1

Consider the stochastic process $(X_{t})_{t\in \mathbb R_{+}}$ on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_{t}^{X})_{t\in \mathbb R_{+}},\mathbb P)$ where $\mathcal{F}_{t}^{X}=\sigma (X_{s}:0\leq s\leq t) $. Let $T$ be a $(\mathcal{F}_{t}^{X})_{t\in \mathbb R_{+}}-$stopping time such that there is a pair $\omega, \overline{\omega} \in \Omega$ where:

$ X_{t}(\omega)=X_{t}(\overline{\omega})$ for all $t \in [0,T(\omega)]\cap [0,\infty)$

Then show that $T(\omega) = T(\overline{\omega})$

My failed attempt:

Perhaps a contradiction may help. Assume that $T(\omega) < T(\overline{\omega})$, then $\overline{\omega}\notin \{T\leq t\} \in \mathcal{F}_{t}^{X}$ for any $t \leq T(\omega)$, in particular:

$\overline{\omega}\notin \{T\leq T(\omega)\} \in \mathcal{F}_{T(\omega)}^{X}$ but how will this help me? Any ideas/tips?

MinaThuma
  • 998
  • 6
  • 16

1 Answers1

2

The $\sigma$-fields $\mathcal F^X_t$ ($t>0$) have a "saturation" property that is the key: If $\omega\in A\in\mathcal F^X_t$ and if $\overline\omega$ is such that $X_s(\omega) = X_s(\overline\omega)$ for all $0\le s\le t$ then also $\overline\omega\in A$.

Now suppose we have $\omega$ and $\overline\omega$ such that $X_s(\omega)=X_s(\overline\omega)$ for all finite $s$ in $[0,T(\omega)]$. Consider first the case $T(\omega)<+\infty$ and apply the saturation property with $t=T(\omega)$ and $A=\{T=t\}$. The conclusion is that $\overline\omega\in A$ as well; that is, $T(\overline\omega) = t$.

To deal with the possibility that $T(\omega)$ might be $+\infty$, apply the preceding to the stopping time $T_n:=T\wedge n$ ($n=1,2,\ldots$) to conclude that $T(\overline\omega)\wedge n=T_n(\overline\omega) = T_n(\omega) =n$ for each $n$. Consequently, $T(\overline\omega)=+\infty=T(\omega)$.

John Dawkins
  • 25,733