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I think I have some problem understanding markov chains, because we defined them as abstract objects but our professor does proofs with them as if they where just elementary conditional probabilities.

This is our definition of a markov chain: Given prob. space $(\Omega, \mathcal{A}, \mathbb{P})$, standard borel space $(S, \mathcal{S})$ and a sequence of random variables $X_n: \Omega \to \mathcal{S}, n \in \mathbb{N}$. $(X_n)_{n\in \mathbb{N}}$ is called markov chain if $$\forall B \in \mathcal{S}: \mathbb{E}[\mathbb{1}_B(X_n)|\sigma(X_0, ..., X_{n-1})] = \mathbb{E}[\mathbb{1}_B(X_n) |\sigma(X_{n-1})]$$


So far, so good. But now we've got the following preposition:

Given $(\xi_i)_{i\in \mathbb{N}}$ iid rv on $\mathbb{R}^d$ and random variable $X_0$ independent of $(\xi_i)_{i \in \mathbb{N}}$ (also on $\mathbb{R}^d$) we define $X_n := X_0 + \sum_{i=1}^n \xi_i$. Then $(X_n)_{n \in \mathbb{N}_0}$ is a markov chain.

Proof: $$P(X_n \in B | X_0 = x_0, ..., X_{n-1} = x_{n-1}) = $$

$$= P(\xi_n + x_{n-1} \in B | X_0 = x_0, X_0 + \xi_1 = x_1, ..., X_0 + \sum_{i=1}^{n-1}\xi_i = x_{n-1}) = $$

$$=_{independce} P(\xi_n + x_{n-1} \in B) = ... = P(X_n \in B | X_{n-1} = x_{n-1})$$ Why do we start treating these conditional expectations just like elementary conditional probability for events?

Sorry for the awful formatting of the proof

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  • $\begingroup$ What do you mean by "treating conditional expectations just like conditional probability for events"? What's written up there in the definition are conditional probabilities for events. $\endgroup$ – saz May 3 '15 at 17:48
  • $\begingroup$ @saz can you elaborate on that? the definition I gave is about conditional expectations as I understand and wrote. you get the conditional expectation through the radon-nikodym theorem (so in a quite "abstract way"). though the conditional probabilities for events for events are a similar concept, they have not much to do with the definition of the conditional expectation. by the "conditional prob. for events" I mean the cond. expect. defined by formula: $P(A|B) = \frac{P(A\cap B)}{P(B)}$. Please correct me anywhere possible $\endgroup$ – fubal May 3 '15 at 17:52

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