62.10. CONTINUOUS SUBMARTINGALE CONVERGENCE THEOREM 2101

Next let

τ = min(inf{t : X (t)<−λ} ,T )

then as before, X (0) ,X (τ) ,X (T ) is a submartingale and so∫[τ<T ]

X (τ)dP+∫[τ=T ]

X (τ)dP =∫

Ω

X (τ)dP≥∫

Ω

X (0)dP

Now for ω ∈ [τ < T ] ,X (t)(ω)<−λ for some t <T and so by right continuity, X (τ)(ω)≤−λ . therefore,

−λ

∫[τ<T ]

dP≥−∫[τ=T ]

X (T )dP+∫

Ω

X (0)dP

If X∗ (T ) < −λ , then from the definition given above, there exists t < T such that X (t) <−λ and so τ < T. If τ < T, then by definition, there exists t < T such that X (t)<−λ andso X∗ (T )<−λ . Hence [τ < T ] = [X∗ (T )<−λ ] . It follows that

P([X∗ (T )<−λ ]) = P([τ < T ])

≤ 1λ

∫[τ=T ]

X (T )dP− 1λ

∫Ω

X (0)dP

≤ 1λ

E (|X (T )|+ |X (0)|)

and this proves 62.9.34.Finally, combining the above two inequalities,

P([sup{|X (s)| : s < T}> λ ])

= P([X∗ (T )<−λ ])+P([X∗ (T )> λ ])

≤ 2λ

E (|X (T )|+ |X (0)|) .

62.10 Continuous Submartingale Convergence TheoremIn this section, {Y (t)} will be a continuous submartingale and a < b. Let

X (t)≡ (Y (t)−a)++a

so X (0)≥ a. Then X is also a submartingale. It is an increasing convex function of one. IfY (t) has an upcrossing of [a,b] , then X (t) starts off at a and ends up at least as large as b.If X (t) has an upcrossing of [a,b] then it must start off at a since it cannot be smaller andit ends up at least as large as b. Thus we can count the upcrossings of Y (t) by consideringthe upcrossings of X (t).