arXiv:1311.2002v1 [math.PR] 8 Nov 2013
Simulation of multivariate distributions with fixed
marginals and correlations
Mark Huber∗ and Nevena Mari´c†
November 11, 2013
Abstract
Consider the problem of drawing random variates (X1, . . . , Xn)
from a distribution where the marginal of eachXi is specified, as well
as the correlation between every pairXi andXj. For given marginals,
the Fr´echet-Hoeffding bounds put a lower and upper bound on the correlation betweenXi andXj. Hence any achievable correlation can
be uniquely represented by a convexity parameter λij ∈ [0,1] where
1 gives the maximum correlation and 0 the minimum correlation. We show that for a given convexity parameter matrix, the worst case is when the marginal distribution are all Bernoulli random variables with parameter 1/2 (fair 0-1 coins). It is worst case in the sense that given a convexity parameter matrix that is obtainable when the marginals are all fair 0-1 coins, it is possible to simulate from any marginals with the same convexity parameter matrix. In addition, we characterize com-pletely the set of convexity parameter matrices for symmetric Bernoulli marginals in two, three and four dimensions.
1
Introduction
Consider the problem of simulating a random vector (X1, . . . , Xn) where for
allithe cumulative distribution function (cdf) of Xi isFi, and for alliand
j the correlation betweenXi and Xj should be ρij ∈[0,1]. The correlation
here is the usual notion
Corr(X, Y) = E[(X−E(X))(Y −E(Y))]
SD(X)SD(Y) =
E[XY]−E[X]E[Y] SD(X)SD(Y) ,
∗Claremont McKenna College (email: mhuber@cmc.edu)
for standard deviations SD(X) and SD(Y) that are finite.
Let Ω denote the set of matrices with entries in [−1,1], all the diagonal entries equal 1, and are nonnegative definite. Then it is well known that any correlation matrix (ρij) must lie in Ω.
This problem, in different guises, appears in numerous fields: physics [16], engineering [11], ecology [4], and finance [12], to name just a few. Due to its applicability in the generation of synthetic optimization problems, it has also received special attention by the simulation community [9], [8].
It is a very well studied problem with a variety of approaches. When the marginals are normal and the distribution is continuous with respect to Lebesgue measure, this is just the problem of generating a multivariate normal with specified correlation matrix. It is well known how to accomplish this (see, for instance [6], p. 223) for any matrix in Ω.
For marginals that are not normal, the question is very much harder. A common method is to employ families of copulas (see for instance [15]), but there are very few techniques that apply to general marginals. Instead, dif-ferent families of copulas typically focus on difdif-ferent marginal distributions. Devroye and Letac [3] showed that if the marginals are beta distributed with equal parameters at least 1/2, then when the dimension is three it is possible to simulate such a vector where the correlation is any matrix in Ω. This set of beta distributions includes the important case of uniform [0,1] marginals, but they have not been able to extend their technique to higher dimensions.
Chagnanty and Joe [1] characterized the achievable correlation matrices when the marginals are Bernoulli. When the dimension is 3 their charac-terization is easily checkable, in higher dimensions they give a number of inequalities that grows exponentially in the dimension.
For the case of general marginals, in statistics there is a tradition of using transformations of mutivariate normal vectors dating back to Mardia [14] and Li and Hammond [13]. This approach relies heavily on developing usable numerical methods. In this paper we approach the same problem using exclusively probabilistic techniques.
The convexity matrix approach. Any two random variables X and Y have correlation in [−1,1], but if the marginal distributions of X and Y are fixed, it is generally not possible to build a bivariate distribution for any correlation in [−1,1]. For instance, for X and Y both exponentially distributed, the correlation must lie in [1−π2
/6,1]. The range of achievable correlations is always a closed interval.
For two dimensions it is well known how to find the minimum and max-imum correlation. These come from the inverse transform method, which works as follows. First, given a cdfF, define the pseudoinverse of the cdf as
F−1= inf{x:F(x)≥u}. (1)
When U is uniform over the interval [0,1] (write U ∼Unif([0,1])), F−1(U)
is a random variable with cdfF (see for instance p. 28 of [2]). SinceU and 1−U have the same distribution, both can be used in the inverse transform method. The random variablesU and 1−U areantitheticrandom variables. Of course Corr(U, U) = 1 and Corr(U,1−U) = −1, so these represent an easy way to get minimum and maximum correlation when the marginals are uniform random variables.
The following theorem comes from work of Fr´echet [7] and Hoeffding [10].
Theorem 1 (Fr´echet-Hoeffding bound). For X1 with cdf F1 and X2 with cdf F2, and U ∼Unif([0,1]):
Corr(F1−1(U), F −1
2 (1−U))≤Corr(X1, X2)≤Corr(F− 1 1 (U), F
−1 2 (U)).
In other words, the maximum correlation betweenX1andX2is achieved
when the same uniform is used in the inverse transform method to gener-ate both. The minimum correlation between X1 and X2 is achieved when
antithetic random variates are used in the inverse transform method. Now consider (X1, . . . , Xn), where eachXihas cdfFi, and the correlation
betweenXi and Xj is ρij. Then let
ρ−ij = Corr(Fi−1(U)Fj−1(1−U)) andρ+ij = Corr(Fi−1(U)Fj−1(U)).
For the ρij to be achievable, it must be true that ρij ∈ [ρ−ij, ρ +
ij], so there
existsλij ∈[0,1] so that
ρij =λijρ+ij+ (1−λij)ρ−ij.
Given a correlation matrix (ρij) we will refer to (λij) as the convexity
Theorem 2. Suppose that for a convexity parameter matrix Λ, it is possible to simulate (B1, . . . , Bn) where for all i, P(Bi = 1) =P(Bi = 0) = 1/2 and
(∀i, j)(P(Bi=Bj) =λij).
Then for any specified set of marginals F1, . . . , Fn it is possible to simulate
in linear time from a multivariate distribution on (X1, . . . , Xn) such that
(∀i)(Xi ∼Fi) and(∀i, j)(Corr(Xi, Xj) =λijρ+ij + (1−λij)ρ−ij).
The next result characterizes when such a multivariate Bernoulli exists in two, three, and four dimensions, and gives necessary conditions for higher dimensions.
Theorem 3. Suppose (B1, B2, . . . , Bn) are random variables with P(Bi =
1) = P(Bi = 0) = 1/2 for all i. When n = 2, it is possible to
simu-late (B1, B2) for any λ12 ∈ [0,1]. When n = 3, it is possible to simulate
(B1, B2, B3) if and only if
1 + 2 min{λ23, λ12, λ13} ≥λ23+λ12+λ13≥1.
When n= 4, it is possible to simulate (B1, B2, B3, B4) if and only if
1−(1/2)ℓ≤(1/2)(u−1)
where
ℓ= min(λ14+λ14+λ13+λ23, λ14+λ34+λ12+λ23, λ24+λ34+λ12+λ13)
u= min
{i,j,k}λij +λjk+λik.
2
The algorithm
To present our algorithm, we begin by defining a matrix that measures the chance that any two components are equal.
Definition 1. For a random vector (X1, . . . , Xn) the concurrence matrix
A= (aij) is defined asaij =P(Xi=Xj).
Let Bern(p) denote the Bernoulli distribution with parameter p, so for X∼Bern(p), P(X= 1) =p andP(X= 0) = 1−p.
Proposition 1. For a random vector(B1, . . . , Bn) withBi∼Bern(1/2) for
all i, the concurrence matrix and convexity matrix are the same matrix.
Proof. Let (B1, . . . , Bn) be a random vector with marginals Bern(1/2), and
i6=j be elements of {1, . . . , n}.
Let 1(expression) denote the indicator function that is 1 if the expres-sion is true and 0 otherwise. The inverse transform method for generating
Bern(1/2) random variables isB =1(U >1/2) whereU ∼Unif([0,1]). From
the Fr´echet-Hoeffding bound (Theorem 1), the correlation between Bi and
Bj can range anywhere from -1 to 1. Hence
Corr(Bi, Bj) =λij(1) + (1−λij)(−1)⇒λij = (1/2)(Corr(Bi, Bj) + 1).
Now denoteP(Bi =a, Bj =b) bypab. Then
P(Bi =Bj) =p11+p00=aij
P(Bi = 1) =p10+p11= 1/2
P(Bj = 1) =p01+p11= 1/2
X
a,b
P(Bi =a, Bj =b) =p00+p01+p10+p11= 1
Adding the top three equations and then using the fourth gives
3p11+p10+p01+p00=aij+ 1 = 2p11+ 1.
Hencep11=aij/2.
That means
Corr(Bi, Bj) = E
[BiBj]−E[Bi]E[Bj]
SD(B1)SD(B2)
= (1/2)aij −(1/2)(1/2)
Given the ability to generate from a multivariate Bernoulli with given convexity/concurrence matrix, generating from a multivariate with general marginals with the same convexity matrix turns out to be easy. Suppose thenby nconvexity matrix Λ = (λij) and cdf’s F1, . . . , Fn are given.
Generating multivariate distributions with given convexity matrix
1) Draw U uniformly from [0,1].
2) Draw (B1, . . . , Bn) (Bi∼Bern(1/2)) with concurrence matrix Λ = (λij).
3) For all i∈ {1, . . . , n}, letXi←F− 1
i (U Bi+ (1−U)(1−Bi)).
That is, ifBi= 1 then chooseXi with the inverse transform method and
U, otherwise use the inverse transform method and 1−U. Theorem 2 can now be restated in terms of this algorithm.
Theorem 4. The above algorithm generates (X1, . . . , Xn) with Xi ∼Fi for
all i and convexity matrix Λ.
Proof. Each Xi can be written as BiF− 1
i (U) + (1−Bi)F− 1
i (1−U). That
is, it is a mixture of F−1
i (U) and F− 1
i (1−U) with weights equal to 1/2.
Since Fi−1(U) andFi−1(1−U) have the same distribution with cdf Fi, the
random variable Xi also shares this distribution. Hence they all have the
same mean (call itµi) and standard deviation (call itσi.)
That means the correlations satisfy
Corr(Xi, Xj)σiσj =E[(Xi−µi)(Xj−µj)]
=E[E[(Xi−µi)(Xj −µj)]|Bi, Bj]
= X
a∈{0,1} b∈{0,1}
P(Bi =a, Bj =b)E[(Xi−µi)(Xj −µj)|Bi=a, Bj =b].
Note when Bi = Bj = 0, then Xi =Fi−1(U) and Xj = Fj−1(U) which
are maximally correlated. So
E[(Xi−µi)(Xj −µj)|Bi= 0, Bj = 0] =ρ+ijσiσj.
Following this same logic further gives
E[(Xi−µi)(Xj −µj)|Bi = 1, Bj = 0] =ρ−ijσiσj
E[(Xi−µi)(Xj −µj)|Bi = 0, Bj = 1] =ρ−ijσiσj
Hence
Corr(Xi, Xj) =ρ+ijP(Bi=Bj) +ρ−ijP(Bi 6=Bj) =ρ+ijaij +ρ−ij(1−aij).
Theorem 2 is an immediate consequence.
The complete algorithm for generating multivariate distributions where componenti has cdfFi and Corr(Xi, Xj) =ρij is as follows.
Generating multivariate distributions with given correlation matrix
1) For every {i, j} ⊆ {1, . . . , n} do (whereA∼Unif([0,1]))
2) ρ+
ij ←Corr(F− 1
i (A), F− 1
i (1−A)), ρ +
ij ←Corr(F− 1
i (A), F− 1
i (1−A)).
3) λij ←(ρij−ρ−ij)/(ρ+ij −ρ−ij)
4) Draw U ←Unif([0,1])
5) Draw (B1, . . . , Bn) (Bi∼Bern(1/2)) with concurrence matrix Λ = (λij).
6) For all i∈ {1, . . . , n}, letXi←F− 1
i (U Bi+ (1−U)(1−Bi)).
Line 2 just generates the maximum and minimum correlation bounds from Theorem 1. Line 3 is set up so thatρij =λijρ+ij + (1−λij)ρ−ij.
Note that when the marginals for i and j are the same distribution, ρ+
ij = 1. However, even for equal marginals, the minimum correlation is
often strictly greater than −1. See [5] and the references therein for the minimum correlation for a variety of common distributions.
This algorithm is only applicable when it is actually possible to simulate from the multivariate symmetric Bernoulli distribution with specified con-currence matrix. The next section gives necessary and sufficient conditions for this to be possible for two, three, and four dimensions.
3
Attainable convexity matrices
In this section we prove Theorem 3 which gives necessary and sufficient conditions for dimensions two, three, and four for when it is possible to create a multivariate Bernoulli 1/2 with a given convexity matrix. We break the Theorem into three parts.
Lemma 1. For any λ12 ∈[0,1], there exists a unique joint distribution on
{0,1}2
such that (B1, B2)with this distribution has B1, B2 ∼Bern(1/2) and
Proof. Letpij =P(B1 =i, B2 =j). Then the equations that are necessary
and sufficient to meet the distribution and convexity conditions are:
p10+p11= 0.5, p01+p11= 0.5, p11+p00=λ12, and p00+p01+p10+p11= 1.
This system of linear equations has full rank, so there exists a unique solu-tion. Given there exists a unique solution, it is easy to verify that
p00= (1/2)λ12, p01= (1/2)[1−λ12], p10= (1/2)[1−λ12], p11= (1/2)λ12
satisfies the equations.
This provides an alternate algorithm to that found in [5] for simulating from bivariate distributions with correlation betweenρ−1,2 and ρ
+ 1,2.
Lemma 2. A random vector (B1, B2, B3) with Bi ∼Bern(1/2) exists (and
is possible to simulate from in a constant number of steps) if and only if the concurrence matrix satisfies
1≤λ23+λ12+λ13≤1 + 2 min{λ12, λ13, λ23}
Proof. Letting pijk =P(B1 =i, B2 =j, B3 =k), there are three conditions
from the marginals.
X
j,k∈{0,1}
p1jk= 0.5,
X
i,k∈{0,1}
pi1k= 0.5,
X
ij∈{0,1}
pij1 = 0.5.
Then there are three conditions from the correlations
X
k∈{0,1}
p00k+p11k=λ12,
X
j∈{0,1}
p0j0+p1j1 =λ13,
X
i∈{0,1}
pi00+pi11=λ23.
A seventh condition is P
i,j,kpijk = 1. Since eight equations are needed,
suppose that p111=α.
This 8 by 8 system of equations has full rank, so there is a unique solu-tion. It is easy to verify that the solution is
p000 = (1/2)(λ12+λ13+λ23−1)−α
p001 = (1/2)(1−(λ13+λ23)) +α
p010 = (1/2)(1−(λ12+λ23)) +α
p011 = (1/2)λ23−α
p100 = (1/2)(1−(λ12+λ13)) +α
p101 = (1/2)λ13−α
p110 = (1/2)λ12−α
In order for this solution to yield probabilities, all must lie in [0,1]. The p011, p101,and p110 equations give the restriction
α≤(1/2) min{λ12, λ23, λ13}. (2)
Of course, α≥0 so the p111 equation can be satisfied.
The p000 equation requires that
α≤(1/2)(λ13+λ12+λ23−1). (3)
With these two conditions, equationsp001,p010, andp100 give the constraint
α≥(1/2)(λ13+λ12+λ23−min{λ13, λ12, λ23} −1). (4)
The lower bound in (4) is below the upper bound in equation (3), but not necessarily below the upper bound in equation (2). So this gives another restriction:
(1/2)(λ13+λ12+λ23−min{λ13, λ12, λ23}−1) ≤(1/2) min{λ13, λ12, λ23}. (5)
As long as (3) and (5) are satisfied, their will exist a solution. If satisfied with strict inequality, there will be an infinite number of solutions.
Note that not all positive definite correlation matrices are attainable with Bern(1/2) marginals. For instance, if λ12 = λ13 = λ23 = 0.3, then ρ12=ρ13=ρ23=−0.4.With diagonal entries 1, theρvalues form a positive
definite matrix, but it is impossible to build a multivariate distribution with
Bern(1/2) marginals with these correlations.
4
Asymmetric Bernoulli distributions
Now consider the problem of drawing a multivariate Bernoulli (X1, . . . , Xn)
where Xi ∼ Bern(pi) where i is not necessarily 1/2, and the concurrence
matrix is given.
Lemma 3. Anndimensional multivariate Bernoulli distribution where the marginal of component iis Bern(pi) and concurrence matrix Λexists if and
only if an n+ 1dimensional multivariate Bernoulli distribution exists with
Bern(1/2) marginals and concurrence matrix
Λ
p..1 .
pn
p1· · ·pn 1
Proof. Suppose such ann+ 1 dimensional distribution exists withBern(1/2)
marginals and specified concurrence matrix. Let (B1, . . . , Bn+1) be a draw
from this distribution. Then set Xi = 1(Bi = Bn+1). The concurrence
matrix givesP(Xi = 1) =pi, and fori6=j,P(Xi=Xj) =P(Bi=Bj) =λij.
Conversely, suppose such an n dimensional distribution with Bern(pi)
marginals exists. Let Bn+1 ∼ Bern(1/2) independent of the Xi, and set
Bi =Bn+1Xi+(1−Bn+1)(1−Xi).ThenP(Bi = 1) = (1/2)pi+(1/2)(1−pi) =
1/2, and P(Bi = Bn+1) = pi, the correct concurrence parameter. Finally,
fori6=j,
P(Bi =Bj) =P(Xi =Xj) =λij.
This result gives an alternate way of deriving Lemma 2 using the Fr´echet-Hoeffding bounds. Consider a bivariate asymmetric Bernoulli (X, Y) where X ∼Bern(p), and Y ∼Bern(q). The pseudoinverse cdf of Bern(p) is given
byF−1
p (U) =1(U >1−p). Hence the minimum and maximum correlation
ofX andY can be directly found using Theorem 1:
ρ−(X, Y) = (p+qp−1)1(p+q >1)−pq
pq(1−p)(1−q) ρ+(X, Y) = pmin(p, q)−pq
pq(1−p)(1−q).
Suppose that we want to simulate X and Y so that P(X =Y) =r. For whatr is this possible?
Letpij =P(X=i, Y =j). Then
P(X =Y) =r =p11+p00
P(X= 1) =p=p10+p11
P(Y = 1) =q =p01+p11
Adding the three equations above we have
p+q+r=p11+p00+p10+p11+p01+p11= 1 + 2p11
and thereforep11= (p+q+r−1)/2. On the other hand
E(XY) =E(1(X= 1, Y = 1)) =P(X= 1, Y = 1) =p11.
The Fr´echet-Hoeffding Theorem provides the condition
E(F−1 p (U)F−
1
q (1−U))≤E(XY)≤E(F− 1 p (U)F−
which is equivalent to
(p+q−1)1(p+q >1)≤(p+q+r−1)/2≤min(p, q). Solving the inequalities for r we get the following condition
|1−(p+q)| ≤r≤2 min(p, q) + 1−(p+q).
Lettingp=λ13,q=λ23, andr=λ12 we obtain an equivalent condition
to the one in Lemma 2.
As a further consequence of this new proof technique, we obtain a simple method for simulating from the 3 dimensional symmetric Bernoulli distri-bution for given concurrence probabilities when possible.
Generating trivariate symmetric Bernoulli
Input: 3 by 3 matrix (λij) with 1≤λ12+λ13+λ23≤1 + 2 min{λ12, λ13, λ23}. Output: (B1, B2, B3) where (∀i)(Bi ∼Bern(1/2)) and (∀i, j)(P(Bi =Bj) =λij)
1) Draw B3←Bern(1/2) andU ←Unif([0,1]).
2) X1←1(U ∈[0, λ13])
3) X2←1(U ∈[(1 +λ13−λ23−λ12)/2,(1 +λ13+λ23−λ12)/2])
4) B1←B3X1+ (1−B3)(1−X1)
5) B1←B3X2+ (1−B3)(1−X2)
Proposition 2. The above algorithm simulates (B1, B2, B3) where Bi ∼
Bern(1/2) for all iand P(Bi=Bj) =λij for alli and j.
Proof. Note 1≤λ12+λ13+λ23≤1 + 2 min{λ12, λ13, λ23} if and only if
0≤ 1 +λ13−λ23−λ12
2 ≤λ13≤
1 +λ13+λ23−λ12
2 ≤1.
Hence
P(X1 = 1) =λ13−0 =λ13.
P(X2 = 1) =
1 +λ13+λ23−λ12
2 −
1 +λ13−λ23−λ12
2 =λ23.
P(X1=X2) =
λ13−
1 +λ13−λ23−λ12
2
+
1−1 +λ13+λ23−λ12
2
=λ12.
Lemma 3 also allows a characterization of the 4 dimensional symmetric Bernoulli concurrence (and so also convexity) matrices.
Lemma 4. A random vector (B1, B2, B3, B4) with Bi ∼ Bern(1/2) exists
(and is possible to simulate in a constant number of steps) if and only if for
ℓ= min(λ14+λ14+λ13+λ23, λ14+λ34+λ12+λ23, λ24+λ34+λ12+λ13)
u= min
{i,j,k}λij +λjk+λik,
it is true that
1−(1/2)ℓ≤(1/2)(u−1)
Proof. By using Lemma 3, the problem is reduced to finding a distribution for (X1, X2, X3) whereXi ∼Bern(λi4) and the upper 3 by 3 minor of Λ is the
new concurrence matrix. Just as in Lemma 2, this gives eight equations of full rank with a single parameter α. Letting qijk=P(X1 =i, X2 =j, X3 =
k), the unique solution is
q000= (1/2)(λ12+λ13+λ23)−(1/2)−α
q001=−(1/2)(λ14+λ24+λ13+λ23) + 1 +α
q010=−(1/2)(λ14+λ34+λ12+λ23) + 1 +α
q011= (1/2)(λ24+λ34+λ23)−(1/2)−α
q100=−(1/2)(λ24+λ34+λ12+λ13) + 1 +α
q101= (1/2)(λ14+λ34+λ13)−(1/2)−α
q110= (1/2)(λ34+λ24+λ12)−(1/2)−α
q111=α.
All of these right hand sides lie in [0,1] if and only if 1−(1/2)ℓ ≤
(1/2)(u−1), andα is chosen to lie in [1−(1/2)ℓ,(1/2)(u−1)].
As with the 3 dimensional case, this proof can be used to simulate a 4 dimensional multivariate symmetric Bernoulli: generate (X1, X2, X3) using
anyα∈[1−(1/2)ℓ,(1/2)(u−1)] and theqdistribution, then generateB4 ∼
Bern(1/2), and then setBi to beB4Xi+ (1−B4)(1−Xi) fori∈ {1,2,3,4}.
5
Conclusions
for higher dimensions the correlation matrix can be written as a convex-ity matrix whose parameters indicated where on the line from the lower to the upper bound the correlation lies. With a simple algorithm, we have shown that when viewed as a convexity matrix problem, the worse case for marginals is when they are all symmetric Bernoulli random variables. It is worst case in the sense that if it is possible to build a symmetric Bernoulli multivariate distribution with the given convexity matrix, then it is possi-ble to build a multivariate distribution with that convexity matrix for any marginal distributions. For symmetric Bernoulli marginals, the convexity matrix is also the concurrence matrix that gives the probabilities that any pair of random variables are the same.
For three and four dimensions, the set of convexity matrices that yield a symmetric Bernoulli distribution is characterized completely. For five or higher dimensions, every subset of three and four have these characteriza-tions as necessary condicharacteriza-tions.
Acknowledgement Support from the National Science Foundation (grant DMS - 1007823) is gratefully acknowledged.
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