62.10. CONTINUOUS SUBMARTINGALE CONVERGENCE THEOREM 2103

Let

UnM[a,b] ≡ lim

ε→0

n

∑k=0

X (τ2k+1)−X (τ2k)

ε +X (τ2k+1)−X (τ2k)

Note that if an upcrossing occurs after τ2k on [0,M], then τ2k+1 > τ2k because there existst such that

(X (t ∨ τ2k)−X (τ2k))+ = b−a

However, you could have τ2k+1 > τ2k without an upcrossing occuring. This happens whenτ2k < M and τ2k+1 = M which may mean that X (t) never again climbs to b. You break thesum into those terms where X (τ2k+1)−X (τ2k) = b− a and those where this is less thanb− a. Suppose for a fixed ω, the terms where the difference is b− a are for k ≤ m. Thenthere might be a last term for which X (τ2k+1)−X (τ2k)< b−a because it fails to completethe up crossing. There is only one of these at k = m+1. Then the above sum is

≤ 1b−a

m

∑k=0

X (τ2k+1)−X (τ2k)+X (M)−a

ε +X (M)−a

≤ 1b−a

n

∑k=0

X (τ2k+1)−X (τ2k)+X (M)−a

ε +X (M)−a

≤ 1b−a

n

∑k=0

X (τ2k+1)−X (τ2k)+1

Then UnM[a,b] is clearly a random variable which is at least as large as the number of

upcrossings occurring for t ≤M using only 2n+1 of the stopping times. From the optionalsampling theorem,

E (X (τ2k))−E (X (τ2k−1)) =∫

Ω

X (τ2k)−X (τ2k−1)dP

=∫

Ω

E(X (τ2k) |Fτ2k−1

)−X (τ2k−1)dP

≥∫

Ω

X (τ2k−1)−X (τ2k−1)dP = 0

Note that, X (τ2k) = a while X (τ2k−1) = b so the above may seem surprising. However,the two stopping times can both equal M so this is actually possible. For example, it couldhappen that X (t) = a for all t ∈ [0,M].

Next, take the expectation of both sides,

E(

UnM[a,b]

)≤ 1

b−a

n

∑k=0

E (X (τ2k+1))−E (X (τ2k))+1

≤ 1b−a

n

∑k=0

E (X (τ2k+1))−E (X (τ2k))+1

b−a

n

∑k=1

E (X (τ2k))−E (X (τ2k−1))+1

=1

b−a(E (X (τ1))−E (X (τ0)))+

1b−a

n

∑k=1

E (X (τ2k+1))−E (X (τ2k−1))+1