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$n > k > k_1 > \dots > k_r > 0$ are natural numbers, let $A_1, A_2, \dots, A_m$ be events such that $A_i = (X_{k_1} = a_{i1}, \dots X_{k_r} = a_{ir})$, where X_n is Markov chain and $\not\exists i, j: A_i = A_j$. $j_i$ are different numbers.

I already proved that $P(X_n = x | X_k = j \cap \bigcup_{i = 1}^m A_i) = P(X_n = x | X_k = j)$. I managed to prove $P(X_n = x | \bigcup_{i = 1}^g (X_k = j_i) \cap \bigcup_{i = 1}^m A_i) = \sum_{i = 1}^g (P(X_n = x | X_k = j_i)\cdot P(X_k = j_i, \bigcup_{i = l}^m A_l))\frac{1}{P(\bigcup_{i = 1}^g X_k = j_i \cap \bigcup_{l = 1}^m A_l)}$

but can't go any further.

Main idea I was using is $P(A|B) = P(B|A)\frac{P(A)}{P(B)}$

mokrota21
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  • Are $X_k=j_i$ distinct events for different $i$? – Diger Oct 06 '22 at 14:03
  • And do we also have $$P\left((X_k = j) \cap \bigcup_{i = 1}^m A_i\right)=P(X_k=j) , ?$$ – Diger Oct 06 '22 at 19:13
  • @Diger Yes, $X_k = j_i$ are different events (edited in post as well).

    Don't see why second is true though, as I know Markov chain only implies $P(X_n = x| X_k = j, X_{k_1} = a_1, \dots X_{k_m}) = P(X_n = x | X_k = j)$ where $n > k > k_1 > \dots > k_m$.

     – mokrota21 Oct 07 '22 at 14:29
  • Just wondering. Do we know anything about $$P\left(\bigcup_{i = 1}^m A_i \Bigg| \bigcup_{i=1}^g (X_k=j_i)\right) , ?$$For example if it is independent on the k-th step or on $g$? Similarly, is e.g. $$\frac{P\left((X_k = j_i) \cap \bigcup_{i = 1}^m A_i\right)}{P\left(\bigcup_{i=1}^g(X_k = j_i) \cap \bigcup_{i = 1}^m A_i\right)}$$ independent on $\bigcup_{i = 1}^m A_i$?. – Diger Oct 07 '22 at 20:29
  • Did you find a solution for the missing bit? – Diger Oct 09 '22 at 15:44

1 Answers1

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This is just an idea (not yet an answer though) with $g=2$. It should be straight-forward to generalize though.

Let $A,B_1,B_2$ be events and consider the addition law of probability with the definition of the conditional probability $$P\left( B_1\cup B_2 | A \right)P\left( A \right)=P\left( (B_1 \cup B_2) \cap A \right)=P\left( (B_1\cap A)\cup(B_2\cap A) \right) \\ =P\left( B_1\cap A \right)+P\left( B_2\cap A \right)-P\left( (B_1\cap B_2) \cap A \right) \\ =P\left( B_1|A \right)P(A)+P\left( B_2|A \right)P(A)-P\left( B_1\cap B_2|A \right)P(A)$$ and canceling $P(A)\neq 0$, we arrive at $$P\left(B_1 \cup B_2|A\right)=P\left( B_1|A \right)+P\left( B_2|A \right)-P\left( B_1\cap B_2|A \right) \, .$$

Now consider Bayes theorem $$P\left( (B_1\cup B_2)\cap A \right)=P\left( A|B_1\cup B_2 \right)P(B_1\cup B_2) =P\left( B_1\cup B_2|A \right)P(A)\\ =\left\{ P\left( B_1|A \right) + P\left( B_2|A \right) - P\left( B_1\cap B_2|A \right) \right\}P(A) \\ =P\left( A | B_1 \right)P(B_1) + P\left( A | B_2 \right)P(B_2) - P\left( A | B_1\cap B_2 \right)P(B_1\cap B_2)$$ and so $$P\left( A|B_1\cup B_2 \right)=\frac{P(B_1)}{P(B_1\cup B_2)} \, P(A|B_1) + \frac{P(B_2)}{P(B_1\cup B_2)} \, P(A|B_2) \\ - \frac{P(B_1 \cap B_2)}{P(B_1\cup B_2)} \, P(A|B_1\cap B_2)\,. \tag{1}$$ If you set $$A\equiv (X_n=x) \\ B_1\equiv(X_k=j_1)\cap \bigcup_{i=1}^m A_i \\ B_2\equiv(X_k=j_2)\cap \bigcup_{i=1}^m A_i$$ you can use your result $P(A|B_1)=P(A|X_k=j_1)$ already obtained (similarly for $B_2$). Since $X_k=j_1$ and $X_k=j_2$ can not be true at the same time (since $j_1$ and $j_2$ are distinct), the last term in (1) with $P(B_1\cap B_2)$ vanishes. It remains an idea for my above comment.

Diger
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