Simple question regarding stopping times.

Question

I have this exercise regarding stopping times, but I am not able to solve it.

You have a probability space $(\Omega, \mathcal{F},P)$, with a filtration $\{\mathcal{F}_t\}$. You have two stopping times where $S,T$, where $S\le T $ a.s.

I am supposed to show that $\mathcal{F}_S\subseteq\mathcal{F}_T$.

Here is what I have so far:

We have that by definition $\mathcal{F}_S=\{F \in \mathcal{F}: F\cap\{S\le t\}\in \mathcal{F}_t~\forall t\}$,$\mathcal{F}_T=\{F \in \mathcal{F}:F\cap\{T\le t\}\in \mathcal{F}_t~\forall t\}$.

So assume that $F \in \mathcal{F}_S$. For any $t,$ I then have that $F\cap\{S \le t\}\in \mathcal{F}_t$, and I must show that $F\cap\{T\le t $}$\in \mathcal{F}_t$.

Call $\{S \le t$} event $S_1$, $\{T \le t$} event $T_1$. Then $S,T$ are stopping times, so these events are in $\mathcal{F}_t$. So in summary I have $S_1\in \mathcal{F}_t, T_1 \in \mathcal{F}_t, S_1 \cap F \in \mathcal{F}_t$. I do not know if $F \in \mathcal{F}_t$. But what I at least get is that $S_1\cap T_1\cap F \in \mathcal{F}_t$.

What I have not yet used is that $S\le T$ a.s. My only idea is then to split the event $T_1$ in different cases. But I am not sure how I am supposed to split it up. Can you please help me?

Answer 1

Since $S \leq T$, we have

$$\{T \leq t\} = \{S \leq t\} \cap \{T \leq t\}.$$

Hence,

$$F \cap \{T \leq t\} = \underbrace{(F \cap \{S \leq t\})}_{\in \mathcal{F}_t} \cap \underbrace{\{T \leq t\}}_{\in \mathcal{F}_t} \in \mathcal{F}_t$$

for any $F \in \mathcal{F}_S$ and $t \geq 0$.

Answer 2

The key here is that $\{S\le t\} \subset \{T \le t\}$ always (by the assumptions made in the problem). You already have that $S_1 \cap T_1 \cap F$ is in T's sigma algebra, now just note that $S_1 \cap T_1 = S_1$.