# Backward representation of the general Markov process

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

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 $V$.