Let $B=(B_t)_{t\ge 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal A,\operatorname{P})$, i.e. $B$ is a real-valued stochastic process with

  • $B_0=0$ almost surely
  • $B$ has independent and stationary increments
  • $B_t\sim\mathcal{N}_{0,\;t}$
  • $B$ is almost surely continuous

Here, stationary increments means, that $$B_{s+t}-B_t\sim B_s\;\;\;\text{for all }s,t\ge 0\;.$$ It's easy to verify, that $B$ has the following property:

$(B_{s+t}-B_t)_{s\ge 0}$ is a Brownian motion independent of $(B_s)_{s\le t}$, for all $s<t$.

Some people call this property Markov property. However, the definition of the elementary Markov property, that I know, is as follows:

Let $I\subseteq\mathbb{R}$ and $E$ be a Polish space. A stochastic process $X=(X_t)_{t\in I}$ with values in $E$ has the elementary Markov property $:\Leftrightarrow$ $$\operatorname{P}\left[X_t\in A\mid\mathcal{F}_s\right]=\operatorname{P}\left[X_t\in A\mid X_s\right]\;\;\;\text{for all }A\in\mathcal{B}(E)\text{ and }s,t\in I\text{ with }s\le t\;,$$ where $(\mathcal{F}_t)_{t\in I})$ is the filtration generated by $X$ and $\mathcal{B}(E)$ is the Borel $\sigma$-algebra on $E$.

In the case of the Brownian motion $B$: Can we show, that both properties are equivalent or one implies the other?

  • $\begingroup$ I don't agree with your definition of the Markov Property. You would need to impose the additional condition that $s<t$, right? $\endgroup$ – Shalop Jul 3 '15 at 17:34
  • $\begingroup$ @Shalop Of course. $\endgroup$ – 0xbadf00d Jul 3 '15 at 19:10

It is true that the second property can be deduced from the first one. Indeed, the first property implies that $(B_t)$ has stationary, independent increments. Hence, the following statement would give you the implication in the forward direction:

Proposition: Any real-valued stochastic process with stationary, independent increments has the elementary Markov Property.

Proof: We just need to show that $E[f(X_t)|\mathcal F_s] = E[f(X_t)|X_s]$ for any $s<t$ and any bounded, measurable function $f: \Bbb R \to \Bbb R$. It's straightforward to show that this is equivalent to the above definition.

Notice that if $t>s$, then for any (bounded-measurable) function $f: E \to \mathbb{R}$, we can write $$E\big[f(X_t)\big|\mathcal{F}_s\big] = E\big[f(X_t-X_s+X_s)\big|\mathcal{F}_s\big] = E\big[g(X_t-X_s,X_s)\big|\mathcal{F}_s\big]$$ where $g(x,y) := f(x+y)$. We know that $X_s$ is $\mathcal{F}_s$-measurable, and that $X_t-X_s$ is independent of $\mathcal{F}_s$ (this is just independence of increments). Therefore, using this question (Conditional Expectation of Functions of Random Variables satisfying certain Properties), we know that $$E\big[g(X_t-X_s,X_s)\big|\mathcal{F}_s\big] = E\big[g(X_t-X_s,X_s)\big|X_s\big] = E\big[f(X_t)\big|X_s\big]$$It follows that $(X_t)_t$ is a Markov Process. $\Box$

  • $\begingroup$ It's not that straightforward for me to see, that what you've shown implies the elementary Markov property. Could you provide a proof? $\endgroup$ – 0xbadf00d Jul 3 '15 at 19:19
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    $\begingroup$ Just take $f=1_A$, and you have that $ P(X_t \in A | \mathcal F_s)=E[f(X_t)|\mathcal F_s] = E[f(X_t)|X_s] = P(X_t \in A | X_s)$. Since $1_A(X_t) = 1_{\{X_t \in A \}}$, (note that the characteristic function on the left is on $\Bbb R$ and the one on the right is on $\Omega$). $\endgroup$ – Shalop Jul 3 '15 at 19:33
  • $\begingroup$ Sorry, I've meant the other direction. $\endgroup$ – 0xbadf00d Jul 3 '15 at 19:36
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    $\begingroup$ You don't need the other direction for this particular problem. But just in case you're interested, suppose that $f: \Omega \to \Bbb R$ is nonnegative and measurable, and that the definition written in your problem statement holds true. Then you can find a sequence of simple functions $f_n \uparrow f$. The claim is true for simple functions by the defining condition of the Markov Property, so by MCT it will be true for $f$. If $f$ is not nonnegative, consider positive and negative parts. $\endgroup$ – Shalop Jul 3 '15 at 19:43

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