60.6. SUBMARTINGALE CONVERGENCE THEOREM 1973

=

odds − evens︷ ︸︸ ︷n−1

∑k=0

((XT ′2k+1

−a)+−(

XT ′2k−a)+)

+

evens − odds︷ ︸︸ ︷n

∑k=1

((XT ′2k−a)+−(

XT ′2k−1−a)+)

.

Now denote by UN[a,b] the number of upcrossings. When T ′k is such that k is odd,

(XT ′k−a)+

is above b− a and when k is even, it equals 0. Therefore, in the first sum XT ′2k+1−XT ′2k

≥b− a and there are UN

[a,b] terms which are nonzero in this sum. (Note this might not be nbecause many of the terms in the sum could be 0 due to the definition of T ′k .) Hence

(XN−a)+− (X0−a)+ = (XN−a)+

≥ (b−a)UN[a,b]+

n

∑k=1

((XT ′2k−a)+−(

XT ′2k−1−a)+)

. (60.6.15)

Now UN[a,b] is a random variable. To see this, let Zk (ω) = 1 if T ′2k+1 > T ′2k and 0 otherwise.

Thus UN[a,b] (ω) = ∑

n−1k=0 Zk (ω) . Therefore, it makes sense to take the expected value of both

sides of 60.6.15. By the optional sampling theorem,{(

XT ′k−a)+}

is a submartingale andso

E((

XT ′2k−a)+−(

XT ′2k−1−a)+)

=∫

Ω

E((

XT ′2k−a)+|FT ′2k−1

)dP−

∫Ω

(XT ′2k−1

−a)+

dP≥ 0.

Therefore,

E((XN−a)+

)≥ (b−a)E

(UN[a,b]

). (60.6.16)

This proves most of the following fundamental upcrossing estimate.

Theorem 60.6.9 Let {Xn} be a real valued submartingale such that Xn is Fn measurable.Then letting UN

[a,b] denote the upcrossings of {Xn} from a to b for n≤ N,

E(

UN[a,b]

)≤ 1

b−aE((XN−a)+

).

Proof: The estimate 60.6.16 was based on the assumption that X0 (ω) ≤ a. If this isnot so, modify X0. Change it to min(X0,a) . Then the inequality holds for the modifiedsubmartingale which has at least as many upcrossings. Therefore, the inequality remains.

Note this theorem holds if the submartingale starts at the index 1 rather than 0. Justadjust the argument.