5
$\begingroup$

I am reading the book Random perturbation of dynamical sustem of Freidlin and Wantzell (2nd edition). On page 20, they define a Markov process as follow:

Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and $(X,\mathcal B)$ the state space. Let $(\mathcal F_t)$ a filtration. Let $(\mathbb P_x)_{x\in X}$ a familly of probability measure. Define the function $p$ as $$p(t,x,\Gamma )=\mathbb P_x\{X_t\in \Gamma \},\quad \Gamma \in \mathcal B, t\in [0,T],x\in X.$$ Then $X=(X_t)_{t\leq T}$ is a Markov process in $X$ if:
a) $X$ is adapted to the filtration.
b) $x\mapsto p(t,x,\Gamma )$ is measurable wrt $\mathcal B$.
c) $p(0,x, X\setminus \{x\})=0$.
d) $\mathbb P_x\{X_u\in Γ\mid \mathcal F_t\}=p(u-t,X_t,\Gamma )$ for all $t,u\in [0,T]$, $t\leq u$, $x\in X$ and $\Gamma \in \mathcal B$.

I am not sure how to interpret c) and d). Would these be correct?

Q1) For c), is it $\mathbb P_x\{X_0=x\}=1$?

Q2) For d), is it $$\mathbb P_{X_0=0}\{X_{t+h}\in \Gamma \mid X_t=k\}=\mathbb P_{X_0=k}\{X_h\in \Gamma \}?$$

But I don't really know how to interpret it.

$\endgroup$
1
+50
$\begingroup$

$\def\Γ{{\mit Γ}}$For Q1, since$$ p(0, x, X \setminus \{x\}) = P_x(X_0 \in X \setminus \{x\}) = 1 - P_x(X_0 = x), $$ so $p(0, x, X \setminus \{x\}) = 0 \Leftrightarrow P_x(X_0 = x) = 1$. The process under $P_x$ can be regarded as starting from $x$.

For Q2, your identity is only a corollary of d) and not equivalent since $\mathscr{F}_t$ might be larger that $σ(X_t)$. To put d) in a clearer form, it is$$ P_x(X_u \in \Γ \mid \mathscr{F}_t) = P_{X_t}(X_{u - t} \in \Γ). $$ In other words, d) says that for any $0 \leqslant t < u$, if one knows the information of time $t$, which corresponds to expectation conditioning on $\mathscr{F}_t$, then the probability of an event in the future, i.e. $\{X_u \in \Γ\}$, with the process starting from $x$ is the same as the probability of $\{X_{u - t} \in \Γ\}$ with the process starting from $X_t$. This simply means that what happens before time $t$ does not matter to the evolution of the process as long as the information at time $t$, i.e. $\mathscr{F}_t$, is known.

$\endgroup$
0
$\begingroup$

To put it in simple words, Q1: You begin in state $x$ with probability 1.

Q2: (The Markov property: ) Given the entire history upto time $t$ (that's what $|\mathcal{F_t}$ means), the present sample $X_u$ is dependent only on the value of the most recent sample $X_t$.

Also in d: I think you wrote $\mathcal{G}$ instead of $\Gamma$.

$\endgroup$

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.