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For a stopping time T, define $\mathcal{F}_T$ by

$\mathcal{F}_T={A \in \mathcal{F}:A \cap \{T \le t\} \in \mathcal{F}_t, \text{for each t.}}$

Verify that $\mathcal{F}_T$ is a $\sigma$-algebra.

To prove this I know we must show that $\mathcal{F}_T$ satisfies the three properties of $\sigma$-algebras, ie that

A $\sigma$-algebra $\mathcal{F}$ of subsets of a set X is a collection $\mathcal{F}$ of subsets satisfying the following properties;
(1) $\emptyset$ $\in$ $\mathcal{F}$
(2) if A $\in$ $\mathcal{F}$ then it's compliment $A^{c}$ is also in $\mathcal{F}$
(3) if $A_1,A_2,...$ is a countable collection of sets in $\mathcal{F}$ then their union $\cup^{\infty}_{n=1}A_n \in \mathcal{F}$

I have attempted this myself but am concerned I am not being precise enough. Any hints or comments are gladly welcome.

For (3):
Suppose $A_1,A_2,...$ is a countable collection of sets in $\mathcal{F}_T$
Then $A_n \in \mathcal{F}$ : $A_n \cap \{T \le t\} \in \mathcal{F}_t$ for all n by definition
Then $\cup^{\infty}_{n=1}(A_n \cap \{T \le t\})\in \mathcal{F}_t$
Then $(\cup^{\infty}_{n=1}A_n) \cap (\{T \le t\})\in \mathcal{F}_t$ Then $(\cup^{\infty}_{n=1}A_n) \in \mathcal{F}_T$

For(2) I am less convinced
My reasoning would be that if $A \in \mathcal{F}_T$ then $A \in \mathcal{F}: A \cap \{T \le t\} \in \mathcal{F}_t$
We know that $A \cap \{T \le t\} \in \mathcal{F}_t$
Now $A^c \in \mathcal{F}$ since $A^c = \mathcal{F}/A$
Then $A^{c} \cap \{T \le t\} = \mathcal{F} \cap \{T \le t\} - A \cap \{T \le t\}$
Since $\mathcal{F} \cap \{T \le t\} = \mathcal{F}_t$
Then $A^{c} \cap \{T \le t\} = \mathcal{F}_t - A \cap \{T \le t\}$
ie $A^{c} \in \mathcal{F}_T$
However for this to hold we must consider the situation where $A \cap \{T \le t\} = \mathcal{F}_t$
In other words here $A^c$ would be $\emptyset$ and this is where I'm stuck.

If I could prove $\emptyset \in \mathcal{F}_T$ (2) would follow so any hints to get me started would be really appreciated as I really don't know where to start, or if I am completely wrong that would also be good to know! Thanks in advance!

user325403
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  • You seem to refer to $\mathcal F_T$ in your definition of $\mathcal F_T$. That makes it unclear to me. – drhab Mar 23 '16 at 18:25
  • Yes I thought that strange as well when reading the problem for the first time but it is the standard definition for this type of filtration with stopping time T – user325403 Mar 23 '16 at 18:31
  • No, it should read $A \cap {T \leq t} \in \mathcal{F}_t$. – saz Mar 23 '16 at 18:49
  • You're both completely correct, my apologies, it's a typo on my problem sheet and I didn't notice the subtle difference when checking the definition online, stay with me while i check if any of my work still makes sense or if it's just utter nonsense. – user325403 Mar 23 '16 at 18:59
  • As far as i can tell it hasn't change my reasoning, but obviously my reasoning could be flawed from the beginning so feel free to tell me where I'm going wrong! Thank you both very much for the correction though, I never would have spotted it! – user325403 Mar 23 '16 at 19:10
  • Could you add the definition of $\mathcal F_t$ for a fixed $t$? – drhab Mar 24 '16 at 09:31
  • I recently answered a similar question here: maybe this is helpful. – Cettt Dec 21 '16 at 08:24

2 Answers2

3

\begin{multline} 1) \emptyset : \emptyset \cap \{T\leq n\} =\emptyset \in \mathcal{F}_n \Rightarrow \emptyset \in \mathcal{F}_n \end{multline}

\begin{multline} 2) \Omega : \Omega \cap \{T\leq n\} =\{T\leq n\} \in \mathcal{F}_n \Rightarrow \Omega \in \mathcal{F}_n \end{multline}

\begin{multline} 3) A \in \mathcal{F}_n : A \cap \{T\leq n\} \in \mathcal{F}_n \forall n \end{multline}

We have to prove that $A^c \cap \{T\leq n\}\in \mathcal{F}_n$ \begin{multline} 4) A^c \cap \{T\leq n\} = \underbrace{(A^c \cap \{T\leq n\})^c}_{\in \mathcal{F}_n} \cap \underbrace{\{T\leq n\}}_{\in \mathcal{F}_n}\in \mathcal{F}_n \end{multline}

\begin{multline} 5) A_k \in \mathcal{F}:(\bigcup^{\infty}_{k=1}A_k) \cap \{T\leq n\} = \bigcup^{\infty}_{k=1} \underbrace{(A_k \cap \{T\leq n\})}_{\in \mathcal{F}_n}\in \mathcal{F}_n \end{multline}

Koval Boris
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1

Your argument for (2) doesn't make sense - in particular, the statement $A^c=\mathcal{F}-A$ is not defined: On the lefthand side is a set; on the righthand side is a collection of sets.

Start with $A^c=\Omega-A$ (if $\Omega$ is the set on which the filtration is taken), and note $A=(A\cap\{T\le t\})\cup(A^c\cap \{T\le t\})$.

manofbear
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