Looking for an example of stochastic processes with "very" different trajectories, but still modifications of each other

126 Views Asked by At

Could anyone offere an example of two stochastic processes that are modifications of each other, i.e.

$P(A_t)=1\qquad \forall t\in [0,T]$,$\qquad$ with $A_t:=\{\omega\in\Omega:X_t=Y_t\}$,

and such that

$P(\omega\in\Omega: \mu(I_{\omega})>0)>0$, with $I_{\omega}:=\{t\in [0,T] : \omega \in A^c_t \}$, and where $\mu$ is the Lebesgue measure on $[0,T]$?

Let me give you a bit of context. I am trying to understand how different could be, in principle, the trajectories of two versions of the same process. So far, all I could find is the typical textbook example of a modification that is not indistinguishable, which is simple but not satisfactory as the trajectories only differ at one time.

My intuition (which could be wrong of course) is that a modification like the one above should exist, but I struggle to find a formal argument to prove it.

Thanks in advance!

1

There are 1 best solutions below

2
On

As $\mathbb{P}(A_t^c)=0$ for each $t \geq 0$, we have by Tonelli's theorem

$$\mathbb{E} \left( \int_0^{T} 1_{A_t^c} \, dt \right) = \int_0^{T} \mathbb{P}(A_t^c)=0,$$

and so

$$\int_0^{T} 1_{A_t^c} \, dt = 0 \quad \text{a.s.}$$

Since $1_{A_t^c}(\omega) = 1_{I_{\omega}}(t)$ this implies that

$$\int_0^T 1_{I_{\omega}}(t) \, dt=0 \quad \text{a.s.},$$

i.e. $\mu(I_{\omega})=0$ almost surely. This means that there doesn't exist a modification with "very different" sample paths.