When is the reverse time stochastic process Markov?

Question

I'm reading the paper "Time Reversal of Diffusions" by Haussman and Pardoux. The authors begin with a diffusion process in $[0, 1]$ via the SDE $dX_t = b(t, X_t) dt + a(t, X_t) dw_t$ where $w$ is a Wiener process and $a, b$ are sufficiently nice determinisitic functions.

They define $\bar X_{t} = X_{1 - t}$ and in the first sentence of their main proof on page 1190 they say "... since it is already known that $\bar X_t$ is a Markov process..."

Edit: The comment below made me realize I was confusing Markov and Martingale properties. This was my original confused comment.

How do they know $\bar X_t$ is a Markov process? In fact, isn't the statement false under the filtration generated by $\bar X_t$? If we set $X_0 = 0$ deterministically, we know that this pins the terminal point of $\bar X$, in which case, $E[\bar X_1 | \bar X_t ] = 0$ for all $t \in [0, 1]$, violating the Markov condition.

Now with the correct definition of Markov property, let me update the question.

Let $\mathcal{F}_t$ be the filtration generated by process $X_t$. Similarly, let $\bar{ \mathcal{F}_t}$ be the filtration generated by $\bar X_t = X_{1 - t}$. The authors' claim seems to be, for $t_0 \le t$ that $E[\bar X_t | \bar {\mathcal{F}}_{t_0} ] = E[\bar X_t | \bar X_{t_0}]$. Does this follow from $X_t$ being Markov?

Edit 2: I'm seeing that the authors say in the very first sentence "It is well known that a Markov process remains a Markov process under a time reversal." I guess it's just always true, but I don't see it from the definition of Markov process.

For reference, here is the definition of Markov process that appears in Karatzas and Shreve.

Answer 1

This follows from the very definition:

$X$ is Markov if its future values and the past values are independent given the present value.

It is symmetric with respect to time reversal, so a time-reversed Markov process is Markov.

Having a fixed value in "the future" does not violate the Markov condition, as a constant is independent from anything. For example, Brownian bridge is a Markov process, moreover, a diffusion.