Could anyone help me to understand Remark 6.2 on page 190 by some example?
http://janroman.dhis.org/finance/Books%20Notes%20Thesises%20etc/Probability%20Theory/PROB.ch6.pdf
One way to generate markov processes with multiple invariant measures is to start with two markov processes with transition probabilities $ \pi_1$, $ \pi_2$,
$$\pi(x,A)= \begin{cases} \pi_1(x,A\cap X_1) & ;x\in X_1,\\ \pi_2(x,A\cap X_2) & ;x\in X_2. \\ \end{cases}$$
$X=X_1\cup X_2$, A is some Borel set in X, So what is $X_1, X_2$ anyway? Can $X_1=X_2=(0,1)$
Why the new chain $\pi(x, A)$ has several invariant measures, I didn't understand, could anyone be of little more explicit with some example, please? Thanks.
Can I say from this example that if I switch randomly two Markov chains with a unique invariant probability measure, the resulting chain may not possess a unique invariant measure?
$X_1$ and $X_2$ are the state spaces of the two Markov processes we're starting with; they should be disjoint for this example to work.
Let's take a discrete example: $X_1 = \{1,2\}$ and $X_2 = \{3,4\}$ where $\pi_1(1,1) = \pi_1(1,2) = \pi_1(2,1) = \pi_1(2,2) = \frac12$ and $\pi_2(3,3) = \pi_2(3,4) = \pi_2(4,3) = \pi_2(4,4) = \frac12$.
Then $\pi(x,y)$ is equal to $\pi_1(x,y)$ if $x,y \in X_1$, it is equal to $\pi_2(x,y)$ if $x,y \in X_2$, and is equal to $0$ otherwise. If we start in $X_1$, the combined random process is going to stay in $X_1$; if we start in $X_2$, it will stay in $X_2$.
The two processes have invariant measures $\mu_1(1) = \mu_1(2) = \frac12$ and $\mu_2(3) = \mu_2(4) = \frac12$, which we can extend to measures on $X_1 \cup X_2$ by setting $\mu_1(3) = \mu_1(4) = 0$ and $\mu_2(1) = \mu_2(2) = 0$. These are still invariant measures in the combined process; in general, if we have $\mu(1) = \mu(2) = a$ and $\mu(3) = \mu(4) = \frac12 - a$, then $\mu$ is an invariant measure.