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Let $S$ and $T$ two stopping time w.r.t. the filtration $(\mathcal F_t)_t$ such that there is $M>0$ s.t. $$S\leq T\leq M\quad a.s.$$ Show that if $B\in \mathcal F_S$, we have that $$\sigma =T\boldsymbol 1_B+M\boldsymbol 1_{B^c},$$ is a stopping time where $$\mathcal F_S=\{A\in \mathcal F_\infty\mid A\cap \{T\leq t\}\in \mathcal F_t,\forall t\geq 0 \}.$$

Try

I try to show that $\{\sigma \leq t\}^c\in \mathcal F_t$. $$\{\sigma >t\}=\{T\boldsymbol 1_B+M\boldsymbol 1_{B^c}>t\}=\bigcup_{r\geq t, r\in \mathbb Q^+}\{T\boldsymbol 1_B>r\}\cap \{M\boldsymbol 1_{B^c}>t-r\}.$$

How can I continue ? I guess that $$\{T\boldsymbol 1_B>r\}=B\cap \{T>r\}\quad \text{and}\quad \{M\boldsymbol 1_{B^c}>t-r\}$$ and thus $$\{T\boldsymbol 1_B>r\}\cap \{M\boldsymbol 1_{B^c}>t-r\}=(B\cap B^c)\cap \{T>r\}\cap \{M>t-r\}=\emptyset,$$ so probably something is wrong...

John
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1 Answers1

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Note that when $r>t$ we get $\{MI_{B^{c}}>t-r\}=\Omega$ because $MI_{B^{c}} \geq 0 >t-r$. BTW you also have to include $r=t$ in the union. The union is $B \cap (T>t)$.

Kavi Rama Murthy
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  • Oh, I see, so at the end, we get $...=\bigcup_{r>t, r\in \mathbb Q^+}B\cap {T>r}$. At then end ${T>r}\in \mathcal F_r\supset \mathcal F_t$ and $B\in \mathcal F_S$. How can I conclude ? – John Aug 25 '19 at 12:16
  • Also, Isn't it true that ${M\boldsymbol 1_{B^c}>r}=B^c\cap {M>t-r}$ ? – John Aug 25 '19 at 12:17
  • When you take the union of $B \cap (T>r)$ over $r>t$ you get $B \cap (T>t)$. This set belongs to $\mathcal F_t$ because $B \in \mathcal F_s \subset \mathcal F_t$. – Kavi Rama Murthy Aug 25 '19 at 12:19
  • Did you read the part of my answer where I said $MI_{B^{c}} \geq 0$ hence $>t-r$ for every $\omega$? – Kavi Rama Murthy Aug 25 '19 at 12:22
  • Oh, no sorry. I see :) Last thing is I don't really understand why you need $r\geq t$ in the union ? – John Aug 25 '19 at 12:23
  • At the end, if $t\in \mathbb Q$, then if $r=t$ then ${M\boldsymbol 1_{B^c}>t-r}=B^c\neq \Omega $... no ? – John Aug 25 '19 at 12:26
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    Yes, that is why I mentioned that $r=t$ also has to be considered. But here you use the fact that $B \cap B^{c}$ is empty, so this part of the union does not contribute anything to the union. – Kavi Rama Murthy Aug 25 '19 at 12:28