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In our reading we had the following Theorem concerning Markov chains:

Take $0<n<N$ and $(X_n)_{n\in\mathbb{N}}$ a Markov chain. Then for all $a_n\in E$ (where $E$ is the state space) and all subsets $G\subseteq E^n, F\subseteq E^{N-n}$ we have that $$ \mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a_n,(X_{n-1},...,X_0)\in G)\\=\mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a). $$

Now the following question:

Does the statement of the theorem remain true if $\left\{X_n=a\right\}$ is replaced by $\left\{X_n\in A\right\}$ for some arbitrary $A\subseteq E$?

I tried to prove that in the following way using the theorem and the $\sigma$-additicity of $\mathbb{P}$ and $\left\{X_n\in A\right\}=\biguplus_{a_n\in A}(X_n=a_n)$:

$$ \mathbb{P}((X_{n+1},...,X_N)\in F|X_n\in A,(X_{n-1},...,X_0)\in G)\\=\mathbb{P}((X_{n+1},...,X_N)\in F|\biguplus_{a_n\in A}(X_n=a_n)\cap(X_{n-1},...,X_0)\in G)\\ =\mathbb{P}((X_{n+1},...,X_N)\in F|\biguplus_{a_n\in A}[(X_n=a_n)\cap (X_{n-1},...,X_0)\in G])\\=\frac{\mathbb{P}[(X_{n+1},...,X_N)\in F]\cap\biguplus_{a_n\in A}[(X_n=a)\cap (X_{n-1},...,X_0)\in G])}{\mathbb{P}(\biguplus_{a_n\in A}[(X_n=a_n)\cap (X_{n-1},...,X_0)\in G])}\\=\frac{\sum_{a_n\in A}\mathbb{P}([(X_{n+1},...,X_N)\in F]\cap [ (X_n=a_n)\cap (X_{n-1},...,X_0)\in G])}{\mathbb{P}(\biguplus_{a_n\in A}[(X_n=a_n)\cap (X_{n-1},...,X_0)\in G])}\\=\frac{\sum_{a_n\in A}\mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a_n,(X_{n-1},...,X_0)\in G)\cdot\mathbb{P}(X_n=a_n,(X_{n-1},...,X_0)\in G)}{\mathbb{P}(\biguplus_{a_n\in A}[(X_n=a_n)\cap (X_{n-1},...,X_0)\in G])} $$ Now using the theorem I can write the summands as $$ \mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a_n,(X_{n-1},...,X_0)\in G)\cdot\mathbb{P}(X_n=a,(X_{n-1},...,X_0)\in G) =\mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a_n)\mathbb{P}(X_n=a,(X_{n-1},...,X_0)\in G), $$

so I now have $$ =\frac{\sum_{a_n\in A}\mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a_n)\cdot \mathbb{P}(X_n=a_n, (X_{n-1},...,X_0)\in G)}{\mathbb{P}(\biguplus_{a_n\in A}[(X_n=a_n)\cap (X_{n-1},...,X_0)\in G])} $$

Do not know how to continue...

Maybe I am not right up to here or maybe the statement does not remain true..

mathfemi
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1 Answers1

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Take $0<n<N$ and $(X_n)_{n\in\mathbb{N}}$ a Markov chain. Then for all $a_n\in E$ (where $E$ is the state space) and all subsets $G\subseteq E^n, F\subseteq E^{N-n}$ we have that $$ \mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a_n,(X_{n-1},...,X_0)\in G)\\=\mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a). $$ Does the statement of the theorem remain true if $\{X_n = a\}$ is replaced by $\{X_n \in A\}$ for $A \subset E$?

This is, perhaps surprisingly, not true, even in trivial cases.

Consider the Markov chain on $\{0,1\}$ with transition kernel given by $$P(X_{n+1} = 1 \;|\ X_n = 0) = 1, \quad P(X_{n+1} = 0 \;|\ X_n = 1) = 1.$$ Let $P(X_0 = 0) = P(X_0 = 1) = 1/2$. Let $A = \{0, 1\}$. Then \begin{gather*} P(X_2 \in \{0\} \;|\ X_1 \in A, X_0 \in \{0\}) = 1, \end{gather*} since $X_0 = 0$ implies $X_1 = 1$ and $X_2 = 0$. On the other hand, \begin{gather*} P(X_2 \in \{0\} \;|\ X_1 \in A) < 1. \end{gather*}

snar
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  • What are $n$ and $N$ in your counter example? – mathfemi Oct 24 '14 at 20:09
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    $N = 2, n = 1$, $F = G = {0} \subset {0,1}$. If I've understood your question correctly, I don't think the counter-example is hard to interpret. – snar Oct 24 '14 at 22:10
  • Could you please explain me why the second probability is <1? I cannot see it. – mathfemi Oct 25 '14 at 13:05
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    It's actually $1/2$; I wrote $1$ because I thought it would be easier to see. Note that since $A = E$, the whole state space, $P(X_1 \in A) = 1$. Therefore, $P(X_2 = 0 | X_1 \in A) = \frac{P(X_2 = 0, X_1 \in A)}{P(X_1 \in A)} = P(X_2 = 0, X_1 \in A) = P(X_2 = 0)$. This last expression can be expanded as $P(X_2 = 0) = \sum_{i=0}^1 P(X_2 = 0 | X_0 = i) P(X_0 = i) = 1\cdot \frac{1}{2} + 0\cdot \frac{1}{2} = 1/2$. – snar Oct 25 '14 at 16:38
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    Informally, you can think of it as, "if $X_1 \in A$, there is a non-zero probability that $X_1 = 0$, in which case $X_2 = 0$ is impossible," hence $P(X_2 = 0 ;|\ X_1 \in A) < 1$. – snar Oct 25 '14 at 20:03
  • Why is $P(X_2=0,X_1\in A)=P(X_2=0)$? – mathfemi Oct 26 '14 at 13:47
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    Because if $A \subset B$, then $P(A \cap B) = P(A)$. Observe that ${X_2 = 0} \subset {X_1 \in A}$, because ${X_1 \in A}$ doesn't imply anything at all. It just says "$X_1$ will land in the state space" and that always happens. – snar Oct 26 '14 at 14:06
  • Now I think I got it. Thank you very much for your patience. – mathfemi Oct 26 '14 at 14:13