Sklar’s Theorem and the Rüschendorf transform revisited ... fileSklar’s Theorem and the...

Sklar’s Theorem and the Rüschendorftransform revisited - An analysis of right

quantiles

Frank Oertel

Copulas and Their ApplicationsTO COMMEMORATE THE 75TH BIRTHDAY OF

PROFESSOR ROGER B. NELSENUniversity of Almería

July 3 - 5, 2017

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Content

1 Sklar’s Theorem

2 Rüschendorf’s proof of Sklar’s Theorem revisited

3 The distribution of FV(X) for arbitrary distribution functions F

4 The distribution of FV(X) in the case of F = FX

5 A quotation - There is some truth to it !

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1 Sklar’s Theorem





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Sklar’s TheoremLet F be a n-dimensional distribution function with marginalsF1, . . . ,Fn. Then there exists a copula CF, such that for all(x1, . . . , xn) ∈ Rn we have

F(x1, . . . , xn) = CF(F1(x1), . . . ,Fn(xn)) .

If in addition F is continuous, the copula CF is unique.Conversely, for any univariate distribution functions H1, . . . ,Hn,and any copula C, the composition C (H1, . . . ,Hn) defines an-dimensional distribution function with marginals H1, . . . ,Hn.

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F(x1, . . . , xn) = CF(F1(x1), . . . ,Fn(xn)) .

If in addition F is continuous, the copula CF is unique.

Conversely, for any univariate distribution functions H1, . . . ,Hn,and any copula C, the composition C (H1, . . . ,Hn) defines an-dimensional distribution function with marginals H1, . . . ,Hn.

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F(x1, . . . , xn) = CF(F1(x1), . . . ,Fn(xn)) .

If in addition F is continuous, the copula CF is unique.Conversely, for any univariate distribution functions H1, . . . ,Hn,and any copula C, the composition C (H1, . . . ,Hn) defines an-dimensional distribution function with marginals H1, . . . ,Hn.

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1 Sklar’s Theorem





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The Rüschendorf transform

In the following we fix an arbitrary distribution functionF : R −→ [0, 1].

DefinitionLet λ ∈ [0, 1] and x ∈ R. Put

RF(x, λ) := Fλ(x) : = F(x−) + λ∆F(x)

= (1− λ)F(x−) + λF(x)

Given its importance in Rüschendorf’s proof of Sklar’s Theoremwe denote the function RF : R× [0, 1] −→ [0, 1] as Rüschendorftransform.In particular, for all (x, λ) ∈ R× [0, 1] the following inequalityholds:

F(x−) ≤ Fλ(x) ≤ F(x) .

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RF(x, λ) := Fλ(x) : = F(x−) + λ∆F(x)

= (1− λ)F(x−) + λF(x)

Given its importance in Rüschendorf’s proof of Sklar’s Theoremwe denote the function RF : R× [0, 1] −→ [0, 1] as Rüschendorftransform.

In particular, for all (x, λ) ∈ R× [0, 1] the following inequalityholds:

F(x−) ≤ Fλ(x) ≤ F(x) .

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RF(x, λ) := Fλ(x) : = F(x−) + λ∆F(x)

= (1− λ)F(x−) + λF(x)

Given its importance in Rüschendorf’s proof of Sklar’s Theoremwe denote the function RF : R× [0, 1] −→ [0, 1] as Rüschendorftransform.In particular, for all (x, λ) ∈ R× [0, 1] the following inequalityholds:

F(x−) ≤ Fλ(x) ≤ F(x) .

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Reminder on right- and left-quantilefunctions

Consider the generalised inverse function F∧ : (0, 1) −→ R,given by

F∧(α) := minx ∈ R : F(x) ≥ α .

Then

F∧(α) ≤ infx ∈ R : F(x) > α = supx ∈ R : F(x) ≤ α =: F∨(α)

andF(F∧(α)−

)≤ α ≤ F

(F∧(α)+

)= F

(F∧(α)

)for all α ∈ (0, 1).

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Rüschendorf’s proof of Sklar’sTheorem I

Rüschendorf’s proof of Sklar’s Theorem (cf. [5]) is completelybuilt on the following result:

Theorem (Rüschendorf 2009)Let X,V be two random variables, both defined on the sameprobability space (Ω,F ,P), such that V ∼ U(0, 1) and V isindependent of X. Let F = FX be the distribution function of X.Let

FV(X) := RF(X,V) = F(X−) + V (F(X)− F(X−))

be the distributional transform of X.(i) FV(X) ∼ U(0, 1) is a uniformly distributed random variable.(ii) Moreover,

X = F∧(FV(X)

)= F∧

(F(X−) + V (F(X)− F(X−))

)P-a.s.

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Theorem (Rüschendorf 2009)Let X,V be two random variables, both defined on the sameprobability space (Ω,F ,P), such that V ∼ U(0, 1) and V isindependent of X. Let F = FX be the distribution function of X.

LetFV(X) := RF(X,V) = F(X−) + V (F(X)− F(X−))


X = F∧(FV(X)

)= F∧

(F(X−) + V (F(X)− F(X−))

)P-a.s.

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FV(X) := RF(X,V) = F(X−) + V (F(X)− F(X−))

be the distributional transform of X.

(i) FV(X) ∼ U(0, 1) is a uniformly distributed random variable.(ii) Moreover,

X = F∧(FV(X)

)= F∧

(F(X−) + V (F(X)− F(X−))

)P-a.s.

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FV(X) := RF(X,V) = F(X−) + V (F(X)− F(X−))

be the distributional transform of X.(i) FV(X) ∼ U(0, 1) is a uniformly distributed random variable.

(ii) Moreover,

X = F∧(FV(X)

)= F∧

(F(X−) + V (F(X)− F(X−))

)P-a.s.

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FV(X) := RF(X,V) = F(X−) + V (F(X)− F(X−))


X = F∧(FV(X)

)= F∧

(F(X−) + V (F(X)− F(X−))

)P-a.s.

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Rüschendorf’s proof of Sklar’sTheorem II

Proof of SklarLet i ∈ 1, 2, . . . , n and Vi ∼ U(0, 1). On 0 < Vi ≤ 1 put

Ui := RFi(Xi,Vi) = Fi(Xi−) + Vi∆Fi(Xi) .

According to Rüschendorf’s theorem on the distributionaltransform there exist null sets M1,M2, . . . ,Mn ∈ F , such that onΩ \Mi ⊆ 0 < V ≤ 1 Zi := Fi

∧(Ui) is well-defined and satisfiesXi = Zi for every i ∈ 1, 2, . . . , n. Thus, P(M) = 0, whereM :=

⋃ni=1 Mi. Clearly, Xi = Zi holds on Ω \M for all i

“altogether”.

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Rüschendorf’s proof of Sklar’sTheorem III

Proof (ctd).Consider the copula

CF(γ1, . . . , γn) := P(U1 ≤ γ1, . . . ,Un ≤ γn) ,

where (γ1, . . . , γn)> ∈ [0, 1]n.

Since

u ∈ (0, 1) : u ≤ Fi(xi) = u ∈ (0, 1) : Fi∧(u) ≤ xi

for all i ∈ 1, 2, . . . , n and P(M) = 0, it consequently follows

CF(F1(x1), . . . ,Fn(xn)) = P(Z1 ≤ x1, . . . ,Zn ≤ xn ∩ R \M

)= P

(X1 ≤ x1, . . . ,Xn ≤ xn ∩ R \M

)= F(x1, . . . , xn) .

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CF(γ1, . . . , γn) := P(U1 ≤ γ1, . . . ,Un ≤ γn) ,

where (γ1, . . . , γn)> ∈ [0, 1]n. Since

u ∈ (0, 1) : u ≤ Fi(xi) = u ∈ (0, 1) : Fi∧(u) ≤ xi



)= P

(X1 ≤ x1, . . . ,Xn ≤ xn ∩ R \M

)= F(x1, . . . , xn) .

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CF(γ1, . . . , γn) := P(U1 ≤ γ1, . . . ,Un ≤ γn) ,

where (γ1, . . . , γn)> ∈ [0, 1]n. Since

u ∈ (0, 1) : u ≤ Fi(xi) = u ∈ (0, 1) : Fi∧(u) ≤ xi



)

= P(X1 ≤ x1, . . . ,Xn ≤ xn ∩ R \M

)= F(x1, . . . , xn) .

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CF(γ1, . . . , γn) := P(U1 ≤ γ1, . . . ,Un ≤ γn) ,

where (γ1, . . . , γn)> ∈ [0, 1]n. Since

u ∈ (0, 1) : u ≤ Fi(xi) = u ∈ (0, 1) : Fi∧(u) ≤ xi



)= P

(X1 ≤ x1, . . . ,Xn ≤ xn ∩ R \M

)

= F(x1, . . . , xn) .

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CF(γ1, . . . , γn) := P(U1 ≤ γ1, . . . ,Un ≤ γn) ,

where (γ1, . . . , γn)> ∈ [0, 1]n. Since

u ∈ (0, 1) : u ≤ Fi(xi) = u ∈ (0, 1) : Fi∧(u) ≤ xi



)= P

(X1 ≤ x1, . . . ,Xn ≤ xn ∩ R \M

)= F(x1, . . . , xn) .

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1 Sklar’s Theorem





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The distribution of FV(X) I

Next, we will recognise that part (i) of Rüschendorf’s theoremon the distributional transform could be viewed as a corollary ofa more general statement which is of its own interest.

To thisend we fix an arbitrary distribution function F and α ∈ (0, 1) andconsider the following non-empty subset of the “strip” R× (0, 1]:

Aα ≡ Aα[F] := (x, λ) ∈ R× (0, 1] : RF(x, λ) ≤ α .

The following important specification of Aα[F] - stronglyinvolving the right-quantile function F∨ - will help usconsiderably to calculate the interesting probabilityP((X,V) ∈ Aα[F]

).

Note that in our approach F don’t need to coincide with thedistribution function of X!

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The distribution of FV(X) I

Next, we will recognise that part (i) of Rüschendorf’s theoremon the distributional transform could be viewed as a corollary ofa more general statement which is of its own interest.To thisend we fix an arbitrary distribution function F and α ∈ (0, 1) andconsider the following non-empty subset of the “strip” R× (0, 1]:

Aα ≡ Aα[F] := (x, λ) ∈ R× (0, 1] : RF(x, λ) ≤ α .

The following important specification of Aα[F] - stronglyinvolving the right-quantile function F∨ - will help usconsiderably to calculate the interesting probabilityP((X,V) ∈ Aα[F]

).

Note that in our approach F don’t need to coincide with thedistribution function of X!

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The distribution of FV(X) II

LemmaLet α ∈ (0, 1) and F an arbitrary distribution function. Putξ := F∧(α), η := F∨(α), q := F(ξ−), β := ∆F(ξ) andγ := ∆F(η). Then

Aα =

(−∞, η]× (0, 1] if γ = 0(−∞, η)× (0, 1] if ξ < η and γ > 0(−∞, ξ)× (0, 1] ·∪ ξ ×

(0, α−q

β

]if ξ = η and γ(= β) > 0

The proof of this lemma is built on the following accuratedescription of all “flat pieces” of F.

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The distribution of FV(X) III

LemmaLet 0 < λ ≤ 1 and α ∈ (0, 1). Put ξ := F∧(α) and η := F∨(α).

(i) If ξ < η then F(η−) = F(ξ) = α = F (F∧(α)), and

∅ 6= x ∈ R : F(x) = α =

[ξ, η)

if F(η) > α[ξ, η]

if F(η) = α

(ii) If ξ = η, then

x ∈ R : F(x) = α =

∅ if F(η) > α

ξ if F(η) = α

Observationξ < η if and only if there are x1 6= x2 such thatF(x1) = α = F(x2).

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The distribution of FV(X) IVLet 0 < α < 1 and X,V two random variables, both defined onthe same probability space (Ω,F ,P), such that V ∼ U(0, 1) andV is independent of X.

By construction

P(FV(X) ≤ α

)= P (FV(X) ≤ α and V ∈ (0, 1]) = P

((V,X) ∈ Aα

).

Hence,

PropositionLet F : R −→ [0, 1] be an arbitrary distribution function. Letα ∈ (0, 1). Put ξ := F∧(α), η := F∨(α) and β := ∆F(ξ). Let X,Vbe two random variables, both defined on the same probabilityspace (Ω,F ,P), such that V ∼ U(0, 1) and V is independent ofX.

(i) If ξ < η then

P(FV(X) ≤ α

)= P

(X < η

)+ 11F=α(η)P

(X = η

)= P

(X < ξ

)+ P

(F(X) = α

)

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The distribution of FV(X) IVLet 0 < α < 1 and X,V two random variables, both defined onthe same probability space (Ω,F ,P), such that V ∼ U(0, 1) andV is independent of X. By construction

P(FV(X) ≤ α

)= P (FV(X) ≤ α and V ∈ (0, 1]) = P

((V,X) ∈ Aα

).

Hence,


(i) If ξ < η then

P(FV(X) ≤ α

)= P

(X < η

)+ 11F=α(η)P

(X = η

)= P

(X < ξ

)+ P

(F(X) = α

)

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P(FV(X) ≤ α

)= P (FV(X) ≤ α and V ∈ (0, 1]) = P

((V,X) ∈ Aα

).

Hence,


(i) If ξ < η then

P(FV(X) ≤ α

)= P

(X < η

)+ 11F=α(η)P

(X = η

)

= P(X < ξ

)+ P

(F(X) = α

)

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P(FV(X) ≤ α

)= P (FV(X) ≤ α and V ∈ (0, 1]) = P

((V,X) ∈ Aα

).

Hence,


(i) If ξ < η then

P(FV(X) ≤ α

)= P

(X < η

)+ 11F=α(η)P

(X = η

)= P

(X < ξ

)+ P

(F(X) = α

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The distribution of FV(X) V

Proposition (ctd.)

(ii) If ξ = η and β = 0 then

P(FV(X) ≤ α

)= P

(X ≤ ξ

).

(iii) If ξ = η and β > 0 then

P(FV(X) ≤ α

)= P

(X ≤ ξ

)− F(ξ)− α

βP(X = ξ

).

ObservationIn any of the three cases the right side of the representation ofP(FV(X) ≤ α

)is completely independent of the choice of the

random variable V ∼ U(0, 1).

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1 Sklar’s Theorem





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The special case F = FX ILet ξ < η or β = 0. Then

F(ξ) = α = F(η−) .

Hence, if F = FX coincides with the distribution function of X itfollows that

P(X ≤ ξ) = α = P(X < η) .

In particular, if ξ < η :

P (F(X) = α) = P(X < η)− P(X < ξ) = P (X = ξ) .

CorollaryLet α ∈ (0, 1). Let X,V be two random variables, both definedon the same probability space (Ω,F ,P), such that V ∼ U(0, 1)and V is independent of X. Let F = FX be the distributionfunction of X. Then FV(X) is uniformly distributed.

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F(ξ) = α = F(η−) .


P(X ≤ ξ) = α = P(X < η) .


P (F(X) = α) =

P(X < η)− P(X < ξ) = P (X = ξ) .


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F(ξ) = α = F(η−) .


P(X ≤ ξ) = α = P(X < η) .


P (F(X) = α) = P(X < η)− P(X < ξ)

= P (X = ξ) .


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F(ξ) = α = F(η−) .


P(X ≤ ξ) = α = P(X < η) .


P (F(X) = α) = P(X < η)− P(X < ξ) = P (X = ξ) .


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The special case F = FX II

Corollary (ctd.)Moreover, if F has at least one “flat piece” then

P(F(X) ≤ α

)= α = P

(X ≤ F∧(α)

)on the set

α ∈ (0, 1) : F∧(α) < F∨(α)

.

ObservationLet F = FX be a distribution function of a given random variableX. TFAE:

(i) F is continuous.(ii) F(X) is uniformly distributed over (0, 1).

Consequently, if F had non-zero jumps FV(X) still would beuniformly distributed over (0, 1), as opposed to F(X).

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The special case F = FX III

Proof.By contradiction assume that (ii) holds and (i) is false. Thenthere exists x0 ∈ R such that

0 < ∆F(x0) = F(x0)−F(x0−) = P(X = x0) ≤ P(F(X) = F(x0)) = 0

- which is a contradiction.

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The special case F = FX IV

TheoremLet X,V be two random variables, defined on the sameprobability space (Ω,F ,P), such that V ∼ U(0, 1) and V isindependent of X. Let F = FX be the distribution function of therandom variable X. Then

X = F∧(FV(X)

)= F∧

(F(X−) + V∆F(X)

)P-almost surely.

If in addition P(0 < F(X) < 1

)= 1 (for example, if F is

continuous), then

X = F∧(F(X)

)P-almost surely .

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The special case F = FX V

Proof (Sketch).Since V ∼ U(0, 1), it follows that on B

((0, 1]

)the image

measure µFV := P(V ∈ ·) coincides with the Lebesgue measurem.

Thus, by applying the Fubini-Tonelli Theorem to non-negative(or bounded) Borel functions on

(R2,B(R2), µFV ⊗ µFX

)we firstly

recognise that there exists an m-null set L ∈ B((0, 1]

)such that

for all λ ∈ A := (0, 1] \ L there exists a Borel set Nλ ⊆ R suchthat P (X ∈ Nλ) = 0 and for any x ∈ R \ Nλ the value F∧

(Fλ(x)

)is well-defined and satisfies F∧

(Fλ(x)

)= x.

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The special case F = FX V

Proof (Sketch).Since V ∼ U(0, 1), it follows that on B

((0, 1]

)the image

measure µFV := P(V ∈ ·) coincides with the Lebesgue measurem.Thus, by applying the Fubini-Tonelli Theorem to non-negative(or bounded) Borel functions on

(R2,B(R2), µFV ⊗ µFX

)we firstly

recognise that there exists an m-null set L ∈ B((0, 1]

)such that

for all λ ∈ A := (0, 1] \ L there exists a Borel set Nλ ⊆ R suchthat P (X ∈ Nλ) = 0 and for any x ∈ R \ Nλ the value F∧

(Fλ(x)

)is well-defined and satisfies F∧

(Fλ(x)

)= x.

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The special case F = FX VI

Proof (Sketch) - ctd.Hence, since

P(X ∈ NV and V ∈ A

)=

∫AP(X ∈ Nλ

)m(dλ) = 0 ,

it consequently follows

X = F∧(FV(X)

)= F∧

(F(X−) + V∆F(X)

)on the set Ω \ N ⊆ 0 < V ≤ 1, whereN := V /∈ A ·∪X ∈ NV and V ∈ A.

The second part of the claim follows from

X(ω) = F∧ (F(X(ω)−) + V(ω)∆F(X(ω))) ≤ F∧(F(X(ω))) ≤ X(ω)

outside of a P-null set N, satisfying N ⊆ N.

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The special case F = FX VI

Proof (Sketch) - ctd.Hence, since

P(X ∈ NV and V ∈ A

)=

∫AP(X ∈ Nλ

)m(dλ) = 0 ,

it consequently follows

X = F∧(FV(X)

)= F∧

(F(X−) + V∆F(X)

)on the set Ω \ N ⊆ 0 < V ≤ 1, whereN := V /∈ A ·∪X ∈ NV and V ∈ A.The second part of the claim follows from

X(ω) = F∧ (F(X(ω)−) + V(ω)∆F(X(ω))) ≤ F∧(F(X(ω))) ≤ X(ω)

outside of a P-null set N, satisfying N ⊆ N.

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1 Sklar’s Theorem





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Not “confusing”. Rather “quite hidden” Iwould suggest

- W. F. Darsow, B. Nguyen and E. T. Olsen (1996)

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A very few references I

[1] P. Embrechts and M. Hofert.A note on generalized inverses.Mathematical Methods of Operations Research, 77 (3),423-432 (2013).

[2] T. S. Ferguson.Mathematical Statistics: A Decision Theoretic Approach.Probability and Mathematical Statistics, vol. 1., AcademicPress, New York (1967).

[3] J. F. Mai and M. Scherer.Simulating copulas.Imperial College Press, London (2012).

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A very few references II

[4] F. Oertel.An analysis of the Rüschendorf transform - with a viewtowards Sklar’s Theorem.Dependence Modelling Vol. 3, No. 1, 113-125 (2015).

[5] L. Rüschendorf.On the distributional transform, Sklar’s theorem, and theempirical copula process.Journal of Statistical Planning and Inference 139 (11),3921-3927 (2009).

[6] B. Schweizer and A. Sklar.Operations on distribution functions not derivable fromoperations on random variables.Studia Mathematica 52, 43-52 (1974).

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Thank you for your attention!

Are there any questions, comments or remarks?

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