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I just would like to evaluate $P(X_{3}=j,X_{1}=i|X_{2}=k)$

So I tried:

$P(X_{3}=j,X_{1}=i|X_{2}=k) =P(X_{3}=j|X_{2}=k)P(X_{1}=i|X_{2}=k)$

The trouble is evaluating $P(X_{1}=i|X_{2}=k)$

I tried $P(X_{1}=i|X_{2}=k)=P(X_{1}=i)$ since I think the past doesnt depend from future and I applied total probability theorem so:

$P(X_1=i)=\sum_{l \in S}P(X_1=i|X_0=l)$ where $S$ is the state space.

But I'm not sure if this independence holds if not what should I do?

PS: Assume transition matrix is known so $\sum_{l \in S}P(X_1=i|X_0=l)$ is known.

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What happened in the future variables clues what had happened in the past, so the assumption $p(x_1|x_2)=p(x_1)$ is not correct. You should use Bayes theorem to calculate it: $$p(x_1|x_2)=\frac{p(x_2|x_1)p(x_1)}{p(x_2)}$$

You can calculate $p(x_i)$ from total probability theorem, or alternatively do the same calculations using some matrix multiplications as described in this notes, page 10.

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  • $\begingroup$ I voted up and go it as the answer as bayes theorem will solve the problem regardless indepence holds or not but It doesnt make sense, ok!, a finite markov chain will reset but index time not so $X_2$ is the first second transition ever, I`m looking over if the results will be the same. $\endgroup$
    – Davi Américo
    Nov 20, 2021 at 22:19
  • $\begingroup$ I looked over and the results are not the same, how weird. $\endgroup$
    – Davi Américo
    Nov 20, 2021 at 22:41
  • $\begingroup$ @DaviAmérico I apologize for the typo, I corrected it. But, I couldn't follow your objection in your first message. Can you clarify a bit more about what feels wrong? $\endgroup$
    – gunes
    Nov 20, 2021 at 22:44
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    $\begingroup$ I misread What happened in the future variables clues what had happened in the past, eng is not my main language, now past depending from future makes a little bit sense but thats another discussion thank you. $\endgroup$
    – Davi Américo
    Nov 20, 2021 at 22:49
  • $\begingroup$ There is no way to get through this by total probability I was using the wrong theorem, in right way we would have $P(x_1)=\sum_{i \in S}P(x_1|x_0=i)P(x_0=i)$ how would I get $P(X_0=i)$? $\endgroup$
    – Davi Américo
    Nov 20, 2021 at 23:22

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