2
$\begingroup$

I have a question about the definition of a Markov process and a Markov family by Karatzas/Shreve in the book "Brownian Motion and Stochastic Calculus" (cf. p. 74). Let me first recall their definitions:

Markov process: Let $d$ be a positive integer and $\mu$ a probability measure on $(\mathbb{R}^d, \mathcal{B}(\mathbb{R}^d))$. An adapted, $d$-dimensional process $X=\{X_t,\mathcal{F}_t;t\geq 0 \}$ on some probability space $(\Omega, \mathcal{F},P^{\mu})$ is said to be a Markov process with initial distribution $\mu$, if

(i) $P^{\mu}(X_0 \in \Gamma ) = \mu(\Gamma), \forall \Gamma \in \mathcal{B}(\mathbb{R}^d)$.

(ii) for $s,t \geq 0$ and $\Gamma \in \mathcal{B}(\mathbb{R}^d)$: $P^{\mu} (X_{t+s} \in \Gamma | \mathcal{F}_s) = P^{\mu} (X_{t+s} \in \Gamma | X_s)$

Markov Family: Let $d$ be a positive integer. A $d$-dimensional Markov Family is an adapted process $X=\{X_t,\mathcal{F}_t;t\geq 0 \}$ on some $(\Omega, \mathcal{F})$, together with a family of probability measures $(P^x)_{x\in \mathbb{R}^d}$ on $(\Omega,\mathcal{F})$, such that

(a) for each $F\in \mathcal{F}$, the mapping $x\mapsto P^x(F)$ is universally measurable;

(b) $P^x(X_0=x)=1, \forall x \in \mathbb{R}^d$;

(c) for $x\in \mathbb{R}^d$, $s,t \geq 0$ and $\Gamma \in \mathcal{B}(\mathbb{R}^d)$: $P^x(X_{t+s} \in \Gamma | \mathcal{F}_s ) = P^x(X_{t+s} \in \Gamma | X_s ), P^x-a.s.$;

(d) for $x\in \mathbb{R}^d$, $s,t \geq 0$ and $\Gamma \in \mathcal{B}(\mathbb{R}^d)$: $P^x(X_{t+s} \in \Gamma | X_s=y ) = P^y(X_{t} \in \Gamma), P^xX_s^{-1}-a.e. y$;

Now my question: To me it looks like the definition of a Markov Family serves to have a Markov process start at different starting points, and for each different starting point $x$, we have one measure $P^x$. Why do we introduce this new definition instead of considering a family of processes $(X^x_t)_{x\in \mathbb{R}^d, t\geq 0}$ with $X^x_t = X_t+x$, and $(X_t)_{t\geq 0}$ a Markov process starting in zero with respect to one measure $P^0$. In this case we would have a family of processes but only one measure, whereas in the aforementioned definition we have one process but a family of measures. What is the advantage of the aforementioned definition?

Thank you in advance, Luke

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.