# Equivalence of Markov Property

Suppose we are given an $E$ valued stochastic process $(X_t)_{t\in T}$. The time set $T$ is in this context equal $[0,\infty)$. Then we define the canonical realization on the path space $E^T$ as the coordinate process $Y_t$, i.e. $Y_t(v):=v(t)$, where $v:T\to E$. An let $\mathcal{Y}_t^0$ be the raw filtration on the path space, i.e. $\mathcal{Y}_t^0:=\sigma (Y_s;s\le t )$. Furthermore let $\phi$ be the shift operator on $E^T$, i.e. $\phi_t(v)(s):=v(t+s)$.The distribution of $X$ under $P$ on the path space is denoted by $\mathbb{P}_\nu$, where $X_0$ is $\nu$ distributed (called the initial distribution).

I have proved the Markov property for the canonical process, i.e.:

for all $t\ge 0$ and an positive $\mathcal{Y}_\infty^0$ measurable random variable $U$ on $E^T$ we have $$\mathbb{E}_\nu[U\circ\phi_t|\mathcal{Y}_t^0]=\mathbb{E}_x[Z]|_{x=Y_t} \tag{1}$$

I know that $U\circ\phi_t$ is $\sigma (Y_s;s\ge t)$ measurable, the future sigma algebra after time $t$. In addition this is equivalent that $X$ is Markov and the RHS of $(1)$ is $\sigma (Y_t)$ measurable. Now there is a claim that we can also formulate $(1)$ on an abstract probability space $(\Omega,\mathcal{F},P)$, i.e. $$E[\tilde{V}_t|\mathcal{F}_t^0]=E[\tilde{V}_t|\sigma (X_t)]\tag{2}$$ where $\mathcal{F}_t^0$ is just the raw filtration of $X$ and $\tilde{V}\ge 0$ and measurable with respect to the future sigma algebra after time t, i.e. $\sigma (X_s;s\ge t)$. Why are $(1)$ and $(2)$ equivalent? I am not able to show this.

My thoughts so far: By definition the canonical process $Y$ has the same law under $\mathbb{P}_\nu$ as $X$ under $P$. It seems to be correct but a rigorous proof would be appreciated. In addition, is it true (and if why) that $$P[X_{t+h}\in A|\mathcal{F}_t^0]=\mathbb{P}_\nu[Y_{t+h}\in A|\mathcal{Y}_t^0]$$

hulik

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