Equality in Conditional Distribution for a Stationary Process Suppose I have a stationary process $X_t$, defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Then we know that for a finite index set $t_1,t_2,...,t_n$ and a shifted index set $t_1+s,t_2+s,...,t_n+s$ for $s>0$ that the joint distribution of the function indexed by the first list is equal to the joint distribution of the shifted index list. Can I conclude that the conditional distributions $X_{t+s}|X_s=x$ for $t>s$  viewed as a function of $s$ are equal (ie. $(X_{t+s}|X_s=x) \stackrel{d}{=}(X_{t+s}|X_s=x$))?
I wanted to try and show this via the definition of conditional expectation but I'm not sure how to go...For $A \in \mathcal{F},$
$$\mathbb{E}[\mathbb{E}[X_{t+s}|X_s]\mathbb{1}_A]=\mathbb{E}[X_{t+s}\mathbb{1}_{A}]=\mathbb{E}[X_{t+s'}\mathbb{1}_{A}]=\mathbb{E}[\mathbb{E}[X_{t+s'}|X_{s'}]\mathbb{1}_A]$$ where the middle equality is because of stationarity.
So $\mathbb{E}[X_{t+s}|X_s]=\mathbb{E}[X_{t+s'}|X_{s'}]$ a.s., but what now, assuming everything I have done is all fine? Maybe there is a different approach to this, but it intuitively seems to be true.
 A: If $X_t$ is stationary then for any measurable function $g:\mathbb{R}\rightarrow\mathbb{R}$ and any fixed time $s>0$, $g(X_s)$ has the same distribution as $g(X_0)$. 
Fix $t > 0$. Let $B$ be a measurable subset of $\mathbb{R}$ that is of interest.  Let $g(X_0)$ be any version of the conditional expectation of interest: 
$$ g(X_0) = E[1_{\{X_{t} \in B\}} | X_0]$$
Now fix a new starting time $s>0$. we want to show $g(X_s)$ is a version of the conditional expectation: 
$$ g(X_s) = E[1_{\{X_{s+t} \in B\}} | X_s]$$
This will show that the same function $g(\cdot)$ can be used to define the conditional expectation, regardless of the starting time $s$.
Proof: We know $g(X_0)$ is a measurable function of $X_0$. So $g(X_s)$ is a measurable function of $X_s$. Now for any measurable set $A \subseteq \mathbb{R}$ we have: 
\begin{align}
\int_{X_s \in A} 1_{\{X_{s+t} \in B\}} dP &\overset{(a)}{=}\int_{X_0 \in A} 1_{\{X_{t} \in B\}}dP \\
&\overset{(b)}{=}\int_{X_0 \in A} g(X_0) dP \\
&\overset{(c)}{=}\int_{X_s \in A} g(X_s) dP
\end{align}
where (a) and (c) hold by stationarity; (b) holds because 
$g(X_0)$ is a version of the conditional expectation $E[1_{\{X_{t} \in B\}} | X_0]$.  So $g(X_s)$ is a version of the conditional expectation $E[1_{\{X_{s+t} \in B\}} | X_s]$. $\Box$
