59.10. CONDITIONAL EXPECTATION 1889

Corollary 59.10.6 U[a,b] (ω) ≤ the number of unbroken strings of ones in the sequence,{Yk (ω)} there being at most one unbroken string of ones which produces no upcrossing.Also

Yi (ω) = ψ i

({X j (ω)

}i−1j=1

), (59.10.16)

where ψ i is some function of the past values of X j (ω).

Lemma 59.10.7 (upcrossing lemma) Let {Xi}ni=1 be a submartingale and suppose

E (|Xn|)< ∞.

Then

E(U[a,b]

)≤ E (|Xn|)+ |a|

b−a.

Proof: Let φ (x)≡ a+(x−a)+. Thus φ is a convex and increasing function.

φ (Xk+r)−φ (Xk) =k+r

∑i=k+1

φ (Xi)−φ (Xi−1)

=k+r

∑i=k+1

(φ (Xi)−φ (Xi−1))Yi +k+r

∑i=k+1

(φ (Xi)−φ (Xi−1))(1−Yi).

The upcrossings of φ (Xi) are exactly the same as the upcrossings of Xi and from Formula59.10.16,

E

(k+r

∑i=k+1

(φ (Xi)−φ (Xi−1))(1−Yi)

)

=k+r

∑i=k+1

∫Ri(φ (xi)−φ (xi−1))

(1−ψ i

({x j}i−1

j=1

))dλ (X1,···,Xi)

=k+r

∑i=k+1

∫Ri−1

∫R(φ (xi)−φ (xi−1))·(

1−ψ i

({x j}i−1

j=1

))dλ Xi|x1···xi−1dλ (X1,···,Xi−1)

=k+r

∑i=k+1

∫Ri−1

(1−ψ i

({x j}i−1

j=1

)).

∫R(φ (xi)−φ (xi−1))dλ Xi|x1···xi−1dλ (X1,···,Xi−1)

By Jensen’s inequality, Problem 10 of Chapter 15,

≥k+r

∑i=k+1

∫Ri−1

(1−ψ i

({x j}i−1

j=1

))·

[φ (E (Xi|x1, · · ·,xi−1))−φ (xi−1)]dλ (X1,···,Xi−1)