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Let

  • $(\Omega,\mathcal A)$ and $(E,\mathcal E)$ be measurable spaces
  • $I\subseteq[0,\infty)$ be at most countable and closed under addition with $0\in I$
  • $X=(X_t)_{t\in I}$ be a stochastic process on $(\Omega,\mathcal A)$ with values in $(E,\mathcal E)$
  • $\mathbb F=(\mathcal F_t)_{t\in I}$ be the filtration generated by $X$
  • $\tau$ be a $\mathbb F$-stopping time
  • $f:E^I\to\mathbb R$ be bounded and $\mathcal E^{\otimes I}$-measurable

Clearly, $$Y_s:=1_{\left\{\tau=s\right\}}\operatorname E\left[f\circ\left(X_{s+t}\right)_{t\in I}\mid\mathcal F_\tau\right]$$ is $\mathcal F_s$-measurable. Thus,

\begin{equation} \begin{split} \operatorname E\left[f\circ\left(X_{\tau+t}\right)_{t\in I}\mid\mathcal F_\tau\right]&=&\sum_{s\in I}Y_s\\&=&\sum_{s\in I}\operatorname E\left[Y_s\mid\mathcal F_s\right]\\&\color{red}=&\color{red}{\sum_{s\in I}\operatorname E\left[1_{\left\{\tau=s\right\}}\operatorname E\left[f\circ\left(X_{\tau+t}\right)_{t\in I}\mid\mathcal F_s\right]\mid\mathcal F_\tau\right]}\;, \end{split} \end{equation}

but I don't understand why the $\color{red}{\text{red}}$ part is true. It looks like the tower property, but we shouldn't be able to use it unless $\mathcal F_\tau\subseteq\mathcal F_s$, which is obviously wrong. So, how do we need to argue?

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    $\begingroup$ This does not look like the last line of this argument. What is being proved here? $\endgroup$ – zhoraster Sep 30 '15 at 19:21
  • $\begingroup$ @zhoraster What's really being proved is exactly the theorem in the following question: math.stackexchange.com/questions/1278716/… $\endgroup$ – 0xbadf00d Sep 30 '15 at 19:24
  • $\begingroup$ Seems that the second line is unnecessary. Or should there be $\mathcal F_\tau$? Will think a little... $\endgroup$ – zhoraster Sep 30 '15 at 19:29
  • $\begingroup$ @zhoraster No, it's necessary in order to apply the weak Markov property. $\endgroup$ – 0xbadf00d Sep 30 '15 at 19:55
  • $\begingroup$ @zhoraster Why have you deleted your answer? I don't know what you meant by "some limit procedure". Please take a look at the following question: math.stackexchange.com/questions/1456339/…. If $Y$ is $\mathcal F_\tau$ measurable, $1_{\left\{\tau=s\right\}}Y$ is $\mathcal F_s$-measurable and one doesn't need any limit argument in order to prove that. $\endgroup$ – 0xbadf00d Sep 30 '15 at 19:57
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I wouldn't regard John Dawkin's answer as a valid argument, but he is right.

Denote $\eta = f\circ\left(X_{\tau+t}\right)_{t\in I}$ and take $A\in \mathcal F_\tau$. Then $A\cap\{\tau = s\}\in \mathcal F_s$, so$$E[\mathbf{1}_A \mathbf{1}_{\{\tau=s\}} E[\eta\mid \mathcal F_s]] = E[\mathbf{1}_A\mathbf{1}_{\{\tau=s\}}\eta].$$ It follows that $$ E[ \mathbf{1}_{\{\tau=s\}} E[\eta\mid \mathcal F_s]\mid\mathcal F_\tau] = E[\mathbf{1}_{\{\tau=s\}} \eta\mid \mathcal F_\tau] = Y_s. $$

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  • $\begingroup$ First of all, you've really helped me so far. However, this is still not exactly the solution. Maybe it's a typo, but you should consider $\eta=f\circ (X_{\color{red}s+t})_{t\in I}$ not $\eta=f\circ (X_{\color{red}\tau+t})_{t\in I}$. Moreover, $$Y_s=1_{\left\{\tau=s\right\}}\operatorname E\left[f\circ\left(X_{s+t}\right)_{t\in I}\mid\mathcal F_{\color{red}\tau}\right]\;,$$ but using the "local property" for your objects yields $\endgroup$ – 0xbadf00d Oct 1 '15 at 9:42
  • $\begingroup$ $$\operatorname E_x\left[1_{\left\{\tau=s\right\}}\operatorname E_x\left[f\circ\left(X_{\tau+t}\right)_{t\in I}\mid\mathcal F_s\right]\mid\mathcal F_\tau\right]=1_{\left\{\tau=s\right\}}\operatorname E_x\left[f\circ (X_{s+t})_{t\in I}\mid\mathcal F_{\color{red}s}\right]\;.$$ Since we obviously need to "swap $\mathcal F_s$ and $\mathcal F_\tau$" it seems like this argumentation doesn't help at all. $\endgroup$ – 0xbadf00d Oct 1 '15 at 10:51
  • $\begingroup$ @0xbadf00d, this does not matter, but I returned it. $\endgroup$ – zhoraster Oct 1 '15 at 12:16
  • $\begingroup$ Do we need $\mathcal F_s$-measurability of $Y_s$ at all? Let $A\in\mathcal F_\tau$. As you mentioned, $$A\cap\left\{\tau=s\right\}\in\mathcal F_\tau\tag 1$$ and $$A\cap\left\{\tau=s\right\}\in\mathcal F_s\tag 2\;.$$ Now, $Y_s$ is $\mathcal F_{\color{red}\tau}$-measurable. If we first use the definition of $\operatorname E_x\left[f\circ (X_{s+t})_{t\in I}\mid\mathcal F_\tau\right]$ together with $(1)$ and then the definition of $\operatorname E_x\left[f\circ (X_{s+t})_{t\in I}\mid\mathcal F_s\right]$ together with $(2)$, we obtain $\endgroup$ – 0xbadf00d Oct 1 '15 at 13:03
  • $\begingroup$ $$\operatorname E_x\left[1_AY_s\right]=\operatorname E_x\left[1_A1_{\left\{\tau=s\right\}}f\circ (X_{s+t})_{t\in I}\right]=\operatorname E_x\left[1_A1_{\left\{\tau=s\right\}}\operatorname E_x\left[f\circ (X_{s+t})_{t\in I}\mid\mathcal F_s\right]\right]$$ and can conclude that $$Y_s=\operatorname E_x\left[1_{\left\{\tau=s\right\}}\operatorname E_x\left[f\circ (X_{s+t})_{t\in I}\mid\mathcal F_s\right]\mid\mathcal F_\tau\right]\;.$$ So, if I'm not terribly wrong, one doesn't need the $\mathcal F_s$-measurability anywhere and I wonder why this was announced to be important in the first place. $\endgroup$ – 0xbadf00d Oct 1 '15 at 13:08
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The red part is equal to the line just above it because (i) the event $\{\tau=s\}$ is both ${\mathcal F}_s$ measurable and ${\mathcal F}_\tau$ measurable, and (ii) those two $\sigma$-algebras coincide on $\{\tau=s\}$, in the sense that if $A$ is an event and $s\in I$ is fixed then $A\cap\{\tau=s\}\in{\mathcal F}_s$ if and only if $A\cap\{\tau=s\}\in{\mathcal F}_\tau$.

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  • $\begingroup$ They why not write this from the very beginning? Does not sound too convincing to me... $\endgroup$ – zhoraster Sep 30 '15 at 19:52
  • $\begingroup$ I don't think that this property is sufficient. Can you prove it? $\endgroup$ – 0xbadf00d Sep 30 '15 at 19:59
  • $\begingroup$ Just use the definition of ${\mathcal F}_\tau$. $\endgroup$ – John Dawkins Sep 30 '15 at 20:10
  • $\begingroup$ I don't get it. What you write is clear to me, but I don't see how we can make use of this property in the definition (or any property) of conditional expectation. $\endgroup$ – 0xbadf00d Sep 30 '15 at 20:29
  • $\begingroup$ For example: $Y_s$ is ${\mathcal F}_s$ measurable, so $E(Y_s|{\mathcal F}_s) =Y_s$; also, if $F$ is bounded then $E(F|{\mathcal F}_\tau) = E(F|{\mathcal F}_s)$ on $\{\tau=s\}$, in the sense of me earlier comment. $\endgroup$ – John Dawkins Sep 30 '15 at 20:39

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