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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(τ,θ) is a stopping time, and moreover for every integrable random variable X,

E(E(X|Fθ)|Fτ)=E(E(X|Fτ)|Fθ)=E(X|Fτθ).

Note that if both τ and θ are deterministic (i.e. constant) then we recover the usual tower property of conditional expectations. Note also that the events {τθ}, {τ=θ}, and {τθ} all belong to FτFθ, and that FτθFτFθ.

Proof. The fact that τθ 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 AFθ, by denoting Y:=E(X|Fτθ),

E(Y1A)=E(E(X|Fτ)1A).

We start by observing that Y1{τθ}=E(X|Fτ)1{τθ} which gives immediately

E(Y1A1{τθ})=E(E(X|Fτ)1A1{τθ}).

On the other hand, we have

E(Y1A1{τ>θ})=k=0E(E(X|Fk)1A{τ>k}{θ=k}).

Now, since Bk:=A{τ>k}{θ=k}FkFτ,

E(E(X|Fk)1Bk)=E(X1Bk)=E(E(X|Fτ)1Bk)

which gives, by summing over k,

E(Y1A1{τ>θ})=E(E(X|Fτ)1A1{τ>θ}).

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