I'm reading Bernt Oksendal's "Stochastic Differential Equations" and "adapted" is one of the concept that I could not understand.
First, "adapted" is defines at Ch3.1, page 25 (sixth edition):
Definition 3.1.3. Let $\{\mathscr{N}_t\}_{t\geq 0}$ be an increasing family of $\sigma$-algebras of subsets of $\Omega$. A process $g(t,\omega):[0,\infty)\rightarrow \mathbb{R}^n$ is called $\mathscr{N}_t$-adapted if for each $t\geq 0$ the funcction
$$\omega\rightarrow g(t, \omega)$$
is $\mathscr{N}_t$-measurable.
Then, it says:
Thus the process $h_1(t,\omega) = B_{t/2}(\omega)$ is $\mathscr{F}_t$-adapted, while
$h_2(t,\omega) = B_{2t}(\omega)$ is not $\mathscr{F}_t$-adapted.
Here $B_t$ is Brownian motion.
I'm lost. Brownian motion means the particle can appear at anywhere, right? Let's just consider $n=1$, a 1-dimension Brownian motion can take any value for x, just the bigger $x$ is, the smaller the probability.
So I thought Brownian motion is just defined on $\mathscr{B}(\mathbb{R})$, the Borel $\sigma$-algebra on $\mathbb{R}$, for any $t$. So both $h_1(t,\omega) = B_{t/2}(\omega)$ and $h_2(t,\omega) = B_{2t}$ are defined on $\mathscr{B}(\mathbb{R})$, always measurable, so always adapted?
Here $\mathscr{F}_t$ is defined earlier:
Definition 3.1.2. Let $B_t(\omega)$ be $n$-dimensional Bownian motion. Then we define $\mathscr{F}_t = \mathscr{F}_t^{(n)}$ to be the $\sigma$-algebra generated by the random variables $B_s(\cdot); s\leq t$. In other words, $\mathscr{F}_t$ is the smallest $\sigma$-algebra containing all sets of the form
$$\{\omega; B_{t_1}(\omega) \in F_1, \cdots, B_{t_k}(\omega) \in F_k\}$$,
where $t_j\leq t$ and $F_j \subset \mathbb{R}^n$ are Borel sets, $j\leq k=1,2,\ldots$ (we assume that all sets of measure zero are included in $\mathscr{F}_t$).
Best Answer
Let $(\Omega,\mathscr{F},(\mathscr{F}_t)_{t\geq 0},P)$ be a filtered probability space. That is $(\Omega,\mathscr{F},P)$ is a probability space and $(\mathscr{F}_t)_{t\geq 0}$ is a filtration in $\mathscr{F}$, i.e. $\mathscr{F}_t\subseteq \mathscr{F}$ is a sigma-algebra for every $t\geq 0$ and for $0\leq s<t$ we have $\mathscr{F}_s\subseteq\mathscr{F}_t$. Then a mapping $X=(X_t)_{t\geq 0}:[0,\infty)\times\Omega\to\mathbb{R}^n$ is called a stochastic process if for every $t\geq 0$ $$ \Omega\ni \omega\mapsto X(t,\omega)=X_t(\omega) $$ is $(\mathscr{F},\mathscr{B}(\mathbb{R}^n))$-measurable.
Since, for a fixed $t\geq 0$, $\mathscr{F}_t\subseteq\mathscr{F}$, the mapping $\omega\mapsto X(t,\omega)$ is not necessarily $(\mathscr{F}_t,\mathscr{B}(\mathbb{R}^n))$-measurable (why?), and hence we need to make this a definition.
Therefore, we say that $(X_t)_{t\geq 0}$ is $(\mathscr{F}_t)_{t\geq 0}$-adapted if the mapping $$ \Omega\ni \omega\mapsto X(t,\omega) $$ is $(\mathscr{F}_t,\mathscr{B}(\mathbb{R}^n))$-measurable for every $t\geq 0$.
I believe that in your setup, we have $(\mathscr{F}_t)_{t\geq 0}$ being the natural filtration induced by the Brownian motion $(B_t)_{t\geq 0}$. That is $$ \mathscr{F}_t=\sigma(B_s\mid 0\leq s\leq t),\quad\text{for all }\;t\geq 0, $$ i.e. $\mathscr{F}_t$ is the smallest sigma-algebra making all $B_s$, $s\leq t$, measurable. Now, it is obvious that $(B_t)_{t\geq 0}$ is adapted to $(\mathscr{F}_t)_{t\geq 0}$ (right?).
To see why, e.g. $h_1(t,\omega)=B_{t/2}(\omega)$ is $(\mathscr{F}_t)_{t\geq 0}$-adapted, we need to show that $\omega\mapsto h_1(t,\omega)$ is $(\mathscr{F}_t,\mathscr{B}(\mathbb{R}))$-measurable for every $t\geq 0$. So let $t\geq 0$ be given. Then $$ \omega\mapsto B_{t/2}(\omega) $$ is $(\mathscr{F}_{t/2},\mathscr{B}(\mathbb{R}))$-measurable and since $\mathscr{F}_{t/2}\subseteq\mathscr{F}_t$ it is also $(\mathscr{F}_{t},\mathscr{B}(\mathbb{R}))$-measurable.