1
$\begingroup$

Assume you have a filtered probability space $(\Omega, \mathcal{F},\mathcal{F}_t,P)$, assume that $B_t$ is a Brownian motion with respect to this filtered probability space.

Define

$$\text{sign}(x)=1, x \ge 0\\ \text{sign}(x)=-1, x <0.$$

It can then be shown that this stochastic integral is a Brownian motion with respect to $\mathcal{F}_t$:

$$X_t=\int_0^t \text{sign}(B_s)dB_s.$$

Since $X_t$ is also a Brownian motion(this must be shown) we can calculate stochastic integrals with respect to it.

It is stated from various sources that we then have

$$dB_s = \text{sign}(B_s)dX_s.$$

But how do we show this last equation? What is it that allows us to move the $\text{sign}(B_s)$ over? It seems that they have the differential and have divided by $\text{sign}(B_s)$, and used that $1/ \text{sign}(B_s)=\text{sign}(B_s)$. But what allows us to do this?

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

It's easier to just write out the right side of the equation. By definition of $X$, $dX_s = \text{sign}(B_s)dB_s$, so $$\text{sign}(B_s)dX_s = \text{sign}(B_s)\text{sign}(B_s)dB_s = dB_s$$ because $\text{sign}(x)^2 = 1$ for all $x \in \mathbb{R}$.

$\endgroup$
2
  • $\begingroup$ But why are you allowed to do that operation with differentials? Is it a known rule? How is it proven? $\endgroup$
    – user394334
    Aug 2, 2021 at 19:55
  • 2
    $\begingroup$ Additionally (in case OP wants to look it up), the name of the property that allows us to do this is called "Associativity" of the Ito integral. en.wikipedia.org/wiki/It%C3%B4_calculus#Properties $\endgroup$
    – nullUser
    Aug 2, 2021 at 19:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .