27.3. CONDITIONAL EXPECTATION, SUB-MARTINGALES 743

=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) (27.5)

By Jensen’s inequality, Lemma 27.3.4 and that this is a sub-martingale, and that φ is in-creasing in addition to being convex,∫

Rφ (xi)dλ Xi|x1···xi−1 ≥ φ

(∫R

xidλ Xi|x1···xi−1

)≥ φ (xi−1) .

Therefore, from 27.5, E(∑

k+ri=k+1 (φ (Xi)−φ (Xi−1))(1−Yi)

)≥

k+r

∑i=k+1

∫Ri−1

(1−ψ i

({x j}i−1

j=1

))(φ (xi−1)−φ (xi−1))dλ (X1,··· ,Xi−1) = 0

Now let the unbroken strings of ones for {Yi (ω)} be

{k1, · · · ,k1 + r1} ,{k2, · · · ,k2 + r2} , · · · ,{km, · · · ,km + rm} (27.6)

where m = V (ω) ≡ the number of unbroken strings of ones in the sequence {Yi (ω)}. ByCorollary 27.3.7 V (ω)≥U[a,b] (ω).

φ (Xn (ω))−φ (X1 (ω))

=n

∑k=1

(φ (Xk (ω))−φ (Xk−1 (ω)))Yk (ω)

+n

∑k=1

(φ (Xk (ω))−φ (Xk−1 (ω)))(1−Yk (ω)).

Summing the first sum over the unbroken strings of ones (the terms in which Yi (ω) = 0contribute nothing), and observing that for x > a,φ (x) = x,

φ (Xn (ω))−φ (X1 (ω))≥U[a,b] (ω)(b−a)+0+

n

∑k=1

(φ (Xk (ω))−φ (Xk−1 (ω)))(1−Yk (ω)) (27.7)

where the zero on the right side results from a string of ones which does not produce anupcrossing. It is here that we use φ (x) ≥ a. Such a string begins with φ (Xk (ω)) = a andresults in an expression of the form φ (Xk+m (ω))−φ (Xk (ω))≥ 0 since φ (Xk+m (ω))≥ a.