Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathscr{F}_t$).

share|improve this question
    
Is $(\mathscr{F}_t)_{t\geq 0}$ the natural filtration induced by $(B_t)_{t\geq 0}$? –  Stefan Hansen Sep 26 '13 at 9:18
    
actually "filtration" is not defined yet, till here. as i updated the question, $\mathscr{F}_t$ is defined as the smallest $\sigma$-algebra generated. but actually i don't understand it too. the "filtration" is what made me quit "Stochastic Calculus" course last time. –  athos Sep 26 '13 at 9:30
    
just to conclude: refer to math.stackexchange.com/questions/508658/… and math.stackexchange.com/questions/506367/… –  athos Sep 30 '13 at 16:53

3 Answers 3

up vote 1 down vote accepted

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.

share|improve this answer
    
thanks for your reply, pls see my reply just added. –  athos Sep 26 '13 at 17:03
    
@athos: I think you should revisit the section in your book/lecture notes on measurability. It seems like you're missing nearly every point I tried to make. For instance: if a random variable $X$ is $(\mathscr{F},\mathscr{B}(\mathbb{R}))$-measurable and $\mathscr{G}$ is a larger sigma-algebra (i.e. $\mathscr{F}\subseteq\mathscr{G}$), then why is $X$ necessarily also $(\mathscr{G},\mathscr{B}(\mathbb{R}))$-measurable? I don't think you should be dealing with filtrations and stochastic processes before you've become familiar with the basics of measure theory (e.g. measurability etc). –  Stefan Hansen Sep 27 '13 at 7:09
    
let me revisit the measurability part again. –  athos Sep 27 '13 at 9:21
    
could you pls have a look at this question? math.stackexchange.com/questions/508658/… –  athos Sep 30 '13 at 12:29

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 $B$.

share|improve this answer

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?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.