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Correlation Matrices with Average Constraints
Jan Tuitman∗Steven Vanduffel†Jing Yao‡
Abstract
We develop an algorithm that makes it possible to generate all corre-
lation matrices satisfying a constraint on their average value. We extend
the results to the case of multiple constraints. These results can be used
to assess the extent to which methodologies driven by correlation matrices
are robust to misspecification thereof.
1 Introduction
Correlation matrices are a crucial model component in many disciplines includ-
ing statistics, risk management, finance and insurance. For instance, they appear
in portfolio optimization models (e.g., Markowitz’ framework), models for set-
ting capital requirements (e.g., Solvency II directive), and credit risk portfolio
management (e.g., KMV’s credit risk portfolio framework). Clearly, it is often
difficult to accurately estimate all pairwise correlations, i.e., full information
is rarely available. For instance, in a credit risk context the pairwise default
correlation between the default events of two obligors is essentially equivalent
to the probability they default together. As defaults are rare events such joint
probability is very hard to estimate with a reasonable degree of precision. Simu-
lating correlation matrices that are consistent with all available, yet incomplete,
information is then useful to provide insight into the sensitivity of the model
used. Indeed, while pairwise correlations might not be (all) known, information
might be available at a more aggregate level. Specifically, the average correla-
tion might be known with sufficient degree of accuracy. In this paper, we aim to
describe the set of all correlation matrices that are consistent with one or more
average constraints. Specifically, we provide an algorithm to randomly simulate
from this set.
We are not the first to describe sets of correlation matrices that satisfy some
constraints. Marsaglia and Olkin (1984) discuss how to generate correlation ma-
trices with given eigenvalues. Joe (2006) proposes a method to generate correla-
tion matrices based on partial correlations, which was extended in Lewandowski,
∗KU Leuven, Belgium
†Vrije Universiteit Brussel, Belgium. R-Code is available from www.stevenvanduffel.com
‡Maxwell Institute for Mathematical Sciences and School of Mathematical and Computer
Sciences, Heriot-Watt University, Edinburgh, U.K.
1
Kurowicka and Joe (2009) using a vine method. To account for the uncertainty
on the input correlations, Hardin, Garcia and Golan (2014) propose to add in
a controlled manner a noisy matrix to the input correlation matrix. Our paper
is closest to the one of H¨uttner and Mai (2019) who propose an algorithm to
generate correlation matrices with Perron-Frobenius property. As this property
is known to often hold in financial data series such algorithm is highly relevant
for applications in finance. In this letter, we propose an algorithm to generate
random correlation matrices with given average correlation. We extend the re-
sults to the case of multiple average constraints. To the best of our knowledge,
we are the first to study this problem.
The rest of the letter is organized as follows. In Section 2, we develop an
algorithm to generate correlation matrices with given average. Section 3 provides
the extension to further account for multiple average constraints. We provide
pseudo-code and numerical examples showing that our algorithms work well.
Section 4 concludes the paper.
2 Correlation Matrices with a Given Average
2.1 Problem formulation
Let n≥2 be an integer. We denote by Mn×n(R) the set consisting of n×n
matrices in R. A correlation matrix C∈Mn×n(R) is a positive semidefinite
matrix that is symmetric and which has ones on the diagonal. We denote by
h·,·i the standard inner product on Rn, i.e., for all x, y ∈Rn,hx, yi=Pn
i=1 xiyi
. Furthermore, k·k denotes the corresponding (Euclidean) norm, i.e., kxk=
phx, xi. In the sequel, we will also refer to the norm of a vector as its length.
For x, y ∈Rn, the so-called triangle inequalities are well-known:
| kxk−kyk | ≤ kx−yk≤kxk+kyk.
Marsaglia and Olkin (1984) prove the following theorem, which provides a
manner to generate correlation matrices matrix C∈Mn×n(R).
Theorem 1 (Marsaglia and Olkin (1984)) A matrix C∈Mn×n(R)is a
correlation matrix if and only if there exists a matrix T∈Mn×n(R)such that
C=T T twhere the rows of Tconsidered as vectors in Rnhave length 1. More-
over, one can assume the matrix Tto be lower triangular, i.e. Tij = 0 for all
i<j, in which case Tis unique.
Theorem 1 is also known as Cholesky decomposition; see e.g. Gentle (2012).
We aim to design an algorithm that makes it possible to generate all correla-
tion matrices that have a certain average correlation value, i.e, for given weights
σi>0 and a given (admissible) value for ρwe wish to generate all correlation
matrices C∈Mn×n(R) such that
Pi<j σiσjCij
Pi<j σiσj
=ρ(1)
2
Clearly, fixing ρis equivalent to fixing Pn
i,j=1 σiσjCij , which can be seen as
the variance of a random sum X1+... +Xnin which the components Xiare
obtained from independent standard normals Nivia the transformation.
Xi=σi
n
X
j=1
Tij Nj, i = 1, ..., n. (2)
We formulate the following problem:
Problem I Given an integer n≥2, positive real numbers σ1, . . . , σnand a non-
negative real number s, generate correlation matrices Csatisfying the variance
constraint
s2=
n
X
i,j=1
Cij σiσj.
In what follows we will call Pn
i,j=1 Cij σiσjthe total variance of the matrix
C. Note that if Cis the solution to Problem I, i.e., when Chas total variance
equal to s2then Chas average correlation ρsgiven as
ρs=s2−Pn
i=1 σ2
i
2Pi<j σiσj
.(3)
The problem we deal with can thus be interpreted as finding matrices Tthat
linearly transform (via (2)) standard independent normal random variables Ni
into jointly normal variables Xithat have zero mean, standard deviation σiand
satisfy the constraint var(X1+X2+... +Xn) = s2. If we denote the rows of T
by ti∈Rn, we can write Cov(Xi, Xj) = σiσjhti, tjiand we obtain that
s2=kσ1t1+σ2t2+... +σntnk2
must hold. In other words, to solve Problem I we only need to generate n
vectors ti∈Rn(rows of T) having length one and satisfying s2=kσ1t1+σ2t2+
... +σitnk2.Clearly, Problem I may not be well posed, for instance when sis
larger than Piσi. We characterize the admissible values of sin Lemma 1.
Lemma 1 (Well-posedness) Problem I has a solution if and only if
max{σimax −X
i6=imax
σi,0} ≤ s≤
n
X
i=1
σi,
where 1≤imax ≤ndenotes an index for which σimax is maximal among all σi.
Proof.
Suppose there exists a solution, then s=kPn
i=1 σitikis clearly non-negative
and by the triangle inequalities satisfies
σimax −X
i6=imax
σi≤s≤
n
X
i=1
σi.
Conversely, it is clear that for all ssatisfying these conditions a solution exists.
3
2.2 The solution (algorithm) to Problem I
To solve Problem 1 we iteratively construct ti∈Rn(i= 1, ..., n) such that
the subsequent partial sums σ1t1+. . . +σitihave an admissible length in that
kσ1t1+. . . +σntnk=scan be attained. We formally define
li=kσ1t1+. . . +σitik, i = 1,2, ..., n. (4)
and note that the lican also be seen as the variances of the partial sums Si=
X1+... +Xi. Furthermore,
l1=σ1and ln=s(5)
must hold and because of the triangle inequalities we also have that
|li−li−1| ≤ σi≤li+li−1,2≤i≤n. (6)
We proceed in two steps. First, we explain how to generate a series of ad-
missible lengths l1, ..., lnin a random way. Using the l1, . . . , lnas input, we are
then able to iteratively generate vectors t1, . . . , tnin Rnin a second step such
that ktik= 1 and kσ1t1+. . . +σitik=lifor all 1 ≤i≤n. In particular,
kσ1t1+. . . +σitnk=s, i.e., C=T T tin which Tis a matrix with rows tisolves
problem I.
Theorem 2 tells us when a partial sequence l1, . . . , lk(k < n) can be extended
to obtain a complete sequence l1, ..., lnand provides a manner to generate C
that solves Problem I.
Theorem 2 Suppose that σ1≤σ2≤. . . ≤σn. Let 1≤k≤n−2be an integer
and l1, . . . , lknon-negative real numbers that satisfy l1=σ1and |li−li−1| ≤
σi≤li+li−1for all 2≤i≤k. Then the following two are equivalent:
1. There exist non-negative real numbers lk+1, . . . , lnwith ln=ssuch that
|li−li−1| ≤ σi≤li+li−1for all 2≤i≤n.
2. |s−lk| ≤ σk+1 +. . . +σnand s+lk≥σn−(σk+1 +. . . +σn−1).
Proof.
(1.⇒2.) If lk+1, . . . , lnsatisfying the conditions exist, then
|s−lk|=|ln−lk| ≤ |lk+1 −lk|+. . . +|ln−ln−1| ≤ σk+1 +. . . +σn,
s+lk= (ln+ln−1)+(lk−lk+1) + . . . + (ln−2−ln−1)≥σn−(σk+1 +. . . +σn−1).
(2.⇒1.) We construct the liinductively for k+ 1 ≤i≤n−1, by choosing
li∈
max{|li−1−σi|, s −
n
X
j=i+1
σj, σn−
n−1
X
j=i+1
σj−s},min{li−1+σi, s +
n
X
j=i+1
σj}
4
arbitrarily. Note that this closed interval has been obtained by imposing all
conditions from 1. and 2. on li. From the assumptions
|s−li−1| ≤
n
X
j=i
σjand s+li−1≥σn−
n−1
X
j=i
σj
it follows that this interval is non-empty. Here we are also using that the σiare
increasing. It follows from the definition of lithat
1. |li−li−1| ≤ σi≤li+li−1,
2. |s−li| ≤ σi+1 +. . . +σn,
3. s+li≥σn−(σi+1 +. . . +σn−1).
Finally, using the definition of ln−1it is easily checked that ln=ssatisfies
|ln−ln−1| ≤ σn≤ln+ln−1.
Note that in Theorem 2 the assumption of increasing σiis made without
loss of generality.
We use Theorem 2 to provide an explicit algorithm for generating feasible
lengths l1, . . . , ln.
Algorithm 1a : Step 1 - Generation of series of admissible lengths
given σ1, . . . , σnincreasing and ssuch that
σn−(σ1+. . . +σn−1)≤s≤σ1+. . . +σn
generate l1, . . . , ln
l1:= 1
for i:= 2 to n−1 do
li:= random in [max{s−Pn
j=i+1 σj, σn−Pn−1
j=i+1 σj−s, |li−1−σi|},
min{s+Pn
j=i+1 σj, li−1+σi}]
end for
ln:= s
return l1, . . . , ln=0
Note that when implementing this algorithm, one still needs to choose a
distribution of the lion the intervals of allowed values for li. A natural first
choice could be to take the uniform distribution on these intervals. However,
other choices could be envisaged as well.
The following lemma tells us how to generate a vector tkbased on lengths
l1, . . . , lkand vectors t1, . . . , tk−1.
Lemma 2 Assume that t1, . . . , tk−1are vectors in Rnsuch that ktik= 1 and
kuik=lifor all 1≤i≤k−1in which ui:= σ1t1+. . . +σiti.Then we have
the following equivalence:
kukk=lk⇔ htk, uk−1i=l2
k−l2
k−1−σ2
k
2σk
(7)
5
Proof. This follows easily from:
kukk2=kuk−1+σktkk2
=kuk−1k2+kσktkk2+ 2hσktk, uk−1i
=l2
k−1+σ2
k+ 2σkhtk, uk−1i
We now provide the pseudo-code for implementing the second step of our
approach, i.e., the generation of row vectors tithat compose a matrix Tsuch
that C=T T tsolves Problem I.
Algorithm 1b : Step 2 - Generation of row vectors ti
generate t1, . . . , tn, given l1, . . . , ln
t1:= random in Rnof length 1
u1:= σ1t1
for i:= 2 to ndo
x:= random in Rnof length 1
y:= x−ui−1hx, ui−1i/kui−1k2
z:= (ui−1/kui−1k2)(l2
i−l2
i−1−σ2
i)/(2σi)
ti:= z+ (p1− kzk2/kyk)y
ui:= ui−1+σiti
end for
return t1, . . . , tn=0
Proof of correctness. Note that in the loop yis orthogonal to ui−1which
is parallel to zand hz, ui−1i= (l2
i−l2
i−1−σ2
i)/(2σi). Since tidiffers from zby
a multiple of y, we have hti, ui−1i=hz, ui−1i. Finally, one easily checks that
hti, tii= 1.
The matrix Tis then composed by the row vectors t1, . . . , tnand the corre-
lation matrix C=T T tis then a solution to Problem I.
We present an example.
Example 1 We provide an example of how to compute a random 6×6corre-
lation matrix Cwith average correlation ρ= 0.2using Algorithms 1a and 1b.
We use uniform distributions when simulating the lengths li. This gives as a
particular solution,
C=
1.0000000 −0.7669058 0.1442050 0.1372040 0.3284773 0.6463319
−0.7669058 1.0000000 −0.6059352 −0.2850269 −0.3895216 −0.6645325
0.1442050 −0.6059352 1.0000000 0.7252801 0.7406292 0.6270028
0.1372040 −0.2850269 0.7252801 1.0000000 0.9604118 0.6341328
0.3284773 −0.3895216 0.7406292 0.9604118 1.0000000 0.7682472
0.6463319 −0.6645325 0.6270028 0.6341328 0.7682472 1.0000000
6
3 Multiple Average Constraints
We now want to generate random correlation matrices with some blocks (sub-
matrices) each satisfying an average constraint. To make this more precise, sup-
pose that n=n1+. . . +nkare all positive integers and that we are given an
ni×nicorrelation matrix Cifor each 1 ≤i≤k, where each Cisatisfies an
average constraint, as discussed in the previous section.
In this section, we specifically denote by Ti∈Mni×ni(R) the lower triangular
matrix such that Ci=TiTt
i. The rows of each Tiare denoted as vectors tij ∈Rni,
j= 1,2, ..., ni. Finally, we define vi∈Rnias
vi=σi1ti1+. . . +σinitini,(8)
and denote
kvik=si,1≤i≤k(9)
i.e., siis the total variance of the correlation matrix Ci.
We first explain how to generate a random n×ncorrelation matrix Cwhich
contains the Cias consecutive blocks along the diagonal. Next, we explain how
to do this in the presence of an additional variance constraint (equivalently, an
average correlation constraint) on C.
3.1 Block structures with given averages
Our problem reads as follows.
Problem II Generate random correlation matrices Ccontaining the Cias con-
secutive blocks along the diagonal.
To solve this problem we will use orthogonal matrices. We first introduce
some notation and definitions.
Definition 1 Let m < n be positive integers. We will always consider Rmto
be a vector subspace of Rnby embedding it onto the first mcomponents, i.e.
by sending the vector (x1, . . . , xm)to the vector (x1, . . . , xm,0,...,0). Note that
this choice of embedding preserves the standard inner product, by which we mean
that hx, yidoes not depend on whether we consider x, y ∈Rmto be elements of
Rmor Rn.
Definition 2 For a positive integer n, we let O(n)denote the orthogonal group
in dimension n, i.e. a matrix Mn×n∈O(n)if and only if it preserves the
standard inner product on Rn, by which hMv, M wi=hv, wifor all vectors
v, w ∈Rn.
The following lemma is crucial in solving Problem II.
Lemma 3 Let m, n be positive integers. Suppose that v1, . . . , vmand w1, . . . , wm
are vectors in Rnsuch that hvi, vji=hwi, wjifor all 1≤i, j ≤m. Then there
exists an orthogonal matrix M∈O(n)such that Mvi=wifor all 1≤i≤m.
7
Proof. Let Vbe the vector space spanned by v1, . . . , vmand Wthe vector
space spanned by w1, . . . , wm. Note that if c1v1+. . . +cmvm= 0 for some
c1, . . . , cm∈Rthen c1w1+. . . +cmwm= 0 as well. This follows from the fact
that a vector u∈Rnis zero if and only if hu, ui= 0. In particular this implies
that dim(V) = dim(W). Thus, there exists a unique linear transformation L:
V→Wsuch that L(vi) = wifor all 1 ≤i≤m. Let e1, . . . , elbe an orthogonal
basis for the orthogonal complement of Vin Rnand f1, . . . , flan orthogonal
basis for the orthogonal complement of Win Rn. We extend Lto all of Rn
by setting L(ei) = fifor all 1 ≤i≤l. By construction, Lpreserves all inner
products on Rn. Therefore, its matrix Mis indeed an element of O(n).
Now we are ready to present the algorithm that solves Problem II.
Algorithm 2
generate Cwith blocks C1, . . . , Ckalong the diagonal
for i:= 1 to kdo
Let Tibe the lower triangular matrix (with rows tij) such that Ci=TiTt
i
Mi:= random in O(n)
τij := Mitij
end for
T:= matrix with rows τij
C:= T T t
return C=0
Remark 1 Note that we have not specified the distribution of a random ele-
ment of O(n). However, in this case (unlike in the previous section) there is a
very natural choice of distribution. Since O(n)is a compact topological group, it
carries a unique (left) translation invariant probability measure called the Haar
measure. Note that there are good algorithms available to generate random or-
thogonal matrices according to this distribution (Stewart (1980)).
Example 2 Suppose we are looking for a 6×6correlation matrix Csuch that
the 3×3upper left corner has average correlation 0.2and the 3×3lower right
corner average correlation 0. Using Algorithms 1a and 1b that we developed in
Section 1 we obtain for the upper left block
C1=
1.0000000 0.1307564 0.8562048
0.1307564 1.0000000 −0.3869612
0.8562048 −0.3869612 1.0000000
and for the lower right block
C2=
1.00000000 0.03949805 0.4043501
0.03949805 1.00000000 −0.4438482
0.40435012 −0.44384816 1.0000000
To find a correlation matrix Cwith C1and C2as blocks along the diagonal, we
8
now apply Algorithm 2 and we obtain
C=
1.0000000 0.1307564 0.85620478 0.73725058 0.51956587 0.0912944
0.1307564 1.0000000 −0.38696122 −0.43638596 0.84869870 −0.7474391
0.8562048 −0.3869612 1.00000000 0.87710025 0.06164776 0.5356547
0.7372506 −0.4363860 0.87710025 1.00000000 0.03949805 0.4043501
0.5195659 0.8486987 0.06164776 0.03949805 1.00000000 −0.4438482
0.0912944 −0.7474391 0.53565467 0.40435012 −0.44384816 1.0000000
3.2 Adding a global average constraint
As before, let n=n1+. . . +nkbe positive integers. we want to generate a
random n×ncorrelation matrix Cwhich contains ni×nicorrelation matrices
Cias consecutive blocks and which also satisfies a global average constraint.
Our problem can be formulated as follows:
Problem III Generate random correlation matrices Ccontaining the Ci,i=
1,2, ..., k as consecutive blocks along the diagonal and satisfying the variance
constraint s2, i.e., for given weights σij ,i= 1,2, ..., k;j= 1,2, ..., niand given
s > 0,Cmust also satisfy
s2=
n
X
i,j=1
Cij σiσj.(10)
Let C=T T tsolve Problem III and denote by τij ∈Rnthe rows of T(first
ordered by 1 ≤i≤kthen 1 ≤j≤ni). In what follows, we also specifically
denote the weights by σij (1 <=i <=k, 1<=j <=ni); that is, σij is equal
to σlwhere l=n1+...ni−1+jwith the convention that n1+n0= 0. Let
wi=σi1τi1+. . . +σiniτini, 1 ≤i≤k. As Csolves Problem II, it holds that
kw1+. . .+wkk=sand kwik=sifor all 1 ≤i≤k. The following theorem shows
how to derive from the given matrices Tia matrix Twith these properties. In
particular, C=T T tthen solves Problem III.
Theorem 3 Let w1, . . . , wk∈Rnbe vectors such that kw1+. . . +wkk=sand
kwik=sifor all 1≤i≤k. Then for all 1≤i≤kthere exists an orthogonal
matrix Mi∈O(n)such that wi=Mivi. For all 1≤i≤kand 1≤j≤nidefine
τij ∈Rnby τij =Mitij . Let Tbe the matrix with rows τij (first ordered by i
then j). Then the matrix C=T T tis a solution to Problem III.
Conversely, suppose that the matrix Cis a solution to Problem III and let
Tbe a matrix such that C=T T t. Let τij ∈Rnbe the rows of T(first ordered
by 1≤i≤kthen 1≤j≤ni). For all 1≤i≤k, let wi∈Rndenote the
vector wi=σi1τi1+. . . +σiniτini. Then kw1+. . . +wkk=sand kwik=sifor
all 1≤i≤k. Moreover, for each 1≤i≤kthere exists an orthogonal matrix
Mi∈O(n)such that Mitij =τij for all 1≤j≤ni.
Proof. If kwik=si, then hvi, vii=hwi, wii. Therefore, by Lemma 3 for all
1≤i≤kthere exists Mi∈O(n) such that Mivi=wi. Since Miis orthogonal
and τij =Mitij the inner products between τij are the same as between tij for
9
any fixed value of 1 ≤i≤k. Therefore, the matrix Ccontains Cias consecutive
blocks along the diagonal and the total variance of Cis equal to
kX
1≤i≤k
1≤j≤ni
σij τij k2=k
k
X
i=1
wkk2=s2,
so that the matrix Cis indeed a solution to Problem III.
For the converse, let Cbe a solution to Problem III and let Tbe a matrix
such that C=T T t. Moreover, let τij be the rows of T(first ordered by 1 ≤i≤k
then 1 ≤j≤ni). Since the matrix Ccontains Cias consecutive blocks along
the diagonal, the inner products between τij are the same as between tij for
any fixed value of 1 ≤i≤k. Therefore, by Lemma 3 for all 1 ≤i≤kthere
exists an orthogonal matrix Mi∈O(n) such that Mitij =τij for all 1 ≤j≤ni.
In particular, we have that kwik=kvik=sifor all 1 ≤i≤k. The equality
kw1+. . . +wkk=sfollows directly from the fact that the matrix Chas total
variance s2.
According to Theorem 3, Problem III can be solved in two steps:
1. Generate vectors w1, . . . , wk∈Rnsuch that kw1+. . . +wkk=sand
kwik=sifor all 1 ≤i≤k.
2. Generate M1, . . . , Mk∈O(n) such that Mivi=wi(this is always possi-
ble). Let Tbe the matrix with rows τij =Mitij (first ordered by 1 ≤i≤k
then 1 ≤j≤ni) and define C=T T t.
Step 1 is similar to approach we used to solve Problem I in Section 1. There
we generated t1, . . . , tn∈Rnsuch that kσ1t1+. . . +σntnk=sand ktik= 1
for all 1 ≤i≤n. Here we want to generate w1, . . . , wk∈Rnwith k≤n
such that kw1+. . . +wkk=Sand kwik=Si. Taking n=k,σi=Siand
ti=wi/Sithis is equivalent, apart from the fact that w1, . . . , wkare to be
contained in Rnand not Rk. However, this last problem is easily solved by
generating w1, . . . , wk∈Rk, embedding them into Rnas in Definition 1 and
multiplying all by a single random orthogonal matrix M∈O(n).
Moreover, in a similar way as in Theorem 1 in Section 2, siand shave to be
admissible to generate the corresponding correlation matrices. Thus, we have
the following theorem.
Theorem 4 Problem III has a solution if and only if
simax −X
i6=imax
si≤s≤
k
X
i=1
si,
where imax is an index for which siis maximal.
Proof. We know from (the proof of) Theorem 3 that Problem III has a
solution if and only if there exist w1, . . . , wk∈Rnsuch that kw1+. . . +wkk=s
10
and kwik=sifor all 1 ≤i≤k. Therefore, the result follows from Theorem 1.
Algorithm 3
generate Cwith blocks C1, . . . , Ckalong the diagonal and total variance s2
generate t1, . . . , tk∈Rkwith ktik= 1 such that ks1t1+. . . +sktkk=s(as
in Section 2) and set wi:= sitifor all 1 ≤i≤k
M:= random in O(n).
for i:= 1 to kdo
wi:= Mwi// consider wias an element of Rn
Compute Ti(with rows tij) from Ci
vi:= σi1ti1+. . . +σinitini
Mi:= random in O(n) such that Mivi=wi
for j:= 1 to nido
τij := Mitij
end for
end for
T:= matrix with rows τij
C:= T T t
return C=0
We present numerical examples with block structure.
Remark 2 The complexity of all our algorithms is driven by the complexity
of the computation of the matrix product C=T∗Tt. This complexity is at
least O(n2)and, by the latest results on fast matrix multiplication, it is at most
O(n2.3728639); see, Le Gall (2014).
Example 3 Generate a correlation matrix C6×6with upper 3×3sub-average
correlation 0.2, lower 3×3sub-average correlation 0.4and total average corre-
lation ρ= 0.25. We show 2 samples of simulation.
1 0.44953864 0.4383173 0.6524330 0.2731052 0.60398060
0.44953864 1 −0.2878560 −0.1556668 −0.1022461 0.08723503
0.4383173 −0.2878560 1 0.3080049 0.3916332 −0.10847894
0.6524330 −0.1556668 0.3080049 1 0.6190522 0.68384052
0.2731052 −0.1022461 0.3916332 0.6190522 1 −0.10289275
0.60398060 0.08723503 −0.10847894 0.68384052 −0.10289275 1
;
1 0.34104551 0.29730654 0.0995247 −0.1172317 0.1399298
0.34104551 1 −0.03835204 0.6223703 0.4797049 0.4089176
0.29730654 −0.03835204 1 −0.5257222 0.4260202 0.4164864
0.0995247 0.6223703 −0.5257222 1 0.2634622 0.2275388
−0.1172317 0.4797049 0.4260202 0.2634622 1 0.7089991
0.1399298 0.4089176 0.4164864 0.2275388 0.7089991 1
.
4 Conclusion
In this letter, we propose an algorithm to generate correlation matrices subject
to a given average correlation. We extend the algorithm to account for correla-
tion matrices with block structures. Our algorithms are useful for stress-testing
11
models. Specific applications to model risk assessment are considered in further
research. Finally, we wish to point out that the problem of finding correlation
matrices with given average that we consider in this paper is strongly related
to the so-called problem of joint mixability, which has been studied in a series
of papers initiated by Wang and Wang (2011) and Wang et al. (2013) with
application to the assessment of worst-case portfolio models under (partial) de-
pendence uncertainty; see e.g., Bernard et al. (2017a), Bernard et al. (2017b),
Embrechts et al. (2013), Puccetti et al. (2017) and Bernard, R¨uschendorf and
Vanduffel (2017). Specifically, A tuple of univariate distributions is said to be
jointly mixable if there exist random variables, with respective distributions,
such that their sum is a constant. The results in this paper make it possible
to describe all these joint mixes when the marginal distributions are normally
distributed and their dependence is required to be gaussian.
Acknowledgement Jing Yao acknowledges the National Natural Science Foun-
dation of China (No.11971506).
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