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Let $(X_{t \in \mathbb{N_0}})$ be a Markov Chain with states $E$. Let $A,B \subseteq E$ with $x_0,x_1 \in E$.

Prove the following:

  1. $P (X_2 ∈ B|X_1 = x_1 , X_0 ∈ A) = P (X_2 ∈ B|X_1 = x_1 )$.
  2. $ P(X_2 ∈ B|X_1 ∈ A, X_0 = x_0 ) = P (X_2 ∈ B|X_1 ∈ A) $

I believe the $1^{st}$ is true but for the $2^{nd}$ I came with a counter example (hopefully right one).

$A=\{x_0,x_1\}, B=\{x_2\}$ enter image description here

$$P(X_2 \in \{x_2\} | X_1 \in \{x_0,x_1\}, X_0=x_0)=0$$ $$P(X_2 \in \{x_2\} | X_1 \in \{x_0,x_1\})=1/2$$

Oleg
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    You need to clarify the distribution at time zero in the second expression that you claim is 1/2. The distribution at time 2 depends on the way that X is distributed at time 1 inside of A. Just knowing it is supported in A isn't enough information. Specifying that the chain starts at x_0 is more information, because then we know at time 1 it is still at x_0. When you got 1/2, you were assuming I think that the chain is at x_1 at time 1. – Elle Najt Oct 20 '18 at 19:37
  • @Lorenzo Are you saying that I should make the 2nd expression more specific like : $$P(X_2 \in {x_2} | X_1 \in {x_0,x_1})=>P(X_2 \in {x_2} | X_1 \in {x_0,x_1}, X_0 \in {x_0, x_1}) ?$$ – Oleg Oct 20 '18 at 22:54
  • what I mean is that the second expression isn't well defined unless you specify a starting distribution. You are implicitly doing this in the first expression by writing that X_0 = x_0. The distribution on (X_0, X_1, ...) Is specified by two pieces of data: 1) the transition matrix (your weighted digraph 2) the initial distribution , i.e. the distribution of X_0. – Elle Najt Oct 20 '18 at 23:50
  • @Lorenzo, you mean $$P(X_2 \in {x_2} | X_1 \in {x_0,x_1})=P(X_2 \in {x_2} | X_1 = x_0, X_0=x_0) + $$ $$P(X_2 \in {x_2} | X_1 = x_1, X_0=x_1) $$ and every other possibility is $0$ – Oleg Oct 20 '18 at 23:56
  • That's not correct... I guess backing up, I think that the second problem to prove isn't well defined. P(X_2 in B | X1 in A) depends on the distribution at time 0, or the distribution at time 1 conditioned on being in A at time 1. Imagine X is the location of a random walker. In order to talk about the distribution of X_t, you need to make a choice about where the random walker starts , or what the distribution of it's starting position is. It's this non welldefinedness that is driving the confusion in your example. – Elle Najt Oct 21 '18 at 04:13
  • I think this is a duplicate of https://math.stackexchange.com/questions/1170628/markov-property-confusion . The main problem seems that conditioning on zero probability events makes things undefined, as I am understanding it ... – Thomas Oct 21 '18 at 07:11
  • I think as well that the initial distribution has to be identified to give sense to the conditionings as Lorenzo says. Nevertheless probably the paradox would remain I think even after doing that... – Thomas Oct 21 '18 at 08:29
  • @Lorenzo I can't assume a reasonable initial distribution on my own ? is that why the 2nd example is not solvable? – Oleg Oct 21 '18 at 09:00
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    @Oleg yes you can specify it on your own. But I'm trying to point out that the first expression specifies the starting distribution, and the second doesn't, and the choice you make for the starting distribution changes the outcome of the computation. – Elle Najt Oct 21 '18 at 16:32

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