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Question: Let X be a stochastic process and T a stopping time of ${\mathcal{F}^{X}_{t}}$. Suppose that for some pair $\omega$, $\omega$' $\in$ $\Omega$, we have $X_{t}(\omega)=X_{t}(\omega')$ for all t $\in [0,T(\omega)]\cap[0,\infty]$. Show that $T(\omega)=T({\omega}')$.

By definition, we have $\omega \in T^{-1}([0,T(\omega)])\subseteq \mathcal{F}^{X}_{T(\omega)}$. And my intuition gives that $\omega$' should belong to $T^{-1}([0,T(\omega)])$ since the value of process coincide on $[0,T(\omega)]$.Then, similar argument can be done to show the equality. Embarrassingly, I am not able to show it. The gap is that if $X_{t}(\omega)=X_{t}(\omega')$ for all t $\in [0,T]$, is there any element in ${\mathcal{F}^{X}_{T}}$ containing only one of them? If not, my approach would work.

Please help me fill in the gap if it is true. Or please provide a proof of the root problem.

swang999
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1 Answers1

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Hint: The collection $\cal C$ of subsets $A$ of $\Omega$ with $1_A(\omega)=1_A(\omega^\prime)$ is a $\sigma$-algebra. If $X_s(\omega)=X_s(\omega^\prime)$ for some $s$, then $\sigma(X_s)\subseteq {\cal C}$. If $X_s(\omega)=X_s(\omega^\prime)$ for all $0\leq s\leq T$, then ${\cal F}_T\subseteq {\cal C}$.