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?
Best Answer
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. $\Box$