62.5. SOME MAXIMAL ESTIMATES 2069

Example 62.4.2 Let Ft be a filtration and let Z be in L1 (Ω,FT ,P) . Then let X (t) =E (Z|Ft).

This works because for s < t, E (X (t) |Fs) = E (E (Z|Ft) |Fs) = E (Z|Fs) = X (s).

Proposition 62.4.3 The following statements hold for a stochastic process defined on theproduct [0,T ]×Ω having values in a real separable Banach space, E.

1. If X (t) is a martingale then ||X (t)|| , t ∈ [0,T ] is a submartingale.

2. If g is an increasing convex function from [0,∞) to [0,∞) and E (g(||X (t)||))< ∞ forall t ∈ [0,T ] then then g(||X (t)||) , t ∈ [0,T ] is a submartingale.

Proof:Let s≤ t

||X (s)|| = ||E (X (s)−X (t) |Fs)+E (X (t) |Fs)||

≤=0 a.e.︷ ︸︸ ︷

||E (X (s)−X (t) |Fs)||+ ||E (X (t) |Fs)||≤ ||E (X (t) |Fs)|| .

Now by Theorem 61.1.1 on Page 1985

||E (X (t) |Fs)|| ≤ E (||X (t)|| |Fs) .

Thus ||X (s)|| ≤ E (||X (t)|| |Fs) which shows ||X || is a submartingale as claimed.Consider the second claim. Recall Jensen’s inequality for submartingales, Theorem

60.1.6 on Page 1948. From the first part

||X (s)|| ≤ E (||X (t)|| |Fs) a.e.

and so from Jensen’s inequality,

g(||X (s)||)≤ g(E (||X (t)|| |Fs))≤ E (g(||X (t)||) |Fs) a.e.,

showing that g(||X (t)||) is also a submartingale. This proves the proposition.

62.5 Some Maximal EstimatesMartingales and submartingales have some very interesting maximal estimates. I willpresent some of these here. The proofs are fairly general and do not require the filtration tobe normal.

Lemma 62.5.1 Let {Ft} be a filtration and let {X (t)} be a nonnegative valued submartin-gale for t ∈ [S,T ] . Then for λ > 0 and any p ≥ 1, if At is a Ft measurable subset of[X (t)≥ λ ] , then

P(At)≤1

λp

∫At

X (T )p dP.