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.

  • $\begingroup$ Isn't this the standard definition of a Markov chain? $\endgroup$ – Did Feb 15 '17 at 6:43

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.