I'm studying Stochastic Processes by Richard F. Bass. Within this book I encountered the definition of a Markov process, which is given as follows:

We are given a separable metric space $S$ endowed with its Borel $\sigma$-field and a measurable space $(\Omega, \mathcal{F})$ together with a filtration $\{\mathcal{F}_t\}$. Then

Definition 19.1 A Markov process $(X_t , \Bbb{P}^x)$ is a stochastic process $$X : [0, \infty) \times \Omega \to S$$ and a family of probability measures $\{\Bbb{P}^x : x \in S\}$ on $(\Omega, \mathcal{F})$ satisfying the following.

  1. For each $t$, $X_t$ is $\mathcal{F}_t$ measurable.
  2. For each $t$ and each Borel subset $A$ of $S$, the map $x \mapsto \Bbb{P}^x (X_t \in A)$ is Borel measurable.
  3. For each $s, t \geq 0$, each Borel subset $A$ of $S$, and each $x \in S$, we have $$ \Bbb{P}^x(X_{s+t} \in A \mid \mathcal{F}_s) = \Bbb{P}^{X_s} (X_t \in A), \quad \Bbb{P}^x − \text{a.s.}$$

Okay, this definition was fine. Then in the next chapter, the author begins with the setting that $(X_t , \Bbb{P}^x)$ is a Markov process with respect to $\mathcal{F}_t^{00} = \sigma(X_s : s \leq t)$ such that its sample path is càdlàg with probability 1 under $\Bbb{P}^x$ for all $x \in S$. With $$ \mathcal{F}_t^{0} = \sigma \left(\mathcal{F}_t^{00} \cup \{ A \subset S : A \text{ is } \Bbb{P}^x\text{-null for all } x \in S\} \right) \quad \text{and} \quad \mathcal{F}_t = \mathcal{F}_{t+}^{0} = \bigcap_{\epsilon > 0} \mathcal{F}_{t+\epsilon}^{0},$$

he proved the following property:

Theorem 20.6 Let $(X_t , \Bbb{P}^x)$ be a Markov process and suppose for all bounded Borel measurable function $f$, $$ \Bbb{E}^x[ f(X_{s+t}) \mid \mathcal{F}_s] = \Bbb{E}^{X_s} [ f(X_t)], \quad \Bbb{P}^x − \text{a.s.} $$ holds. Suppose $Y$ is bounded and measurable with respect to $\mathcal{F}_{\infty} = \bigvee_{s \geq 0} \mathcal{F}_s $. Then $$ \Bbb{E}^x[Y \circ \theta_s \mid \mathcal{F}_s] = \Bbb{E}^{X_s} Y, \quad \Bbb{P}^x − \text{a.s.}$$

Until now, still there was no problem. Then it claims the Blumenthal 0-1 law, which is stated as follows:

Proposition 20.8 Let $(X_t , \Bbb{P}^x)$ be a Markov process with respect to $\{\mathcal{F}_t\}$. If $A \in \mathcal{F}_0$, then for each $x$, $\Bbb{P}^x(A)$ is equal to $0$ or $1$.

Proof. Suppose $A \in \mathcal{F}_0$. Under $\Bbb{P}^x$, $X^0 = x$, a.s., and then $$\Bbb{P}^x(A) = \Bbb{E}^{X_0}\mathbf{1}_A = \Bbb{E}^x[\mathbf{1}_A \circ \theta_0 | \mathcal{F}_0] = \mathbf{1}_A \circ \theta_0 = \mathbf{1}_A \in \{0, 1\}, \quad \Bbb{P}^x - \text{a.s.} $$ since $\mathbf{1}_A \circ \theta_0$ is $\mathcal{F}_0$ measurable. Our result follows because $\Bbb{P}^x(A)$ is a real number and not random.

I was puzzled by the claim of the proof, that $X_0 = x$ a.s. under $\Bbb{P}^x$. Clearly there is no such assumption in Definition 19.1 above. So I tried to prove this using the definition. I succeeded in circumventing this seemingly unjustified claim when there is some $x_0 \in S$ such that $\Bbb{P}^{x}(X_0 = x_0) > 0$, but my approach turned out to be inadequate for a general proof.

So here is my question: Is there an argument which either avoids or proofs the claim $\Bbb{P}^x (X_0 = x) = 1$? Or is there a counter-example, so that we should just embrace this as a part of definition and insert it to the incomplete Definition 19.1?

  • $\begingroup$ $P^x(X_0=x)=1$ is a universal convention (this is what the superscript $x$ is for) which has nothing to do with the rest of your post and is usually mentioned once and for all as soon as possible, for example in Definition 19.1. Sure the author does not state it somewhere before in the book? $\endgroup$ – Did Sep 15 '12 at 4:42
  • $\begingroup$ @did, No, I searched in the eBook version of this textbook with the keyword 'Markov', and found that the definition above is the very beginning of the Markov process theory in the book. Of course, the author mentions two examples - Markov chain and Brownian motion -, which both satisfy the condition $\Bbb{P}^x(X_0 = x) = 1$. But I'm pretty sure that there is no such statement before. $\endgroup$ – Sangchul Lee Sep 15 '12 at 4:49
  • $\begingroup$ OK (although I would rather try the keyword initial or initial state.) Anyway, this is the meaning of $P^x$. $\endgroup$ – Did Sep 15 '12 at 5:06
  • $\begingroup$ @did, I also agree with you. Clearly $\Bbb{P}^x$ is designed to capture the conditional case where $X_0$ starts at $x$, thus $\Bbb{P}^x(X_0 = x) = 1$ is essential. I just wanted to know if the author intended it as a result, or if it is so universal that he unwittingly forgot including it to the definition. $\endgroup$ – Sangchul Lee Sep 15 '12 at 5:14

Definition 19.1 in your post is on page 153 of the book, which is the second page of Chapter 19. The notation $\mathbb P^x$ is defined by equation (19.1), which is on page 152, the first page of Chapter 19.

  • $\begingroup$ I also saw that equation (19.1). But it is just a motivating example confined to Markov chains with stationary transition probabilities with finite states. So it is not impossible, especially for a beginner, to think that the condition $\Bbb{P}^x(X_0 = x)$ is dropped out of Definition 19.1 for the sake of sheer generality, which will be retrieved from the definition later. (And this was the case for me.) $\endgroup$ – Sangchul Lee Sep 20 '12 at 22:22
  • $\begingroup$ Nevertheless, this fully answers your question, no? $\endgroup$ – Did Apr 11 '13 at 22:02

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