60.6. SUBMARTINGALE CONVERGENCE THEOREM 1967

upcrossings, consider the following:

τ0 ≡ min(inf{k : X (k)≤ a} ,M) ,

τ1 ≡ min(inf{

k : (X (k∨ τ0)−X (τ0))+ ≥ b−a},M),

τ2 ≡ min(inf{

k : (X (τ1)−X (k∨ τ1))+ ≥ b−a},M),

τ3 ≡ min(inf{

k : (X (k∨ τ2)−X (τ2))+ ≥ b−a},M),

τ4 ≡ min(inf{

k : (X (τ3)−X (k∨ τ3))+ ≥ b−a},M),

...

As usual, inf( /0)≡ ∞. Are the above stopping times? If α ≥ 0, and τ is a stopping time, isk→ (X (τ)−X (k∨ τ))+ adapted?[

(X (τ)−X (k∨ τ))+ > α]=[(X (τ)−X (k))+ > α

]∩ [τ ≤ k]

Now [(X (τ)−X (k))+ > α

]∩ [τ ≤ k] = ∪k

i=0[(X (i)−X (k))+ > α

]∩ [τ ≤ k] ∈Fk

If α < 0, then [(X (τ1)−X (k∨ τ1))+ > α

]= Ω

and so k→ (X (τ)−X (k∨ τ))+ is adapted. Similarly k→ (X (k∨ τ)−X (τ))+ is adapted.Therefore, all those τk are stopping times.

Now consider the following random variable for odd M, 2n+1 = M

U [a,b]M ≡ lim

ε→0

n

∑k=0

X (τ2k+1)−X (τ2k)

ε +X (τ2k+1)−X (τ2k)≤ 1

b−a

n

∑k=0

X (τ2k+1)−X (τ2k)

Now suppose {X (k)} is a nonnegative submartingale. Then since E (X (2τ) |F2τ−1)≥X (τ2k−1)

E

(n

∑k=1

X (τ2k)−X (τ2k−1)

)≥ 0

Hence

E(

U [a,b]M

)≤ 1

b−a

n

∑k=0

E (X (τ2k+1)−X (τ2k))

≤ 1b−a

n

∑k=0

E (X (τ2k+1)−X (τ2k))+1

b−a

n

∑k=1

E (X (τ2k)−X (τ2k−1))

=1

b−a

n

∑k=0

E (X (τk)−X (τk−1))≤1

b−aE (X (τk))

Now by the optional sampling theorem X (0) ,X (τk) ,X (M) is a submartingale. Therefore,the above is no larger than

1b−a

E (|X (M)|)