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An exercise on stopping times

Dice

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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