MHB "Two step" Markov chain is also a Markov chain.

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A Markov chain on a compact metric space X is defined as a measurable map P that assigns Borel probability measures to points in X. The discussion explores the construction of a new measure ν based on an initial probability measure μ and the Markov chain P, leading to the probability of landing in a set E after two steps. The question posed is whether the composed map P², derived from P and the measure transformation, is also a Markov chain and Borel measurable. There is speculation that the continuity of the measure transformation could affirm this. Technical issues regarding post readability were also mentioned, indicating formatting problems in the discussion.
caffeinemachine
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Let $X$ be a compact metric space and $\mathcal X$ be its Borel $\sigma$-algebra. Let $\mathscr P(X)$ be the set of all the Borel probability measures on $X$. A **Markov chain** on $X$ is a measurable map $P:X\to \mathscr P(X)$. We write the image of $x$ under $P$ as $P_x$. (Here $\mathscr P(X)$ is quipped with the Borel $\sigma$-algebra coming from the weak* topology).

Intuitively, for $E\in \mathcal X$, we think of $P_x(E)$ as the probability of landing inside $E$ in the next step given that we are sitting at $x$ at the current instant.

Now let $\mu$ be a probability measure on $X$. We define a new measure $\nu$ on $X$ as follows:

$$\nu(E) = \int_X P_x(E)\ d\mu(x)$$

Intuitively, suppose in the firt step we land in $X$ according to $\mu$. The probability of landing on $x\in X$ according to the measure $\mu$ is $d\mu(x)$.
Now let the Markov chain $P$ drive us. Then the probability of landing in $E$ in the next step given that we are at $x$ is $P_x(E)$.
So the probability of landing in $E$ in two steps is $\int_X P_x(E)\ d\mu(x)$.

So we have a map $\mathscr P(X)\to \mathscr P(X)$ which takes a probability measure $\mu$ and produces a new measure $\nu$ as defined above.

Composing $P:X\to \mathscr P(X)$ with $\mathscr P(X)\to \mathscr P(X)$ we again get a map $X\to \mathscr P(X)$, which I will denote by $P^2$.

Question. Is $P^2$ aslo a Markov chain, that is, is $P^2$ aslo Borel measurable?

My guess is that the map $\mathscr P(X)\to \mathscr P(X)$ is actually continuous, which would answer the question in the affirmative. But I am not sure.
 
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Congratulations. You have managed to color your post unreadable.
 
Walagaster said:
Congratulations. You have managed to color your post unreadable.
I am not sure what is wrong. It's appearing on my computer just fine.

Perhaps you are viewing it on another device?
 
caffeinemachine said:
I am not sure what is wrong. It's appearing on my computer just fine.

When I quote-reply your original post, I find that each paragraph is inside a set of color-tags with color code #cbd2ac. When I remove those tags, the text becomes readable.
 
caffeinemachine said:
I am not sure what is wrong. It's appearing on my computer just fine.
Screenshot shows what I'm getting.

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