2108 CHAPTER 62. STOCHASTIC PROCESSES

Then the following claim will be important.Claim: E (M (τ)) = 0.Proof of the claim: Let t > 0. Then by the Doob optional sampling theorem,

E (M (τ ∧ t)) = E (E (M (t) |Fτ)) = E (M (t)) (62.11.38)= E (E (M (t) |F0)) = E (M (0)) = 0. (62.11.39)

Observe the martingale Mτ must be bounded because it is stopped when M (t) equals eithera or b. There are two cases according to whether τ = ∞. If τ = ∞, then M (t) never hits aor b so M (t) has values between a and b. In this case Mτ (t) = M (t) ∈ [a,b] . On the otherhand, you could have τ < ∞. Then in this case Mτ (t) is eventually equal to either a or bdepending on which it hits first. In either case, the martingale Mτ is bounded and by themartingale convergence theorem, Theorem 62.10.3, there exists Mτ (∞) such that

limt→∞

Mτ (t)(ω) = Mτ (∞)(ω) = M (τ)(ω)

and since the Mτ (t) are bounded, the dominated convergence theorem implies

E (M (τ)) = limt→∞

E (M (τ ∧ t)) = 0.

This proves the claim.Recall

M∗ (ω)≡ sup{|M (t)(ω)| : t ∈ [0,∞]} .Also note that [τa = τb] = [τ = ∞]. Now from the claim,

0 = E (M (τ)) =∫[τa<τb]

M (τ)dP

+∫[τb<τa]

M (τ)dP+∫[τa=τb]∩[M∗>0]

M (∞)dP (62.11.40)

+∫[τa=τb]∩[M∗=0]

M (∞)dP

The last term equals 0. By continuity, M (τ) is either equal to a or b depending on whetherτa < τb or τb < τa. Thus

0 = E (M (τ)) = aP([τa < τb])

+bP([τb < τa])+∫[τa=τb]∩[M∗>0]

M (∞)dP (62.11.41)

Consider this last term. By the definition, [τa = τb] corresponds to M (t) never hittingeither a or b. Since M (0) = 0, this can only happen if M (t) has values in [a,b] . Therefore,this last term satisfies

aP([τa = τb]∩ [M∗ > 0])

≤∫[τa=τb]∩[M∗>0]

M (∞)dP

≤ bP([τa = τb]∩ [M∗ > 0]) (62.11.42)