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Preface: I'm having trouble with the correct solution.

The Original Question: Given that $\mathscr{F}_t$ is a filtration that satisfies all the usual conditions, and given ${\{\mathscr{F}_t\}_\mathbb{R}}_+$ stopping times $\sigma$ and $\tau$, show that $\{min(\sigma,\tau) \leq t\} \in \mathscr{F}_t$.

The Correct Solution: $\{min(\sigma,\tau) \leq t\} = \{\sigma \leq t\}\cup\{\tau \leq t\} \in \mathscr{F}_t$ by the fact that the filtration is closed under countable unions.

My Issues: I don't understand why $\{min(\sigma,\tau) \leq t\} = \{\sigma \leq t\}\cup\{\tau \leq t\}$.

Consider two sample spaces $\Omega_1$ and $\Omega_2$ with $\omega_1 \in \Omega_1$ and $\omega_2 \in \Omega_2$. Then:

  • $\{ \sigma \leq t\} = \{\omega_1 : \sigma(\omega_1) \leq t\}$

  • $\{ \tau \leq t\} = \{\omega_2 : \tau(\omega_2) \leq t\}$

Therefore, the RHS of the solution is: $\{\sigma \leq t\}\cup\{\tau \leq t\} = \{\omega_1 : \sigma(\omega_1) \leq t\} \cup \{\omega_2 : \tau(\omega_2) \leq t\} := RHS$

Now let's expand the LHS of the solution: $\{min(\sigma,\tau) \leq t\} = \{ \omega_1 : \sigma(\omega_1) \leq \tau(\omega_2) \leq t\}\cup\{ \omega_2 : \tau(\omega_2) \leq \sigma(\omega_1) \leq t\}:=LHS $

By this, it's obvious that $LHS \subseteq RHS$, and is not necessarily equal to what's stated in the solution.

MY Question: Where have I gone wrong? I'm newb! I know that my $LHS$ is wrong but I'm not sure how to improve my confusion.

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1 Answer

up vote 1 down vote accepted

The mistake is in the expansion of the LHS: the information $\min(\sigma(\omega),\tau(\omega))\leqslant t$ just gives that one of the numbers $\sigma(\omega)$ or $\tau(\omega)$ is $\leqslant t$, not that both are. If $a,b$ and $c$ are three real numbers and $\min\{a,b\}\leqslant c$, you can say that either $a\leqslant c$ or $b\leqslant c$, but we may have $a=-100$, $b=100$ and $c=0$, for example.

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You're right! Thanks. –  Jase Nov 8 '12 at 16:16
    
So you're saying that in fact $\{min(\sigma,\tau)\leq t\} =\{\omega_1 : \sigma(\omega_1) \leq t\} \cup \{\omega_2 : \tau(\omega_2) \leq t\} $ –  Jase Nov 8 '12 at 16:23
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Yes (it's exactly what is written in the solution) –  Davide Giraudo Nov 8 '12 at 16:25
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