Date: Mar 12, 2014

Today we shall prove some basic results related to Archimedean copula. We shall start by defining a class of functions G consisting of all functions φ:(0,1]→[0,∞) that are strictly decreasing, convex and φ(1)=0. Any φ∈ G will be a called a generator in this note.

Continuity of generators

The following function is a generator that is not continuous:

A discontinuous generator However, we cannot have any discontinuity in (0,1]. We show this below. We start with a general property of convex functions.

Theorem Any convex function defined on [a,b] is continuous everywhere on (a,b).

Proof: Let φ:[a,b]→ IR be a convex function.

Take any c∈(a,b).

Take any α∈ (a,c) and any β∈ (c,b).

Let the equations of red and blue chords below be called y=RED(x) and y=BLUE(x).

The graph must lie in the grey region We shall show that the graph must lie in the grey region between RED and BLUE.

Pick any t∈ (α,c). Then by definition of convexity

BLUE(t) ≤ φ(t) ≤ RED(t).

Again, if you take t∈ (c,β) then

RED(t) ≤ φ(t) ≤ BLUE(t).

Thus the graph of φ over [α,β] is sandwiched between RED and BLUE. As t→ c we have RED(t)→ φ(c) and BLUE(t)→ φ(c), because they are just continuous lines.

So the sandwich law of limit implies that φ(t)→ φ(c), completing the proof. [Q.E.D]

The above theorem gurantees continuity of a convex function in the interior of the domain. This forces any generator φ to be continuous on (0,1). Indeed, a convex function may be discontinuous at the end points as the following example shows.

A convex function discontinuous at both ends We shall now show that a generator must also be continuous at 1 thanks to its decreasing nature.

Theorem Any convex, decreasing function φ:[a,b]→ IR must be continuous at b.

Proof: Since it is a decreasing function, and φ(b) is defined,

hence φ(b-) must exist finitely.

Let, if possible, φ(b-)≠φ(b).

Then φ(b-) > φ(b).

Let ε = (φ(b-)-φ(b))/2.

Then ∃ δ > 0 such that ∀ x∈ (b-δ,b) we have φ(x) ∈ (φ(b-),φ(b-)+ε).

Then the graph of φ lives above its chord over [b-δ,b], contradicting convexity. [Q.E.D]

Why a generator generates a copula

We have not proved yet that an Archimedean copula is indeed a copula. We shall do so now. Let φ be any generator. Let

C(x,y) = φ^[-1] ( φ(x)+φ(y) ) for x,y∈ [0,1].

Shall show that C(x,y) is a copula. The proof will be done in three steps.

In the first step is to show that satisfies the boundary conditions C(0,y)=C(x,0)=0 and C(1,y)=y and C(x,1)=x.

All these are trivial from the definition of φ^[-1] and are left as an excercise.

Next we need to show that for any x₁ < x₂ ∈ [0,1] and any y₁ < y₂ ∈ [0,1] the function C(x,y) assigns a positive mass on the rectangle [ x₁ , x₂ ] × [ y₁ , y₂ ], i.e.,

C( x₂ , y₂ ) - C( x₁ , y₂ ) - C( x₂ , y₁ ) + C( x₁ , y₁ ) ≥ 0.

We shall prove this first for a special case: y₂ = 1. The general case will then follow easily.

For the special case the inequality becomes

C( x₂ , 1 ) - C( x₁ , 1 ) - C( x₂ , y₁ ) + C( x₁ , y₁ ) ≥ 0,

or, thanks to he first step above,

x₂ - x₁ - C( x₂ , y₁ ) + C( x₁ , y₁ ) ≥ 0.

So in the second step we shall prove this. Since there is no y₂ in the picture anymore, let's rename y₁ to y. Then we are to show

C( x₁ , y ) + C( x₂ , y ) ≥ x₁ - x₂ .

Let

a = φ( x₁ ) , b = φ( x₂ ) , c = φ( y ) .

Now consider the points b ≤ a ≤ a+c:

We can express a as a convex combination of b and a+c. If we write the convex combination as

a = (1-γ) b + γ (a+c)

then convince yourself that

γ = (a-b)/(a+c-b) .

By interchanging the roles of γ and 1-γ we have the following picture:

Thus

b+c = γ b + (1-γ) (a+c).

Now apply φ^[-1] to both sides of both these convex combinations. Since φ^[-1] is a convex function, we have

φ^[-1] (a) ≤ (1-γ) φ^[-1] b + γ φ^[-1] (a+c)

and

φ^[-1] (b+c) ≤ γ φ^[-1] b + (1-γ) φ^[-1] (a+c).

Add these up to complete the second step.

Recall that our aim was to prove

C( x₂ , y₂ ) - C( x₁ , y₂ ) - C( x₂ , y₁ ) + C( x₁ , y₁ ) ≥ 0

for any x₁ < x₂ ∈ [0,1] and any y₁ < y₂ ∈ [0,1]. The second step proved this only for the special case y₂₌₁. In the third step we shall prove it in general. First notice that if y₁ = 0, then the result is easy. So assume that y₁ > 0.

Consider φ( y₁ ) - φ( y₂ ). Thanks to the continuity of φ over (0,1] we can find t∈[0,1] such that

φ( y₁ ) - φ( y₂ ) = φ(t).

Carefully understand this. We are using intermediate value theorem here.

Then

C(x_{2,y_1)} = φ^[-1] (φ(x_{2)+φ(y_1))=φ^[-1]} ( φ(x₂₎ + φ(y₂₎ + φ(t)) = C(C(x_{2,y_2),t)}.

Similarly,

C(x_{1,y_1)} = C(C(x_{1,y_2),t)}.

C(x_{2,y_1)} - C(x_{1,y_1)} = C(C(x_{2,y_2),t)} - C(C(x_{1,y_2),t)} ≤ C(x_{2,y_2)} - C(x_{1,y_2),}

using the special case done in step 2. This completes the proof that an Archimedean copula is indeed a copula.

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