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

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 $B$.

Answer 3

To answer the new questions of OP with more elaborate examples: Yes the mapping can be quite weird, assuming that you meant $X_{t} = _{\omega \in \mathbb{Q}}$ then $X_{t}$ is measurable for the filtration $\mathscr{F}_{t} = \mathscr{B}(\mathbb{R})$, while for the filtration $\mathscr{F}_{t} = \mathscr{B}(\mathbb{Q})$ it is not (since $X_{t}(w) = 0$ might be observed for some $t$ e.g. $w = \pi$ and $ \{\pi\} \notin \mathscr{B}(\mathbb{Q})$).

In this case, we implicitly set $\forall t \mathscr{F}_{t} = \mathscr{F}_{0} = \mathscr{B}(\mathbb{R}) $, which is called the complete filtration, where we know all information about possible future events right from the start and our filtration doesn't change over time. As you correctly noticed then $B_{2t}$ is indeed adapted to the filtration $\mathscr{F}_{t}$. There is just one problem, namely that $\sigma(B_{t})$ is supposed to be the smallest (in the sense of set inclusion) $\sigma$-algebra such that $B_{t}$ is measurable. What is the smallest depends on your actual experiment, so the complete filtration is often too large, except for $\sigma(B_{\infty}) = \mathscr{B}(\mathbb{R})$. Note that the definition of adapted process allows us to ignore the question of what our experiment exactly is, by enforcing that $B_t$ becomes measurable, which makes it sort of "obvious" that mostly $\sigma(B_{2t}) \neq \sigma(B_{t})$ holds.

Let's take a concrete example: Draw uniformly independent an infinite sequence of numbers (not necessarily distinct) $ \{ n_{i} \}^{\infty}_{i = 0}$ from the interval $[0, t]$. Define

$$ Y_i = _{ \lfloor 100n_{i} \rfloor - 10\lfloor 10n_{i} \rfloor \mod 2 = 0} - _{ \lfloor 100n_{i} \rfloor - 10\lfloor 10n_{i} \rfloor \mod 2 = 1} $$ $$ P(Y = 1) = P(Y = -1) = \frac{1}{2} $$ $$ B_t = \lim_{n \to \infty} \frac{1}{\sqrt{n}}\sum_{i = 0}^{\lfloor tn\rfloor} Y_{i} $$ which is a fancy way of saying that $Y = 1$ when $n$'s second decimal digit (the first after the zero wouldn't fit the probabilities) is even and $ -1 $ if odd. Also we require $ B_{0} = 0 $ and $ t \in [0, 1] $ then according to Donskers theorem $B_t$ is a Wiener process (the same model as used in the brownian motion). One corresponding probability space is

$$ (\Omega, \mathscr{F}, \mathbb{P}) = (\mathbb{R}^{\infty}, \mathscr{B}(\mathbb{R}^{\infty}),\mathbb{P}). $$

Though to measure $ B_{t} $, only the $Y_{i}'s$ events (not all future) must be measured, which works for the filtration $\mathscr{F}_{Y_{i}} = \mathscr{B}(\mathbb{[0, t]}) $, giving us the generated $\sigma(B_{t}) = \mathscr{B}(\mathbb{[0, t]}^{\infty}) = \mathscr{F}_{t}$, which is smaller than our original choice. This also implies $\sigma(B_{2t}) \nsubseteq \sigma(B_{t})$.

An example for an event that highlights this issue (assuming that $2t$'s second digit is even), is $w_{0} = (2t, t + 0.01, 2t, t + 0.01, ...) $ $ B(2t, w_{0}) = 0 $ although 0 is a value that also $ B_{t} $ can have, we would asses it's probability (which is derived from $ \mathscr{F}_{t} $) wrong for $B_{2t} $ since $ \{ w_{0} \} \notin \mathscr{F}_{t} $.

Answer 4

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?