Jensen’s and martingale inequalities in Riesz spaces

The Hájek-Rényi-Chow maximal inequality and a strong law of large numbers in Riesz spaces

“…Proof: For p = 1 the result follows from the triangle inequality while for p = 2 the result follows from |f | 2 = f 2 , so we now consider only 1 < p < 2. Taking g(X) = X 2/p and F(X, Y ) = ( 1 2 (X +Y ), 1 2 (X +Y )) in Jensen's inequality of [10] on the Riesz space F := E u ×E u with componentwise ordering, we have that F(g(|a| p , |b| p )) ≥ g(F(|a| p , |b| p )) so |a| 2 + |b| 2 ≥ 2 (p−2)/p (|a| p + |b| p ) 2/p and 2 (2−p)/2 (|a| 2 + |b| 2 ) p/2 ≥ |a| p + |b| p . Setting a = x + y and b = x − y we have…”

“…Applying T 1 to (5.3) and taking the order limit as n → ∞, by (4.1) we have Remark Since (X p n , T n ) is a non-negative submartingale, by [10,Corollary 4.5], taking Y j+1 = j i=1 Z i in the above theorem, we have that (Y j , T j ) is a T 1 -bounded submartingale and Theorem 5.1 follows directly from [14,Theorem 3.5].…”

Completeness for vector lattices

“…It is clear that Proposition 3 is still true for functions defined on X + . Thus one can define x p for all 0 ≤ x ∈ X s and p ≥ 1 (one can also use [12,Corollary 4.3]). Now, using the notations introduced above we can state a generalization of the Chebychev Inequality in vector lattices (see [5,Theorem 3.9]).…”

Near-epoch dependence in Riesz spaces

“…Next, we will prove a variant of the conditional Jensen's inequality. For additional details on conditional Jensen's inequalities in Riesz spaces, see [6].…”