Characterization of Stopping Times with respect to Right-Limit Filtration

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Theorem

Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.

Let $T : \Omega \to \closedint 0 \infty$ be a random variable.

Let $\sequence {\GG_t}_{t \ge 0}$ be the right-limit filtration associated with $\sequence {\FF_t}_{t \ge 0}$.


The following statements are equivalent:

$(1) \quad$ $T$ is a stopping time with respect to $\sequence {\GG_t}_{t \ge 0}$
$(2) \quad$ for each $t \in \hointr 0 \infty$, we have $\set {\omega \in \Omega : \map T \omega < t} \in \FF_t$
$(3) \quad$ for each $t \in \hointr 0 \infty$, the pointwise minimum $T \wedge t$ is $\FF_t$-measurable.


Proof

$(1)$ implies $(2)$

Suppose $T$ is a stopping time with respect to $\sequence {\GG_t}_{t \ge 0}$ and let $t \in \hointr 0 \infty$.

Then, for each $n \in \N$ we have:

$\ds \set {\omega \in \Omega : \map T \omega \le t - \frac 1 n} \in \GG_{t - \frac 1 n}$

Since by the definition of the right-limit filtration, we have:

$\ds \GG_{t - \frac 1 n} = \bigcap_{s > t - \frac 1 n} \FF_s$

we have:

$\GG_{t - \frac 1 n} \subseteq \FF_s$ for all $\ds s > t - \frac 1 n$

and in particular:

$\GG_{t - \frac 1 n} \subseteq \FF_t$

and hence:

$\ds \set {\omega \in \Omega : \map T \omega \le t - \frac 1 n} \in \FF_t$

Note that:

$\ds \set {\omega \in \Omega : \map T \omega < t} = \bigcup_{n \mathop = 1}^\infty \set {\omega \in \Omega : \map T \omega \le t - \frac 1 n}$

Since $\FF_t$ is closed under countable union, we have:

$\ds \bigcup_{n \mathop = 1}^\infty \set {\omega \in \Omega : \map T \omega \le t - \frac 1 n} \in \FF_t$

and hence:

$\set {\omega \in \Omega : \map T \omega < t} \in \FF_t$

$\Box$

$(2)$ implies $(1)$

Suppose that for each $t \in \hointr 0 \infty$, we have $\set {\omega \in \Omega : \map T \omega < t} \in \FF_t$

Let $t \in \hointr 0 \infty$ and $s > t$.

We have:

$\ds \set {\omega \in \Omega : \map T \omega < t + \frac 1 n} \in \FF_{t + \frac 1 n}$

Since:

$\ds t + \frac 1 n \to t$

and the sequence is decreasing, there exists $N_t \in \N$ such that:

$\ds t + \frac 1 n < s$

for $n \ge N_t$.

Then, for $n \ge N_t$, we have:

$\ds \set {\omega \in \Omega : \map T \omega < t + \frac 1 n} \in \FF_s$

since $\sequence {\FF_t}_{t \ge 0}$ is a filtration.

Now note that:

$\ds \set {\omega \in \Omega : \map T \omega \le t} = \bigcap_{n \mathop = N_t}^\infty \set {\omega \in \Omega : \map T \omega < t + \frac 1 n}$

Since $\FF_s$ is closed under countable intersection, we have:

$\ds \set {\omega \in \Omega : \map T \omega \le t} \in \FF_s$ for each $s > t$.

So, by the definition of the right-limit filtration, we have:

$\ds \set {\omega \in \Omega : \map T \omega \le t} \in \GG_t$

So $T$ is a stopping time with respect to $\sequence {\GG_t}_{t \ge 0}$.

$\Box$

$(1)$ implies $(3)$

Suppose that $T$ is a stopping time with respect to $\sequence {\GG_t}_{t \ge 0}$.

Then for each $s < t$, we have:

$\set {\omega \in \Omega : \map T \omega \le s} \in \GG_s$

and hence:

$\set {\omega \in \Omega : \map T \omega \le s} \in \FF_t$

since $\GG_s \subseteq \FF_t$ for $s < t$.

Now note that if $\map T \omega \le s$ for $s < t$, we have:

$\min \set {\map T \omega, t} \le s$

Conversely, if:

$\min \set {\map T \omega, t} \le s < t$

we must have $\map T \omega \le s$.

So, we have:

$\set {\omega \in \Omega : \map {\paren {T \wedge t} } \omega \le s} \in \FF_t$

If $s \ge t$, then:

$\set {\omega \in \Omega : \map {\paren {T \wedge t} } \omega \le s} = \Omega \in \FF_t$

Since $\FF_t$ is a $\sigma$-algebra.

We conclude that $T \wedge t$ is $\FF_t$-measurable.

$\Box$

$(3)$ implies $(2)$

Suppose that for each $t \in \hointr 0 \infty$, the pointwise minimum $T \wedge t$ is $\FF_t$-measurable.

Fix $t \in \hointr 0 \infty$, we want to show that:

$\set {\omega \in \Omega : \map T \omega < t} \in \FF_t$

We have:

$\set {\omega \in \Omega : \map {\paren {T \wedge t} } \omega \le s} \in \FF_t$

for all $s \in \hointr 0 \infty$.

That is:

$\set {\omega \in \Omega : \map T \omega \le s} \in \FF_t$

for $s < t$, so that:

$\ds \set {\omega \in \Omega : \map T \omega \le t - \frac 1 n} \in \FF_t$

for each $n \in \N$.

Since $\FF_t$ is closed under countable union, we have:

$\ds \bigcup_{n \mathop = 1}^\infty \set {\omega \in \Omega : \map T \omega \le t - \frac 1 n} \in \FF_t$

That is:

$\set {\omega \in \Omega : \map T \omega < t} \in \FF_t$

$\blacksquare$


Sources

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