Borel Right Process

In the mathematical theory of probability, a Borel right process, named after Émile Borel, is a particular kind of continuous-time random process.

Let $E$ be a locally compact, separable, metric space.
We denote by $\mathcal E$ the Borel subsets of $E$.
Let $\Omega$ be the space of right continuous maps from $[0,\infty)$ to $E$ that have left limits in $E$,
and for each $t \in [0,\infty)$, denote by $X_t$ the coordinate map at $t$; for
each $\omega \in \Omega$, $X_t(\omega) \in E$ is the value of $\omega$ at $t$.
We denote the universal completion of $\mathcal E$ by $\mathcal E^*$.
For each $t\in[0,\infty)$, let

: $\mathcal F_t = \sigma\left\,$

: $\mathcal F_t^* = \sigma\left\,$

and then, let

: $\mathcal F_\infty = \sigma\left\,$

: $\mathcal F_\infty^* = \sigma\left\.$

For each Borel measurable function $f$ on $E$, define, for each $x \in E$,

: $U^\alpha f(x) = \mathbf E^x\left[ \int_0^\infty e^ f(X_t)\, dt \right].$

Since $P_tf(x) = \mathbf E^x\left[f(X_t)\right]$ and the mapping given by $t \rightarrow X_t$ is right continuous, we see that
for any uniformly continuous function $f$, we have the mapping given by $t \rightarrow P_tf(x)$ is right continuous.

Therefore, together with the monotone class theorem, for any universally measurable function $f$, the mapping given by $(t,x) \rightarrow P_tf(x)$, is jointly measurable, that is, $\mathcal B([0,\infty))\otimes \mathcal E^*$ measurable, and subsequently, the mapping is also $\left(\mathcal B([0,\infty))\otimes \mathcal E^*\right)^$-measurable for all finite measures $\lambda$ on $\mathcal B([0,\infty))$ and $\mu$ on $\mathcal E^*$.
Here,
$\left(\mathcal B([0,\infty))\otimes \mathcal E^*\right)^$ is the completion of
$\mathcal B([0,\infty))\otimes \mathcal E^*$ with respect
to the product measure $\lambda \otimes \mu$.
Thus, for any bounded universally measurable function $f$ on $E$,
the mapping $t\rightarrow P_tf(x)$ is Lebeague measurable, and hence,
for each $\alpha \in [0,\infty)$, one can define

: $U^\alpha f(x) = \int_0^\infty e^P_tf(x) dt.$

There is enough joint measurability to check that $\$ is a Markov chain, Markov resolvent on $(E,\mathcal E^*)$,
which uniquely associated with the Markovian semigroup $\$.
Consequently, one may apply Fubini's theorem to see that

:
U^\alpha f(x) = \mathbf E^x\left \int_0^\infty e^ f(X_t) dt \right

The following are the defining properties of Borel right processes:

* Hypothesis Droite 1:

:For each probability measure $\mu$ on $(E, \mathcal E)$, there exists a probability measure $\mathbf P^\mu$ on $(\Omega, \mathcal F^*)$ such that $(X_t, \mathcal F_t^*, P^\mu)$ is a Markov process with initial measure $\mu$ and transition semigroup $\$.

* Hypothesis Droite 2:

:Let $f$ be $\alpha$-excessive for the resolvent on $(E, \mathcal E^*)$. Then, for each probability measure $\mu$ on $(E,\mathcal E)$, a mapping given by $t \rightarrow f(X_t)$ is $P^\mu$ almost surely right continuous on ,\infty).

Notes

References

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{{DEFAULTSORT:Borel Right Process
Stochastic processes