0

Let $(\Omega, (\mathcal F_t)_{t\geq 0}, \mathcal F)$ be a filtered measurable space and $(X_t)_{t\geq0}$ be a progressively measureable process. Let $T: \Omega \rightarrow [0,\infty]$ be a stopping time and $\mathcal F_T$ be defined by

$$ \mathcal F_T := \{A \in \mathcal F: A \cap \{T \leq t \} \in \mathcal F_t \forall t\geq0 \} $$

What does it mean by $X_T$ defined on $\{ T < \infty\}$ is $\mathcal F_T$-measurable?

This question buzzed my head a lot since $X_T$ is not defined on $\{ T = \infty\}$ but we are considering it as a random variable. However, I dont see on which probability space it is a measurable map. My take is we are viewing

$$X_T: (\{ T< \infty\}, \mathcal F_T \cap \{ T < \infty \}) \longrightarrow (\mathbb R^n, \mathcal B(\mathbb R^n))? $$

Or may be $ X_T 1_{T< \infty} $ is $\mathcal F_T$-measurable as a map from $(\Omega, \mathcal F_T)$ to $(\mathbb R^n, \mathcal B(\mathbb R^n))$?

I am not looking for a proof, but instead a correct interpretation of this please. In fact, it has been asked in this link but people only gave proofs of this and no interpretation was given at all. Here.

I hope to receive a clear explanation please. Thank you very much in advance!

Jeffrey Jao
  • 444
  • 1
    I think the interpretation is: $ X_T 1_{T< \infty} $ is $\mathcal F_T$-measurable as a map from $(\Omega, \mathcal F_T)$ to $(\mathbb R^n, \mathcal B(\mathbb R^n))$ – geetha290krm Jun 26 '23 at 05:01
  • Try and look up progressive measurability. I think there's a section on it in Rene Schillings Brownian Motions book if I recall correctly. – Mr.Gandalf Sauron Jun 26 '23 at 05:10
  • In particular look at page 83 of that book(might vary on different editions) and you'll find exactly what you need. – Mr.Gandalf Sauron Jun 26 '23 at 05:20
  • 1
    So I saw the page you mentioned in first edition of the book, which disappears in later editions. It says that $B_T1_{{ T< \infty}}$ is $\mathcal F_T$ measurable so I think it comes down to what it means by "$Y$ is $\mathcal F_T$-measurable" if $Y$ is a random variable, isn't it...? Could you explain why in page 83 he used $({T \geq t}, \mathcal F_t)$ as a probability space? Clearly $\mathcal F_t$ contains bigger sets than ${ T \leq t}$, shouldn't it be $\mathcal F_t \cap { T\leq t}$? – Jeffrey Jao Jun 26 '23 at 13:56

0 Answers0