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I have a question on find a counter example of Markov Chain. The question states as follows:

Suppose $X_0$, $X_1$ are the Markov chain whose state space is $\mathbb{Z}$, Since we know from Markov property, we know that $$ \mathbb{P}(X_n = i_n | X_1 = i_0, X_2 = i_1, ... X_{n-1} = i_{n-1}) = \mathbb{P}(X_n = i_n | X_{n-1} = i_{n-1}) $$ now we want to find an example such that the following property doesn't holds:

$$ \mathbb{P}(X_n \geq 0 | X_1 \geq 0, X_2 \geq 0, ... X_{n-1} \geq 0) = \mathbb{P}(X_n \geq 0 | X_{n-1} \geq0) $$

My thought is to start from a state space with only three states $1$, $2$ and $3$, and the equality holds when $X_2$ is not depends on $X_1$ and $X_2$ so the inequality will holds. But I don't think my example is good enough. Can you guys come up with a better one? Thanks!

Z-Harlpet
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    A Markov chain involves transition probabilities that depend on state alone. I don't understand what precise example you have in mind, largely because you have said nothing about the transition probabilities. – hardmath May 07 '18 at 21:19
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    If the state space $S$ is a subset of the nonnegative integers (such as your example $S={1, 2,3}$), then the information $X_i\geq 0$ in fact provides no information ("we already know $X_i\geq 0$"). So your state space should include at least one negative integer. I like the idea of using 3 states. – Michael May 07 '18 at 21:30
  • @hardmath I think we need to find a example such that the in which is transition probability fails in this equality – Z-Harlpet May 07 '18 at 22:50

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