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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 (Ω,F,P) be a probability space equipped with a filtration (Fn)n0. Recall that a random variable τ taking values in N={0,1,} is a stopping time when {τ=n}Fn for any nN. We define the σ-field Fτ:={AF:n,{τ=n}AFn}. Remarkably, for any integrable random variable X and for any nN,

E(X|Fτ)1{τ=n}=E(X|Fn)1{τ=n}.

To see it, for every AFn, since A{τ=n}Fτ and {τ=n}Fτ, we get

E(1AX1{τ=n})=E(1A{τ=n}E(X|Fτ))=E(1AE(X1{τ=n}|Fτ)),

and therefore E(X1{τ=n}|Fn)=E(X1{τ=n}|Fτ). Now {τ=n}FnFτ.

A nice property. If τ and θ are stopping times then τθ:=min 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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