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Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and let $T:\Omega \to \mathbb N$ a stoping time. Let $(X_n)_{n\geq 1}$ a stochastic process and $\mathcal F_n=\sigma (X_1,...,X_n)$, and $\mathcal F_\infty =\sigma (X_1,X_2,...)$.

I don't really understand what $$\mathcal F_T=\{A\cap \{T=n\}\in \mathcal F_n\mid A\in \mathcal F_\infty \}.$$ I saw on the internet that it's called the $\sigma -$algebra of previous event, but that doesn't really enlighten me.

Also, what would be $$\mathcal G_T=\{A\cap \{T=n\}\in \mathcal F_n\mid A\in \mathcal F\} \ \ ?$$ is $\mathcal G_T$ interesting ?

user657324
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2 Answers2

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When you are studying a process $\{X_n\}$ you might as well assume that $\mathcal F =\mathcal F_{\infty}$. Sets in $\mathcal F$ which don't belong to $\mathcal F_{\infty}$ don't come into the picture in the analysis of $\{X_n\}$. Se we can take $\mathcal F_T=\mathcal G_T$.

$\mathcal F_T$ consists (intuitively) of events $A$ such that when you restrict your attention to $\{T=n\}$ the events is defined completely in terms of $X_1,X_2,...,X_n$ (and $X_k$ with $k >n$ are not involved).

Kavi Rama Murthy
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  • Can we say that $\mathcal F_T$ is the trace $\sigma -$algebra of $T$ ? If yes, is this the smallest $\sigma -$algebra s.t. $T$ is measurable ? – user657324 May 15 '19 at 14:32
  • $\mathcal F_T$ is not the smallest sigma algebra s.t. $T$ is measurable even when $T$ is a constant. (Every constant $T$ is a stopping time and when $T$ is a constant the smallest sigma algebra s.t. $T$ is measurable is simply ${\emptyset,\Omega}$ whereas $\mathcal F_T$ is the sigma algebra generated by ${X_s: s\leq t})$. – Kavi Rama Murthy May 15 '19 at 23:07
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Note that one can prove the equivalent following definition :

$$\mathcal F_T = \bigvee^\infty_0 \sigma(x_{T \wedge n }) $$ where $x_{T \wedge n } $ represents the process equal to $x_n$ when $n$ hasn't reached the stopping time.

Marine Galantin
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