caffeinemachine
Gold Member
MHB
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- 15
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.
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.
