Given a Markov chain $X_n$ with it's states taking values in the set $S$. $f$ is a function, $f:S\rightarrow\mathbb{R}$. If a $f(X_n)$ is also a Markov chain, prove that either $f$ is injective or $f$ is constant. I'm totally clueless as to how I can proceed.
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asked Nov 6, 2018 at 16:37
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3$\begingroup$ Perhaps try to come up with an example with $S = \{1,2,3\}$ and $f$ is not injective and not constant. $\endgroup$– GEdgarCommented Nov 6, 2018 at 16:40
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2$\begingroup$ The idea I think they're getting at is that if $f(s_1) = f(s_2) = r$, then the transition probabilities out of $r$ depend on whether $r$ is "really" $s_1$ or $s_2$, and thus depend on the chain's history (which could be used to distinguish $s_1$ and $s_2$). Though it seems you could find counterexamples if $f$ "merges" states of the Markov chain with the same transition probabilities. $\endgroup$– Connor HarrisCommented Nov 6, 2018 at 16:47
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1 Answer
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The statement you're trying to prove is false.
For example, let $S =\{1,2,3\}$. Let $X_n$ be a Markov chain with transition matrix $$ \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]. $$ Let $f(x) = 1$ if $x=1$, and $2$ if $x \in \{2,3\}$. Then $f$ is neither injective nor constant, but $f(X_{n})$ is a Markov chain with transition matrix $$ \left[ \begin{array}{cc} 1 & 0\\ 0 & 1\\ \end{array} \right]. $$
The above example was not irreducible. Here's an irreducible example: $$ \left[ \begin{array}{ccc} 1/3 & 1/3 & 1/3 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{array} \right]. $$ We keep the same $f$. Then $f(X_{n})$ has transition matrix $$ \left[ \begin{array}{cc} 1/3 & 2/3\\ 1 & 0\\ \end{array} \right]. $$
Here's a more complicated example: $$ \left[ \begin{array}{ccc} 2/6 & 1/6 & 3/6 \\ 1/4 & 2/4 & 1/4 \\ 1/4 & 1/4 & 2/4 \end{array} \right]. $$ Then $f(X_{n})$ has transition matrix $$ \left[ \begin{array}{cc} 1/3 & 2/3\\ 1/4 & 3/4\\ \end{array} \right]. $$
As suggested in the second comment on the question, it's possible to merge states with the same outgoing probabilities, but in the last example, states 2 and 3 do not have the same outgoing probabilities. They only have the same outgoing probability to state 1.
