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One usually deals with a discrete time Markov process in the following form: given a state space $E$ the Markov process is defined by transition kernel $T(B|x)$ such that $$ \mathsf{P}(X_1\in B|X_0 = x) = T(B|x) $$ for all $x\in E$, $B\in\mathcal{B}(E)$.

So, given a current state we have a distribution of the future state.

On the other hand it can be an interesting problem given a current state to find a distribution of the previous state which "fits" with a transition kernel $T$.

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It's only for the continuous time. – Ilya Apr 18 '11 at 12:05
As an aside note that your sentence so given a state $X_{n+1}$ we know the distribution of $X_n$ is misleading because in the formula just above, $v_n$ is not independent on $X_{n+1}$. – Did Apr 18 '11 at 14:09
Yes, you're right. Thank you for this comment on independence. – Ilya Apr 18 '11 at 14:48
So? You might care to correct this paragraph in your post... – Did Apr 25 '11 at 20:56
up vote 2 down vote accepted

This is called time-reversal of Markov chains and is presented in numerous lecture notes available on the web, see here for example.

Remark As said in the comments, the sentence of the post stating that given a state $X_{n+1}$ we know the distribution of $X_n$ signals a deep misunderstanding of the structure of Markov processes. In the example considered, $v_n$ is not independent on $X_{n+1}$ and one should note that, given any nonzero $a$ and $b$ and any random $X$, any random variable $Y$ can be written as $Y=X/a-(b/a)V$ for a well chosen random variable $V$.

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