Functions operating on several multivariate distribution functions

Definition

f : I → R is n - ↑ (read “ n -increasing”) iff ( Δ h p f ) ( s ) ≥ 0 ∀ s ∈ I , ∀ h ∈ R + d , ∀ p ∈ N 0 d and 0 ≠ p ≤ n , such that s j + p j h j ∈ I j ∀ j ∈ [ d ] .

Definition

Let I 1 , … , I d ⊆ R be non-degenerate intervals, I = I 1 × ⋯ × I d , f : I → R , and k ∈ N . Then, f is called k -increasing (“ k - ↑ ”) iff ∀ j ∈ [ k ] , ∀ h ( 1 ) , … , h ( j ) ∈ R + d , ∀ s ∈ I such that s + h ( 1 ) + ⋯ + h ( j ) ∈ I

(We do not assume f to be continuous.)

Remark 1

Already in 2005, Bronevich [2] introduced k - ↑ functions, calling them “ k -monotone.” Later on, the name “strongly k -monotone” was used [5,8], a terminology usually associated with strict inequalities, therefore not really adequate.

Lemma 1

Let f : [ 0 , 1 ] d → R be 2 - ↑ . Then,

1. f is continuous iff f is continuous in 1 d .

2. f is right-continuous and on [ 0 , 1 [ d continuous.

3. If f ( x 0 ) = f ( 1 d ) for some x 0 ≠ 1 d , hence α ≔ { i ≤ d ∣ x i 0 = 1 } ⊊ [ d ] , then for α ≠ ∅ f depends only on x α ≔ ( x i ) i ∈ α . In case, α = ∅ f is constant.

4. For each y ∈ [ 0 , 1 ] d ⧹ [ 0 , 1 [ d , the limit

   f 0 ( y ) ≔ lim x → y x < y f ( x )

   exists, f 0 ( y ) ≤ f ( y ) , f 0 is a 2 - ↑ and continuous extension of f ∣ [ 0 , 1 [ d . If f is k - ↑ , so is f 0 .

Proof

(i) For x , h ∈ [ 0 , 1 ] d such that also x ± h ∈ [ 0 , 1 ] d , we have (with 1 ≔ 1 d )

from which the claim follows, f being increasing.

(ii) For any x ∈ [ 0 , 1 [ d , the univariate function [ 0 , 1 ] ∋ t ↦ f ( t x ) is convex, hence continuous, also in t = 1 (being defined and convex in a neighborhood of 1). Since f is increasing, f is continuous in x . For x ∈ [ 0 , 1 ] d ⧹ [ 0 , 1 [ d , x ≠ 1 d , let α ≔ { i ≤ d ∣ x i = 1 } . Then, for y ≥ x , also y i = 1 ∀ i ∈ α , and [ 0 , 1 [ α ∁ ∋ z ↦ f ( 1 α , z ) is continuous, in particular in the point x α ∁ , implying f to be right-continuous in x = ( 1 α , x α ∁ ) .

(iii) If α = ∅ , then x 0 = 1 − h 0 for some h 0 ∈ ] 0 , 1 ] d , f ( 1 ) = f ( 1 − h ) ∀ 0 ≤ h ≤ h 0 , and f is constant by ( * ) . For α ≠ ∅ , we define g : [ 0 , 1 ] α ∁ → R + by g ( z ) ≔ f ( 1 α , z ) . Also, g is 2 - ↑ , and

hence g is constant. But then, ∀ y ∈ [ 0 , 1 ] α

showing f ( y , z ) to be independent of z .

(iv) The existence of f 0 ( y ) is clear, f being increasing and bounded. The defining inequalities for f being k - ↑ prevail for f 0 , for any k ≥ 2 . Since f 0 is continuous in 1 , it is everywhere continuous.□

Theorem 1

Let I ⊆ R d be a non-degenerate interval, f : I → R , k , d ∈ N . Then, there are equivalent:

1. f is k - ↑ ,

2. f is n - ↑ ∀ n ∈ N 0 d with 0 < ∣ n ∣ ≤ k ,

3. ∀ m ∈ N , ∀ non-degenerate interval J ⊆ R m , ∀ positive affine φ : R m → R d such that φ ( J ) ⊆ I , also f ∘ φ is k - ↑ ,

4. ∀ m , J , φ as before, and ∀ n ∈ N 0 m with 0 < ∣ n ∣ ≤ k the function f ∘ φ is n - ↑ ,

5. ∀ m , J , φ as before, and ∀ n ∈ { 0 , 1 } m with 0 < ∣ n ∣ ≤ k the function f ∘ φ is n - ↑ .

Remark 2

For k ≥ d ≥ 2 and f ≥ 0 , the aforementioned function f is not only right-continuous (Lemma 1(ii)), but also 1 d - ↑ , hence a d.f. on I ; however, with the extra property that f ∘ φ is a d.f., too, for any positive affine φ : R m → R d . In other words, if f ( x ) = P ( X ≤ x ) for some d -dimensional random vector X , and k ≥ m , then also y ↦ P ( X ≤ φ ( y ) ) is an m -dimensional d.f.

Proof

We show (i) ⇔ (ii) and (i) ⇒ (iii) ⇒ (iv) ⇒ (v) ⇒ (i).

( i ) ⇒ ( i i ) ̲ is clear.

( i i ) ⇒ ( i ) ̲ : We use induction on k , k = 1 being obvious. Let now k ≥ 2 and suppose the case k − 1 is already known. Fix some h ∈ R + d and consider g ≔ Δ h f , i.e., g ( s ) = f ( s + h ) − f ( s ) . With

we have g = g 1 + g 2 + ⋯ + g d . Each g i is n - ↑ for any n ∈ N 0 d with ∣ n ∣ ≤ k − 1 ; hence, ( k − 1 ) - ↑ by assumption, and so is then g . Since h ∈ R + d was arbitrary, f is k - ↑ .

( i ) ⇒ ( i i i ) : ̲ Let j ∈ [ k ] , h ( 1 ) , … , h ( j ) ∈ R + m , x ∈ J such that also x + ∑ i = 1 j h ( i ) ∈ J . Then, with ψ ≔ φ − φ ( 0 ) ,

(note that φ ( x ) + ∑ i = 1 j ψ ( h ( i ) ) = φ ( x + ∑ i = 1 j h ( i ) ) ∈ I ).

( i i i ) ⇒ ( i v ) ⇒ ( v ) ̲ is clear.

( v ) ⇒ ( i ) ̲ : Let j ∈ [ k ] , x ∈ I , h ( 1 ) , … , h ( j ) ∈ R + d such that x + h ( 1 ) + ⋯ + h ( j ) ∈ I . Denote by ψ : R j → R d the linear map whose matrix with respect to the standard bases is ( h ( 1 ) , … , h ( j ) ) ( h ( i ) as column vectors), φ ≔ x + ψ ; obviously, ψ (and φ ) are positive. For J ≔ [ 0 j , 1 j ] ⊆ R j , we have φ ( J ) ⊆ I , since ψ ( e i ) = h ( i ) ∀ i ≤ j , φ ( 0 j ) = x and φ ( 1 j ) = x + ∑ i = 1 j h ( i ) . By assumption,

Corollary 1

Let I ⊆ R d and B ⊆ R be non-degenerate intervals. If g : I → B and f : B → R are both k - ↑ , then so is f ∘ g .

Proof

We show Condition (iv) in Theorem 1 to hold for f ∘ g . We know that g ∘ φ is n - ↑ for ∣ n ∣ ≤ k . A special case of Theorem 12 in [12] implies f ∘ ( g ∘ φ ) to be also n - ↑ , and f ∘ ( g ∘ φ ) = ( f ∘ g ) ∘ φ .□

Theorem 2

Let I ⊆ R d 1 and J ⊆ R d 2 be non-degenerate intervals, f : I → R and g : J → R , both non-negative and k - ↑ . Then, also f ⊗ g is k - ↑ on I × J , and in case I = J , the product f ⋅ g is k - ↑ , too.

Proof

We first apply (ii) of Theorem 1. For 0 ≠ ( m , n ) ∈ N 0 d 1 × N 0 d 2 , ( x , y ) ∈ I × J , h ( 1 ) ∈ R + d 1 , h ( 2 ) ∈ R + d 2 we have

and for ∣ ( m , n ) ∣ = ∣ m ∣ + ∣ n ∣ ≤ k , both factors on the right-hand side are non-negative. Since m = 0 or n = 0 is allowed, only ( m , n ) ≠ 0 being required, we need in fact f ≥ 0 and g ≥ 0 .

For I = J (with d 1 = d 2 ≕ d ), let φ : R d → R 2 d be given by φ ( x ) ≔ ( x , x ) , a positive linear map. Then, φ ( I ) ⊆ I × I , and by Theorem 1 (iii), ( f ⊗ g ) ∘ φ = f ⋅ g is also k - ↑ .□

Examples 1

(a) For a > 0 , the function f ( x , y ) ≔ ( x y − a ) + is 2 - ↑ on R + 2 , since t ↦ ( t − a ) + is 2 - ↑ on R + , by Corollary 1. Its restriction to [ 0 , 1 + a ] 2 is therefore the d.f. of some random vector. In [12] on page 261, it was shown that f is not ( 2 , 1 ) - ↑ (resp. ( 1 , 2 ) - ↑ ), but it is of course ( 1 , 1 ) - ↑ . The tensor product g ( x , y ) ≔ ( x − a ) + ⋅ ( y − b ) + is ( 2 , 2 ) - ↑ ∀ a , b > 0 ; hence, certainly 2 - ↑ , but not 3 - ↑ since x ↦ ( x − a ) + is not.

Similarly, ( x y z − a ) + 2 is 3 - ↑ on R + 3 , for a > 0 , and of course, ( x y − a ) + 2 is 3 - ↑ on R + 2 . We will see later on that x y + x z + y z − x y z is 2 - ↑ on [ 0 , 1 ] 3 , but not 3 - ↑ .

(b) Consider f n ( t ) ≔ t n ∕ ( 1 + t ) for t ≥ 0 . It was shown in [7], Lemma 2.4, that f n is n - ↑ (it is not ( n + 1 ) - ↑ ). So for any non-negative n - ↑ function g on any interval in any dimension, g n ∕ ( 1 + g ) is n - ↑ , too.

If g is “only” an n -dimensional d.f., then so is also g n ∕ ( 1 + g ) – this follows from [11], Theorem 2, but is also a special case of Theorem 6 below.

Theorem 3

Let f : [ 0 , 1 ] d → R have the property that each restriction f ∣ T α for ∅ ≠ α ⊆ [ d ] is continuous. Then,

i.e., the Bernstein polynomials converge pointwise to f everywhere.

Proof

Each x ∈ [ 0 , 1 ] d ⧹ { 1 d } lies in exactly one T α , i.e., x = ( x α , 1 α ∁ ) with x α < 1 α , where ∅ ≠ α ⊆ [ d ] , and is thus a continuity point of f ( α ) ≔ f ∣ T α . As already mentioned, this implies

and we saw also that