1
$\begingroup$

Before asking my question I must first explain the context in which it is formulated.

Let $M\subset \mathbb{R}$, if $M$ is not countable then we have $M$ with Borel $\sigma$-algebra denoted by $\mathcal{B}$. If $M$ is countable then we have $M$ with discrete $\sigma$-algebra (i.e. iduced by $\mathcal{P}(M)$).

Definition: A random recurrent sequence over a set $M$ is a sequence of random variables $(X_{n})$ with values in $M$ (i.e, $X_{n}(\omega)\in M$) defined over a probability space $\left(\Omega,\mathcal{F}, \mathsf{P} \right)$ such that $(X_{n})$ is solution of reccurent equation of the form \begin{equation} \tag1 X_{n+1}(\omega)=f(\theta_{n+1}(\omega),X_{n}(\omega)) \qquad \forall \omega \in \Omega \end{equation} where

  1. $(\theta_{n})$ are random variable iid with values in a measurable space $(\Theta, \mathcal{A})$ (i.e, $\theta_{n}: \Omega \rightarrow \Theta$ measurable)
  2. $\begin{array}{rcl}f: \Theta \times M &\mapsto & M \\ (\theta, x) & \rightarrow & f(\theta,x)=f_{\theta}(x) \end{array}$ is measurable (respect to $\sigma$-algebra $\mathcal{A}\otimes \mathcal{B}$).
  3. $X_{0}$ (the initial condition) is a random variable independent of $(\theta_{n})_{n\in\mathbb{N}}$.

The above definition allows to define a random dynamic system:

Definition: Let $m$ measure in $\Theta$ induced by the distribution of $(\theta_{n})_{n\in\mathbb{N}}$, the pair $(f,m)$ is called random dynamical system (RDS).

Let us suppose that $M$ is countable.

Definition: A transition matrix over $M$ is an aplication $\Pi : M\times M\rightarrow [0,1]$ such that $$\sum_{y\in M} \Pi(x,y) =1 \qquad \forall x\in M.$$

Theorem: Let $(f,m)$ a random dynamical system over $M$. Then $$\Pi(x,y):=m\left( \left\{ \theta \in \Theta \:\: |\:\: f_{\theta}(x)=y\right\} \right)$$ is a transition matrix.

My question: Note that the sets $\left(\left\{ \theta \in \Theta \:\: |\:\: f_{\theta}(x)=y\right\} \right)_{y\in M}$ are disjoint (because $f$ is well-defined). Then $$\Theta= \bigcup_{y\in M} \left\{ \theta \in \Theta \:\: |\:\: f_{\theta}(x)=y\right\}. $$ Therefore $$\sum_{y\in M} \Pi(x,y) = \sum_{y\in M} m\left( \left\{ \theta \in \Theta \:\: |\:\: f_{\theta}(x)=y\right\} \right)=m \left( \bigcup_{y\in M} \left\{ \theta \in \Theta \:\: |\:\: f_{\theta}(x)=y\right\} \right)=m(\Theta)=1.$$

My question is: Why $\left\{ \theta \in \Theta \:\: |\:\: f_{\theta}(x)=y\right\}$ is measurable?

Note: All of the above is taken from the book Promenade Aléatoire, Nicole El Karoui, Michel Benaïm.

  • 0
    Isn't this the standard definition of a Markov chain? – 2017-02-15

1 Answers 1

1

Fix $x$, then define a map $g_x: \Theta \to M$ given by $\theta \mapsto f(\theta,x)$.

This is measurable since it is a composition of measurable maps $\theta \mapsto (\theta,x) \mapsto f(\theta,x)$.

Then $\{\theta \in \Theta: f(\theta,x)=y\} = g_x^{-1}(\{y\})$. Since $g_x$ is measurable and $\{y\}$ is a measurable set, the claim follows.