A stochastic difference equation given in the form
$X_{\{ n+1\}}=T(X_n)+e_{n+1} \;, n\geqslant 0, \; T:\mathbb{R}^m \rightarrow \mathbb{R}^m$. with state space $(\mathbb{R}^m,\mathbb{B}_m, u_m)$.
The random forcing terms on the right-hand side of the stochastic difference equation are of the following form:
$e_n=\left( \begin{array}{l} e_n^{'} \\ 0 \\ \vdots \\ 0 \end{array} \right)$ with $e_n^{'}$ i.i.d., each having an absolutely continuous distribution and p.d.f. $f(.)$ is positif everywhere in $\mathbb{R}$.
Let $A\in \mathbb{B}_m$ and $x\in\mathbb{R}^m$. Let
$x=\left( \begin{array}{l} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right)$
Define
$A_x= \left\{y=\left( \begin{array}{l} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right)\in \mathbb{R}^m : y_i=x_i, i\geqslant 2, \, y\in A \right\}$.
Let $v:\mathbb{R}^m \rightarrow \mathbb{R}$ be the projection map onto the first coordinate, i.e., $v(y)=y_1$.
Then
$\mathbb{P}(x,A)= \displaystyle \int_{v_{(A_{Tx})}-(Tx)_1}f(t) dt$
where $(Tx)_1$ is the first coordinate of $Tx$ and we have written $Tx$ for $T(x)$\
My questions are:
- why $\{X_n\}$ with $\mathbb{P}(x,A)$ forms a Markov chain.
- how we constructed the transition probability.
- I can't understand $v_{(A_{Tx})}-(Tx)_1$.