Show that $\mathcal{F}_{s \wedge t} = \mathcal{F}_s \cap \mathcal{F}_t$ where $\mathcal{F}_{s}, \mathcal{F}_{t}$ and $\mathcal{F}_{s \wedge t}$ are stopping time sigma fields.
Let's suppose that S < T and so $\mathcal{F}_s < \mathcal{F}_t$
$\mathcal{F}_s \cap \mathcal{F}_t \subset \mathcal{F}_s$
This is obvious since, due to the fact that this is an intersection, $\mathcal{F}_s \cap \mathcal{F}_t$ is both in $\mathcal{F}_s$ AND $\mathcal{F}_t$ so it's trivial.
$\mathcal{F}_s \subset \mathcal{F}_s \cap \mathcal{F}_t$
The only thing that we have to show is that $\mathcal{F}_s \subset \mathcal{F}_t$ since obviously is in $\mathcal{F}_s$.
Define $\mathcal{F}_s$:= $\{A \bigcap (S \leq n)\}$
Now, $F_S \bigcap [T \leq n] = \bigcup^n_{m=0} \ \underbrace{\{A \bigcap (S \leq m)\} \bigcap \ [T \leq m]}_{\in \mathcal{F_n}} $
and basically I'm done.
My questions are:
- Is that correct?
- I assumed that S < T. When I prove the second inclusion, do I have to add even this event in the intersections?
I'm not sure what are you trying to do, by example I don't understand why you write $\mathcal{F}_S:=\{A \bigcap (s \leq n)\}$, as $\mathcal{F}_S$ have already a meaning so you can't define it again, and indeed I don't know what $\{A \bigcap (s \leq n)\}$ means.
To show that $\mathcal{F}_S \subset \mathcal{F}_T$ when $S< T$ you just need the definition of these $\sigma $-algebras. If $\mathcal{F}:=\{\mathcal{F}_t\}_{t\in\mathbb{R}_+}$ is a filtration in the probability space $(\Omega ,\mathcal{H},P)$ and $R$ is a stopping time adapted to $\mathcal{F}$ then its defined
$$ \mathcal{F}_R:=\{A\in \mathcal{H}: A\cap \{R\leqslant t\}\in \mathcal{F}_t\text{ for every }t\in \mathbb{R}_+\} $$ Now, using the definition, we want to show that for arbitrary $A\in \mathcal{H}$ $$ \forall t:A\cap \{S\leqslant t\}\in \mathcal{F}_t\implies \forall t:A\cap \{T\leqslant t\}\in \mathcal{F}_t $$ Now note that $$ A\cap \{T\leqslant t\}=A\cap \{S<T\leqslant t\}=A\cap \{S< t\}\cap \{T\leqslant t\} $$ so everything reduces to show that $A\cap \{S<t\}\in \mathcal{F}_t$ when $A\in \mathcal{F}_S$. However note that $\{S<t\}=\bigcup_{n\in \mathbb{N}}\{S\leqslant t-1/n\}$, and as $A\cap \{S\leqslant t-1/n\}\in \mathcal{F}_t$ for every $n$ it follows that $A\cap \{S<t\}\in \mathcal{F}_t$, and we are done.∎
Now suppose that $S$ and $T$ are not necessarily related by a linear order, they are just stopping times adapted to $\mathcal{F}$. Then we can see that $R:=S\,\land\, T$ is also a stopping time and that $R\leqslant S$ and $R\leqslant T$, so from the previous result it follows that $\mathcal{F}_R\subset \mathcal{F}_S \cap \mathcal{F}_T$. Thus it remains to show that $\mathcal{F}_S \cap \mathcal{F}_T\subset \mathcal{F}_R$, this is equivalent to show that
$$ \forall t: A\cap \{S\leqslant t\}\in \mathcal{F}_t \,\land\, A\cap \{T\leqslant t\}\in \mathcal{F}_t \implies \forall t: A\cap \{R\leqslant t\}\in \mathcal{F}_t $$
However when the LHS holds it is equivalent to say that $$ \forall t: A\cap \{S\leqslant t\}\cap \{T\leqslant t\}\in \mathcal{F}_t\\ \iff \forall t: A\cap \{S\leqslant t\,\land\, T\leqslant t\}\in \mathcal{F}_t\\ \iff \forall t: A \cap \{S\,\land\, T\leqslant t\}\in \mathcal{F}_t\\ \iff \forall t: A \cap \{R\leqslant t\}\in \mathcal{F}_t $$ Therefore $\mathcal{F}_S \cap \mathcal{F}_T=\mathcal{F}_{S\,\land\, T}$.∎