Central Bank of Brazil is implementing a Value At Risk (V.A.R.) methodology to establish minimum capital requirements for financial institutions to bear market risk derived from interest rate fluctuations. This article shows that the construction of the correlation matrix of the Brazilian Central Bank's Standard Model for Interest Rate is coherent, in the sense it is positive defined.

José Alvaro Rodrigues Neto

English

Research Papers in Economics

Working Paper Series
ISSN 1518-3548
The Correlation Matrix of the Brazilian Central Bank's
Standard Model for Interest Rate Market Risk
José Alvaro Rodrigues Neto
September, 2000  2 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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   4 
The Correlation Matrix of the Brazilian Central Bank’s 
Standard Model for Interest Rate Market Risk 
 
 
José Alvaro Rodrigues Neto
* 
 
 
Abstract 
 
 
Central Bank of Brazil is implementing a Value At Risk (V.A.R.) 
methodology to establish minimum capital requirements for financial 
institutions to bear market risk derived from interest rate fluctuations. This 
article shows that the construction of the correlation matrix of the Brazilian 
Central Bank’s Standard Model for Interest Rate is coherent, in the sense it 
is positive defined. 
                                                       
* Central Bank of Brazil. The author thanks Alexandre Tombini, André Amante, Fabio Araujo, Guilherme 
Arcoverde, Marta Baltar Moreira and Sérgio Werlang for their help. The remaining errors are only my 
responsability.    5 
The Correlation Matrix of the Brazilian Central Bank’s 
Standard Model for Interest Rate Market Risk 
 
 
1.  Introduction 
 
Brazilian Central Bank recently established capital requirements in order to prevent 
market risk derived from interest rate fluctuations. It was constructed a model to do so, 
named the Standard Model for Interest Rate Market Risk. It is based on a value at risk 
(VaR) methodology. 
 
This article deals with a technical aspect of Brazilian Central Bank’s Standard Model  
for capital requirements for financial institutions to bear market risk. This Standard 
Model has a parameterized structure. Within this structure, the correlation matrix of the 
series of the factors associated to the so called vertexes of the term structure of the 
interest rate has a crucial importance. 
 
This concepts were implemented by the construction of a matrix depending on two 
parameters that captures the observed behavior of the historical correlation between the 
vertexes. 
  
The problem that the construction brought was to guarantee that for all range of the 
parameters the adopted procedures are mathematically correct, that is, the correlation 
matrix must be positive defined, so the calculated value at risk is a well defined real 
positive number. 
 
The Brazilian VaR model for interest rate risk can be found in [1] and [2]. For a general 
treatment of VaR see  [3] or the references there. 
   6 
This work is divided in four sections, one table and 4 graphs. Section 2 explains the 
mathematical problem. Section 3 shows two ideas used to attack the problem, while 
section 4 explains the nature of the difficulty in solving it analytically, argues why it is 
necessary to use careful numerical procedures and explains why it is safe to say that the 
correlation matrix is in fact positive defined for a certain range of the parameters. 
 
 
2.  The Mathematical Problem 
 
The 7x7 matrix of correlation  7 7x A  with coefficients ()
7
1 , = j i ij ρ  is symmetric, with main 
diagonal of ones. The problem is to prove that it is positive defined for a certain range 
of values of its two parameters  k   , ρ . 
 
The definition of the coefficients are:  
 
) , ( ) 1 (
j i m
ij ρ ρ ρ− + = , 
 
where 
k
ij b j i m ) ( ) , ( = , and the matrix  7   , 1 ) ( ≤ ≤ j i ij b  is given by: 
 
= ≤ ≤ 7   , 1 ) ( j i ij b






















1 5 . 1 3 6 12 18 36
5 . 1 1 2 4 8 12 24
3 2 1 2 4 6 12
6 4 2 1 2 3 6
12 8 4 2 1 5 . 1 3
18 12 6 3 5 . 1 1 2
36 24 12 6 3 2 1
 
 
with the parameters  k    , ρ  varying between  9 . 0 1 . 0 ≤ ≤ ρ   and   9 . 0 1 . 0 ≤ ≤ k .    7 
 
Let  [] [] 9 . 0   , 1 . 0 9 . 0   , 1 . 0 × = K . It is clearly a compact set.  
 
It is easy to see that  7 7x A  is symmetric. It is a well known fact that every symmetric 
matrix is equivalent to a diagonal matrix of its eigenvalues, which are real numbers. So 
it will be positive defined if and only if all eigenvalues (and hence the minimum of 
them) are strict positive numbers. 
 
So it is enough to consider the function  ℜ → ℜ ⊂
2 : K f  that takes the parameters  k   , ρ  
in the minimum of the eigenvalues. Nevertheless it is not easy to measure the sensibility 
of  ℜ → ℜ ⊂
2 : K f  with respect to  k    , ρ , that is, to calculate 
k
f f
∂
∂
∂
∂
  ,
ρ
. In fact, it is not 
clear that this function is differentiable or even if it is continuos
1. The figures 1 and 2 
below show that when a parameter of a polynomial is changing smoothly, the minimum 
real zero of the polynomial can change in a non smooth way: 
 
f(x,e)=0.2(x+1.2)(X+0.8)(x-0.8)(x-1.2) +ex
-0.5
0
0.5
1
1.5
2
-3 -2 -1 0 1 2 3
f
(
x
)
 
 
Figure 1:  0 ) 0 , ( = x f  has 4 real roots. The lower one is smaller than –1.   8 
 
 
f(x,e)=0.2(x+1.2)(x+0.8)(x-0.8)(x-1.2)+ex
-0.5
0
0.5
1
1.5
2
-3 -2 -1 0 1 2 3
f
(
x
)
 
 
Figure 2:  0 ) 06 . 0 , ( = − x f  has only two real roots. The smaller one is greater than zero. 
 
Figures 3 and 4 in the end show the complete graph of   ℜ → ℜ ⊂
2 : K f  and its level 
curves. Figures 5 and 6 show zooms at particular region of the graph, where  f  has a 
strange behavior. Notice that  f  is always a continuos function. 
 
 
                                                                                                                                                             
1 In this case f will be continuos.   9 
3.  The Solution 
 
3.1 First Solution 
 
Initially the idea was to use the following well known theorem: 
 
Theorem (Sylvester’s Criterion for a Positive Defined Matrix): 
Let  A be a real symmetric square matrix. So  A is positive defined if and only if 
0 det > A  and  0 det > k A  for all minors  k A , the square sub matrix of  A, of order k , 
obtained from  A by deleting the last rows and columns, that is: 
 
0 11 > ρ ;  0
22 21
12 11 >
ρ ρ
ρ ρ
; ...;  0 > A . 
 
The algebraic expressions for this minors have many parcels and most of them are non 
analytical, so whether all the minors are positive in the desirable range of the parameters  
(9 . 0 1 . 0 ≤ ≤ρ  and  9 . 0 1 . 0 ≤ ≤ k , a continuos range) can only be verified numerically.  
 
A serious problem is that it is necessary to make an infinity number of calculations to 
check it out. 
   10 
3.2 Second Approach – Spectral Decomposition 
 
An different approach could be given by the observation that it is enough to prove that 
for all 
7 ℜ ∈ x  the quadratic formQ defined by: 
 
ℜ → ℜ ℜ
7 7 : x Q  
 
) , ( y x Q ∑ ∑
= =
=
7
1
,
7
1 j
j i j i
i
y x ρ   
 
and associated with the symmetric matrix  A, with coefficients  ij ρ , has the property: 
) , ( x x Q 0   ´ ≥ ⋅ ⋅ = x A x , 
7 ℜ ∈ ∀x , with equality holding if and only if 
7 0 ℜ ∈ = x .  
 
The main idea was the study of the problematic directions, that is, the directions where 
) , ( x x Q  assumes its minimum values. Formally, since  ) , ( y x Q  is a homogeneous 
function of second degree, it is enough to prove the claim for all 
6 S x∈ , the six 
dimensional sphere with unitary radius and center in the origin of 
7 ℜ . 
 
The Signal Matrix 
 
In order to find which parcels of  ) , ( x x Q  are positive it can be defined a 7x7 matrix ∑ , 
whose coefficients are either the positive sign (+) or the negative sign (-). 
   11 
For instance, if  j x  is positive for  4 , 3 , 2 , 1 = j  and negative for  7 , 6 , 5 = j  the matrix 
becomes: 
 
= Σ






















+ + + − − − −
+ + + − − − −
+ + + − − − −
− − − + + + +
− − − + + + +
− − − + + + +
− − − + + + +
 
 
It can be observed that the problematic directions are those corresponding to the vectors 
6 S x∈  with 4 positive coordinates and 3 negative coordinates or vice versa. In this case 
it can be easily checked there are 25 positive parcels and 24 negative ones in  ) , ( x x Q .  
These are the configurations of signs from the  j x , 7 , , 1￿ = j , that gives the minimum 
net amount of positive signs. So, without loss of generality, it can be supposed there are 
4 positive values of  j x  and 3 negative ones. 
 
Let  n l k x x x   ,  x ,   , m  be positive and  r  x ,   , q p x x  be negative, with: 
 
{} { } r q p n m l k , , , , , , 7 , , 1 = ￿  
 
In this case: 
 
= ) , ( x x Q + ∑
=
7
1
2
i
i iix ρ ∑ ∑
− ⊕
+
12
,
9
,
2 2
j i
j i ij
j i
j i ij x x x x ρ ρ  
   12 
where  ∑
⊕ 9
, j i
represents positive parcels, that is,  ) , ( j i  such that  {} n m l k j i , , , , ∈  or 
{} r q p j i , , , ∈  and ∑
− 12
, j i
represents the negative ones ( {} { } r q p j n m l k i , , , , , , ∈ ∈  or vice 
versa). 
 
It was numerically verified that the parameters  1 . 0 = ρ ,  1 . 0 = k  of  A are those which 
make it have the lower minimum eigenvalue. With this parameters it can be observed 
that all the coefficients  ij ρ  are about the same, say  ρ , between 0.96 and 0.99. A table 
with some matrixes for different parameters is presented in the end. 
 
Fix the worst direction, that is, the direction that makes the smaller eigenvalue of the 
matrix A (with fixed parameters  1 . 0 = ρ ,  1 . 0 = k ) assumes its minimum. This direction 
is formally represented by 
6 S x∈ . Associated with this fixed values of  j x  there is an 
unique signal matrix Σ. 
 
In this case: 
 
) , ( x x Q ) , ( . x x R ρ ≅ , where  = ) , ( x x R + ∑
=
7
1
2
i
i x ∑ ∑
− ⊕
+
12
,
9
,
2 2
j i
j i
j i
j i x x x x  
 
So the problem becomes to minimize  ) , ( x x R , with 
6 S x∈ . But R  is an semi-positive 
quadratic form. Its defined by a symmetric matrix (by abuse of notation it is also called 
R ) with eigenvalues   0   , 0   , 0   , 0   , 0   , 0   , 7 . The corresponding eigenvectors forms an 
orthogonal basis of 
7 ℜ . 
   13 
Let  7 v  be the eigenvector corresponding to the eigenvalue 7. It can be seen 
= ) , ( x x Rx ∀    , 0  such that  0 , 7 = 〉 〈 v x . So, there is an hiper-plane H  of co-dimension 
one, such that for all  H x∈ ,  = ) , ( x x R 0. 
 
In that way  ) , ( x x Q  can be extremely close to zero in an set with strict positive 
Lebesgue measure. Whether  ) , ( x x Q  will be greater than zero depends on the 
combination of two factors: the chosen direction and the values  ij ρ  (the parameters). At 
each fixed direction (Σ uniquely defined) the sign of  ) , ( x x Q  will be determinate by the 
sums: 
 
∑
− 12
,
2
j i
j i ij x x ρ   and    + ∑
=
7
1
2
i
i iix ρ ∑
⊕ 9
,
2
j i
j i ij x x ρ . 
 
 
4.  The Results and Conclusions 
 
Within the parameters ( 9 , 0 , 1 , 0 ≤ ≤ k ρ ) the Brazilian Central Bank adopts, the region 
where the lower eigenvalue of the matrix  A has its minimum is a neighborhood
2 of the 
point: 
 



=
=
1 . 0
1 . 0
k
ρ
 
 
being this value approximately 0.0028.  
 
                                                       
2 In fact, this point is the minimum.    14 
For this the numerical values of the function  ℜ → ℜ2 : f  and its partial derivatives 
k
f f
∂
∂
∂
∂
  ,
ρ
 were calculated. By the mean value theorem the desired result follows
3. 
However, it should be clear that it was used a numerical procedure. 
 
So, within the specified parameters  k    , ρ  of  A it is possible to guarantee that  A is 
positive defined. 
 
Nevertheless, the pure analytical prove by Sylvester’s Criterion is not possible because 
the determinants of all the minors  k A  are made of many parcels and a great part of them 
are non analytic. Checking if these expressions are positive for all range of the 
parameters is a problem similar to find a solution for the equation  x
x = 2 . It can only be 
done numerically. 
 
The methodology develop in this study can be easily expanded for similar problems
4. 
It is also interesting to observe the behavior of the function: 
 
[] [] ℜ → × 9 . 0    , 1 . 0 9 . 0    , 1 . 0 : f  (which takes the parameters  k    , ρ  in the minimum 
eigenvalue of  A) in the region: 
 



≤ ≤
≤ ≤
9 . 0 7 . 0
3 . 0 1 . 0
k
ρ
 
 
It looks like it is close to a non continuos point. See figure 5 in the end. 
                                                       
3 In fact, to be more rigors the numerical error of the derivatives should be calculated by the second 
derivatives, being their errors corrected by the third derivatives and so on. But numerical values are 
sufficient to ensure the desired results. 
4 Similar problems with a different formula for the coefficients or with different dimension.   15 
References 
 
[1] Lins Arcoverde, Guilherme “Alocação de Capital para Cobertura do Risco de 
Mercado de Taxas de Juros de Natureza Prefixada”, Dissertação de Mestrado, 
EPGE/FGV, 2000. 
 
[2] Research Department, Central Bank of Brazil  “The Brazilian Central Bank’s Model 
for Interest Rate Market Risk”, Working Paper Series, Banco Central do Brasil. 
 
[3] Jorion, Philippe ”Value at Risk New Benchmark for Controlling Market Risk”, Mc 
Graw Hill. 
   16 
Table 1 
 
m(i,j)
Rho 0.1 1.00 0.99 0.99 0.98 0.97 0.97 0.96 1.00 1.07 1.12 1.20 1.28 1.37 1.43
k0 . 1 0.99 1.00 1.00 0.99 0.98 0.97 0.97 1.07 1.00 1.04 1.12 1.20 1.28 1.34
0.99 1.00 1.00 0.99 0.99 0.98 0.97 1.12 1.04 1.00 1.07 1.15 1.23 1.28
0.98 0.99 0.99 1.00 0.99 0.99 0.98 1.20 1.12 1.07 1.00 1.07 1.15 1.20
0.97 0.98 0.99 0.99 1.00 0.99 0.99 1.28 1.20 1.15 1.07 1.00 1.07 1.12
0.97 0.97 0.98 0.99 0.99 1.00 1.00 1.37 1.28 1.23 1.15 1.07 1.00 1.04
0.96 0.97 0.97 0.98 0.99 1.00 1.00 1.43 1.34 1.28 1.20 1.12 1.04 1.00
m(i,j)
Rho 0.1 1.00 0.92 0.85 0.69 0.47 0.26 0.17 1.00 1.87 2.69 5.02 9.36 17.47 25.16
k0 . 9 0.92 1.00 0.96 0.85 0.69 0.47 0.34 1.87 1.00 1.44 2.69 5.02 9.36 13.48
0.85 0.96 1.00 0.92 0.79 0.60 0.47 2.69 1.44 1.00 1.87 3.48 6.50 9.36
0.69 0.85 0.92 1.00 0.92 0.79 0.69 5.02 2.69 1.87 1.00 1.87 3.48 5.02
0.47 0.69 0.79 0.92 1.00 0.92 0.85 9.36 5.02 3.48 1.87 1.00 1.87 2.69
0.26 0.47 0.60 0.79 0.92 1.00 0.96 17.47 9.36 6.50 3.48 1.87 1.00 1.44
0.17 0.34 0.47 0.69 0.85 0.96 1.00 25.16 13.48 9.36 5.02 2.69 1.44 1.00
m(i,j)
Rho 0.9 1.00 0.98 0.98 0.96 0.95 0.94 0.94 1.00 1.07 1.12 1.20 1.28 1.37 1.43
k0 . 1 0.98 1.00 0.99 0.98 0.96 0.95 0.95 1.07 1.00 1.04 1.12 1.20 1.28 1.34
0.98 0.99 1.00 0.98 0.97 0.96 0.95 1.12 1.04 1.00 1.07 1.15 1.23 1.28
0.96 0.98 0.98 1.00 0.98 0.97 0.96 1.20 1.12 1.07 1.00 1.07 1.15 1.20
0.95 0.96 0.97 0.98 1.00 0.98 0.98 1.28 1.20 1.15 1.07 1.00 1.07 1.12
0.94 0.95 0.96 0.97 0.98 1.00 0.99 1.37 1.28 1.23 1.15 1.07 1.00 1.04
0.94 0.95 0.95 0.96 0.98 0.99 1.00 1.43 1.34 1.28 1.20 1.12 1.04 1.00
m(i,j)
Rho 0.9 1.00 0.91 0.90 0.90 0.90 0.90 0.90 1.00 1.87 2.69 5.02 9.36 17.47 25.16
k0 . 9 0.91 1.00 0.94 0.90 0.90 0.90 0.90 1.87 1.00 1.44 2.69 5.02 9.36 13.48
0.90 0.94 1.00 0.91 0.90 0.90 0.90 2.69 1.44 1.00 1.87 3.48 6.50 9.36
0.90 0.90 0.91 1.00 0.91 0.90 0.90 5.02 2.69 1.87 1.00 1.87 3.48 5.02
0.90 0.90 0.90 0.91 1.00 0.91 0.90 9.36 5.02 3.48 1.87 1.00 1.87 2.69
0.90 0.90 0.90 0.90 0.91 1.00 0.94 17.47 9.36 6.50 3.48 1.87 1.00 1.44
0.90 0.90 0.90 0.90 0.90 0.94 1.00 25.16 13.48 9.36 5.02 2.69 1.44 1.00
0.0067
0.0601
m
i
n
i
m
u
m
 
e
i
g
e
n
v
a
l
u
e
m
i
n
i
m
u
m
 
e
i
g
e
n
v
a
l
u
e
m
i
n
i
m
u
m
 
e
i
g
e
n
v
a
l
u
e
m
i
n
i
m
u
m
 
e
i
g
e
n
v
a
l
u
e
0.0028
0.0328
A(i,j) =
A(i,j) =
A(i,j) =
A(i,j) =
 
 
 
Table 1: matrixes  A and m(i,j), for four sets of the parameters  k    , ρ and its minimum 
eigenvalue.   17 
 
 
 
 
 
 
 
 
 
 
Figure 3: function  ℜ → ℜ ⊂
2 : K f . The vertical axis represents the values of  f . The 
ρ  axis has only one division (at 0.5) and the k  axis has marks at 0.2, 0.4, 0.6 and 0.8.  
 
 
 
 
 
 
 
 
 
 
Figure 4: Level curves of  ℜ → ℜ ⊂
2 : K f . 
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
ρ 
k 
ρ 
k   18 
 
 
 
 
 
 
 
 
 
 
Figure 5: the “strange” region of  ℜ → ℜ ⊂
2 : K f  
 
 
 
 
 
 
 
 
 
 
 
Figure 6: level curves of the previous graph. 
0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88
0.1
0.12
0.14
0.16
0.18
0.2
0.22
0.24
0.26
ρ 
k 
ρ 
k   19 
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Guilherme “Alocação de Capital para Cobertura do Risco de Mercado de Taxas de Juros de Natureza Prefixada”, Dissertação de Mestrado,

Value at Risk New Benchmark for Controlling Market Risk”,

The Correlation Matrix of the Brazilian Central Bank's Standard Model for Interest Rate Market Risk

http://www.bcb.gov.br/pec/wps/ingl/wps08.pdf

The Correlation Matrix of the Brazilian Central Bank's Standard Model for Interest Rate Market Risk

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