# Is filtration necessary for continuous random variables?

Define $(\mathscr{F}_t)_{t\geq 0}$ 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, why is this necessary?

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

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

So, $\forall 0 < s < t, \sigma(B_s) = \sigma(B_t)$.

In the end $\forall 0 < s < t, \sigma(B_s) = \sigma(B_t) = \mathscr{F}_s = \mathscr{F}_t$.

This actually applies to other continuous random variables, filtration is only necessary for discrete random variables. -- is my understanding correct?

• There is no reason why $\sigma(B_s)$ and $\sigma(B_t)$ should be the same for $t\neq s$. To answer your question: your understanding is not correct. – Stefan Hansen Sep 27 '13 at 7:04
• What @Stefan said. // Actually I would be interested to understand how you "deduced" that $\sigma(B_t)=\sigma(B_s)$ from the definition of $\sigma(B_t)$ you correctly recalled. – Did Sep 28 '13 at 8:11
• @Did could you pls have a look at this question? math.stackexchange.com/questions/508658/… – athos Sep 30 '13 at 12:30

## 1 Answer

Refer to question The concept of random variable , consider $\sigma(B_t) = X_t^{-1}[\mathcal{B}(\mathbb{R}^n)]$ in the "pollen room" --

$\mathcal{B}(\mathbb{R}^n)$ are the Borel sets in $\mathbb{R}^n$, while $\sigma(B_t)$ are a grain of pollens that happened to be near the Tulip one at the snapshot time $t$.

Since the Tulip pollen drifts in the room, at each time its neighborhood changes, $\sigma(B_t)$ definitely is different from $\sigma(B_s)$.

A filtration is the joint of $\sigma(B_t)$ for all possible $t$, this is required so that at any time $X_t$ is measurable.