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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
1 / 28
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 !
2 / 28
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 !
3 / 28
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.
4 / 28
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.
4 / 28
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.
4 / 28
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 !
5 / 28
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) .
6 / 28
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) .
6 / 28
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) .
6 / 28
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) .
6 / 28
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) .
6 / 28
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).
7 / 28
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).
7 / 28
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).
7 / 28
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.
8 / 28
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.
LetFV(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.
8 / 28
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.
8 / 28
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.
8 / 28
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.
8 / 28
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”.
9 / 28
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”.
9 / 28
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) .
10 / 28
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) .
10 / 28
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) .
10 / 28
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) .
10 / 28
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) .
10 / 28
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 !
11 / 28
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!
12 / 28
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!
12 / 28
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!
12 / 28
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!
12 / 28
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.
13 / 28
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.
13 / 28
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.
13 / 28
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).
14 / 28
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).
14 / 28
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).
14 / 28
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).
14 / 28
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
)
15 / 28
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
)
15 / 28
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
)
15 / 28
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
)
15 / 28
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
)
15 / 28
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
)15 / 28
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).
16 / 28
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).
16 / 28
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).
16 / 28
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 !
17 / 28
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.
18 / 28
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.
18 / 28
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.
18 / 28
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.
18 / 28
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.
18 / 28
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.
18 / 28
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 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).
19 / 28
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).
19 / 28
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.
20 / 28
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 .
21 / 28
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.
22 / 28
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.
22 / 28
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.
23 / 28
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.
23 / 28
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 !
24 / 28
Not “confusing”. Rather “quite hidden” Iwould suggest
- W. F. Darsow, B. Nguyen and E. T. Olsen (1996)
25 / 28
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).
26 / 28
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?
28 / 28
Thank you for your attention!
Are there any questions, comments or remarks?
28 / 28