Brownian motion and adapted

Question

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 $\mathscr{F}_t$).

Answer 1

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.

Answer 2

not answering my own question, but to make it easier to consult from @stefan-hansen .

For your 1st question:

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.

My understanding is, the mapping $X_t: \omega\mapsto X(t,\omega)$ could be quite weird, so that $X_t$ is not $\mathscr{F}_t$-measurable.

( I don't know how to construct such an $X_t$, but I heard of some weird mappings , such as, take $\Omega$ as $\mathbb{R}$, and $\forall \omega \in \mathbb{R}$, $X_t(\omega) = 0$, if $\omega \in \mathbb{Q}$; $X_t(\omega) = 0$, otherwise. )

Is this right?

For your 2nd question:

we have $(\mathscr{F}_t)_{t\geq 0}$ being the natural filtration induced by the Brownian motion $(B_t)_{t\geq 0}$. That is $$\mathscr{F}_t=\sigma(B_s\mid 0\leq s\leq t), \forall t\geq 0,$$ 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?).

I don't know how to answer. I'm still thinking.

My understanding is, for Brownian motion $B_t$, its smallest $\sigma$-algebra is just the preimage of Borel algebra on $\mathbb{R}^n$:

$$\forall t\geq 0, \sigma(B_t) = X_t^{-1}[ \mathscr{B}(\mathbb{R}^n) ] = \{ F \subset \Omega \, | \, X_t(F) \in \mathscr{B}(\mathbb{R}^n) \}$$

Is this correct?

Answer 3

By definition $B_{t/2}$ is $F_{t/2}$-measurable and since filtrations are monotone (i.e, $F_{t/2} \subseteq F_t$) then it is $F_t$-measurable, whereas $B_{2t}$ is $F_{2t}$-measurable and it could happend that for some borelian $C \subseteq \mathbb{R}$ then $B_{2t}^{-1}(C)$ does not belong to $F_t$.

On the other hand $B_t$ is always adapted to $\sigma(B_s, 0 \leq s \leq t)$ since this includes the smallest sigma algebra that makes the random variable $B_t$ measurable, namely $\sigma(B_t) = B_t^{-1}(\mathcal{B}(\mathbb{R}))$. This is why this filtration is called the natural filtration of $B$.