This post provides the solution to a tiny exercise of probability theory, answering the question asked by a student during the MAP-432 class yesterday. Let ${(\Omega,\mathcal{F},\mathbb{P})}$ be a probability space equipped with a filtration ${{(\mathcal{F}_n)}_{n\geq0}}$. Recall that a random variable ${\tau}$ taking values in ${\mathbb{N}=\{0,1,\ldots\}}$ is a stopping time when ${\{\tau=n\}\in\mathcal{F}_n}$ for any ${n\in\mathbb{N}}$. We define the ${\sigma}$-field ${\mathcal{F}_\tau:=\{A\in\mathcal{F}:\forall n,\{\tau=n\}\cap A\in\mathcal{F}_n\}}$. Remarkably, for any integrable random variable ${X}$ and for any ${n\in\mathbb{N}}$,

$\mathbb{E}(X|\mathcal{F}_\tau)\mathbf{1}_{\{\tau=n\}} =\mathbb{E}(X|\mathcal{F}_n)\mathbf{1}_{\{\tau=n\}}.$

To see it, for every ${A\in\mathcal{F}_n}$, since ${A\cap\{\tau=n\}\in\mathcal{F}_\tau}$ and ${\{\tau=n\}\in\mathcal{F}_\tau}$, we get

$\mathbb{E}(\mathbf{1}_AX\mathbf{1}_{\{\tau=n\}}) =\mathbb{E}(\mathbf{1}_{A\cap\{\tau=n\}}\mathbb{E}(X|\mathcal{F}_\tau)) =\mathbb{E}(\mathbf{1}_A\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_\tau)),$

and therefore ${\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_n) =\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_\tau)}$. Now ${\{\tau=n\}\in\mathcal{F}_n\cap\mathcal{F}_\tau}$.

A nice property. If ${\tau}$ and ${\theta}$ are stopping times then ${\tau\wedge\theta:=\min(\tau,\theta)}$ is a stopping time, and moreover for every integrable random variable ${X}$,

$\mathbb{E}(\mathbb{E}(X|\mathcal{F}_\theta)|\mathcal{F}_\tau) =\mathbb{E}(\mathbb{E}(X|\mathcal{F}_\tau)|\mathcal{F}_\theta) =\mathbb{E}(X|\mathcal{F}_{\tau\wedge\theta}).$

Note that if both ${\tau}$ and ${\theta}$ are deterministic (i.e. constant) then we recover the usual tower property of conditional expectations. Note also that the events ${\{\tau\leq\theta\}}$, ${\{\tau=\theta\}}$, and ${\{\tau\geq\theta\}}$ all belong to ${\mathcal{F}_\tau\cap\mathcal{F}_\theta}$, and that ${\mathcal{F}_{\tau\wedge\theta}\subset\mathcal{F}_\tau\cap\mathcal{F}_\theta}$.

Proof. The fact that ${\tau\wedge\theta}$ is a stopping time is left to the reader. To prove the property on conditional expectations, it suffices by symmetry to prove the second equality, i.e. that for every ${A\in\mathcal{F}_\theta}$, by denoting ${Y:=\mathbb{E}(X|\mathcal{F}_{\tau\wedge\theta})}$,

$\mathbb{E}(Y\mathbf{1}_A) = \mathbb{E}(\mathbb{E}(X|\mathcal{F}_\tau)\mathbf{1}_A).$

We start by observing that ${Y\mathbf{1}_{\{\tau\leq\theta\}}= \mathbb{E}(X|\mathcal{F}_\tau)\mathbf{1}_{\{\tau\leq\theta\}}}$ which gives immediately

$\mathbb{E}(Y\mathbf{1}_A\mathbf{1}_{\{\tau\leq\theta\}}) =\mathbb{E}(\mathbb{E}(X|\mathcal{F}_\tau)\mathbf{1}_A\mathbf{1}_{\{\tau\leq\theta\}}).$

On the other hand, we have

$\mathbb{E}(Y\mathbf{1}_A\mathbf{1}_{\{\tau>\theta\}}) =\sum_{k=0}^\infty\mathbb{E}(\mathbb{E}(X|\mathcal{F}_k)\mathbf{1}_{A\cap\{\tau>k\}\cap\{\theta=k\}}).$

Now, since ${B_k:=A\cap\{\tau>k\}\cap\{\theta=k\}\in\mathcal{F}_k\cap\mathcal{F}_\tau}$,

$\mathbb{E}(\mathbb{E}(X|\mathcal{F}_k)\mathbf{1}_{B_k}) =\mathbb{E}(X\mathbf{1}_{B_k}) =\mathbb{E}(\mathbb{E}(X|\mathcal{F}_\tau)\mathbf{1}_{B_k})$

which gives, by summing over ${k}$,

$\mathbb{E}(Y\mathbf{1}_A\mathbf{1}_{\{\tau>\theta\}}) =\mathbb{E}(\mathbb{E}(X|\mathcal{F}_\tau)\mathbf{1}_A\mathbf{1}_{\{\tau>\theta\}}).$

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