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Let $N$ be a stopping time. i.e $\{N=n\} \in \mathbb{f}_n \forall n$. $\mathbb{f}_n$ is the filtration.

$\mathbb{f}_N=\{A\in \mathbb{f}, A\cap \{N=n\}\in \mathbb{f}_n \forall n\}$ is the sigma algebra of the stopping time N.

If a sequence of coins are flipped and N is the number of coins until the first head. (so we stop once we see a head). I was asked to determine $\mathbb{f_N}$ explicitly. (i.e., list of events that generate it).

Thus, I need to find all possible events $A$, such that $A\cap \{N=n\}\in \mathbb{f}_n$ $\forall n$.

I think that includes all the events except for the ones that contain more information when a head is already seen. (i.e., if $N=4$, $\{TTTHT\}$ would not be a valid event for $\mathbb{f}_N$.)

But I am really stuck on how do define $\mathbb{f}_N$ explicitly.

Did
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Rann
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1 Answers1

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"f" is the limit $\sigma-algebra$, correct? I think your idea is right. Denote $E_n=\{N=n\}=\{\omega: \omega_1=T,\ldots,\omega_{n-1}=T,\omega_n=H\}$, writing an element of the sequence space as $\omega$. If $A$ contains any element of $E_n$, it contains all of $E_n$. Any non-empty proper subset of $E_n$ is of the form $E_n \cap \cap_{n\in I}\{\omega_i=X_i\}\notin f_n$ where $I$ is a non-empty subset of integers $\ge n+1$ and where $X_i\in\{H,T\}$. In words, $E_n$ completely specifies every outcome up to time $n$, so a non-empty proper subset depends on the future. Since every $\omega$ belongs to $E_n$ for some $n$, it follows that $A=\cup_{\{n\in I\}}E_n$ where $I$ is a subset of the natural numbers.

snarfblaat
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