Use of the tower property of conditional expectation

Question

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?

Answer 1

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 $$ E[ \mathbf{1}_{\{\tau=s\}} E[\eta\mid \mathcal F_s]\mid\mathcal F_\tau] = E[\mathbf{1}_{\{\tau=s\}} \eta\mid \mathcal F_\tau] = Y_s. $$

Answer 2

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 $A\cap\{\tau=s\}\in{\mathcal F}_\tau$.