$.
Then $\mathbb E^x[f(X(t+h))\vert\mathcal F^W_T]$ simply means $\mathbb E[f(X(x,t+h))\vert\mathcal F^W_T]$.
Moreover, let $\varphi(x)=\mathbb E^x[f(X(h)]=\mathbb E[f(X(x,h))]$. Then $\mathbb E^{X(t)}[f(X(h))]$ means $\varphi(X(t))$.