Correlation Matrix with Block Structure and Efficient Sampling

Transcription

1 RESEARCH Correlation Matrix with Block Structure and Efficient Sampling Methods September 200 Jinggang Huang Liming Yang, Senior Director, Quantitative Analytical Research Group, , A version of this paper was published in the Journal of Computational Finance, September 200 Work on this paper done while employed by Standard & Poor s

2 Abstract Random sampling from a multivariate normal distribution is essential for Monte Carlo simulations in many credit risk models For a portfolio of N obligors, standard methods usually require O(N 2 ) calculations to get one random sample In many applications, the correlation matrix has a block structure that, as we show, can be converted to a quasi-factor model As a result, the cost to get one sample can be reduced to O(N) Such a conversion also enables us to check whether a user-defined correlation matrix is positive semidefinite and fix it if necessary in an efficient manner Disclaimer: The models and analyses presented here are exclusively part of a quantitative research effort intended to improve the computation time of Monte Carlo simulations when we deal with a correlation matrix that has a block structure The views expressed in this paper are the authors own and do not necessarily represent the views of Standard & Poor s Furthermore, no inferences should be made with regard to Standard & Poor s credit ratings or any current or future criteria or models used in the ratings process for credit portfolios or any type of financial security Table of Contents Contents 2 Introduction 3 2 Correlation Matrix With Block Structure 3 3 Implied Factor Model For Some Special Cases 4 4 A Quasi-Factor Model For General Cases 6 5 Performance Of Our Method In Real Life Problems 6 Acknowledgments 3 2 wwwstandardandpoorscom

3 Introduction A typical portfolio or collateral pool of a structure deal usually consists of more than one obligor It is well known that the overall risk of a portfolio or a structured product will depend not only on single obligor risk, but also on how they are correlated with each other Therefore, the correlation between obligors is an important component in understanding the overall risk of a portfolio In practice, two methods are generally used to specify the correlation structure, that is, using a factor model or by specifying the correlation matrix directly In a factor model, all obligors correlate to each other through some common factors Mathematically, we assume that X, an N vector of standard normal random variables, can be written as: X = BF + ɛ () where B is a constant N d matrix of factor loadings and the common factor F is a d vector of independent standard normal random variables ɛ is the idiosyncratic error term, an N vector of independent normal random variables; the variance of each random variable of ɛ is picked so that the corresponding random variable of X has variance Also F and ɛ are independent of each other It is clear that, in such a setup, once the factor loading matrix B is given, the correlation matrix of X can be determined In a factor model, the common factors can be identified in advance (as in the observable factor model as described in McNeil et al (2005)) or derived from historical data (as in the latent factor model described in McNeil et al (2005)) For the other approach, the correlations between obligors are directly specified by analysts, who usually classify the obligors into groups based on industries, sectors, countries, etc, and then the group-specific correlations are determined As a result, the correlation matrix defined in this way has a block structure, that is, the correlation between any two obligors is determined by the groups that they belong to Examples of this approach can be found in Standard & Poor s (2008) and Fitch Ratings (2005) A detailed discussion of the pros and cons of each approach is beyond the scope of this paper, but we would like to point out one (apparent) advantage of the factor approach over the correlation matrix approach In a factor model, a random sample can be obtained efficiently because of the assumption of independence given the common factors; therefore, the cost of getting one random sample is proportional to N, the number of obligors in the portfolio In this case, Monte Carlo simulations can be performed quickly to get an estimate of the portfolio loss On the other hand, the cost for sampling from a multi-dimensional Gaussian distribution using a standard method (eg, the Cholesky factorization) will be proportional to N 2 Since the estimation of portfolio loss is crucial in practice, it is important to find ways to speed up random sampling In this paper, we show that, if the correlation matrix has a block structure, the model is equivalent to a factor model like structure Hence, the cost of performing the Monte Carlo calculation can be reduced to a level similar to that of a factor model This paper is organized as follows In Section 2, we introduce the concept of a correlation matrix with block structure In Section 3, we consider a special case where a model specified by a correlation matrix with block structure can be transformed to a factor model In Section 4, we consider an efficient sampling method for a general block correlation matrix In Section 5, we present examples that compare the performance of our method with the standard approach 2 Correlation Matrix With Block Structure Assume that X = (X, X 2, X N ) is a vector of Gaussian random variables and each X i is a standard Gaussian random variable with zero mean and unit standard deviation Assume we can divide them into K groups (for example, with respect to industry, region, size or some combinations of these criteria) We rewrite X as X = (X () X (2),, X (K) ) T, where each X (i) = (X (i), X(i) N i ) T is the vector of variables that belong to the same group i We consider the case where correlation between two random variables only depends on the group they belong to That is, for any two groups i, j (including the case i = j), there exists ρ i,j such that for any 3

4 two (non-identical ) random variables X m (i), X n (j) from the two groups, respectively, we have the group-specific correlation: ρ(x (i) m, X (j) n ) = ρ i,j (2) The correlation matrix Σ X has the block structure, as shown below: ρ, ρ, ρ, ρ,2 ρ,2 ρ,2 ρ,2 ρ,k ρ,k ρ,k ρ,k ρ, ρ, ρ, ρ,2 ρ,2 ρ,2 ρ,2 ρ,k ρ,k ρ,k ρ,k ρ, ρ, ρ, ρ,2 ρ,2 ρ,2 ρ,2 ρ,k ρ,k ρ,k ρ,k ρ, ρ, ρ, ρ,2 ρ,2 ρ,2 ρ,2 ρ,k ρ,k ρ,k ρ,k ρ 2, ρ 2, ρ 2, ρ 2, ρ 2,2 ρ 2,2 ρ 2,2 ρ 2,K ρ 2,K ρ 2,K ρ 2,K ρ 2, ρ 2, ρ 2, ρ 2, ρ 2,2 ρ 2,2 ρ 2,2 ρ 2,K ρ 2,K ρ 2,K ρ 2,K ρ 2, ρ 2, ρ 2, ρ 2, ρ 2,2 ρ 2,2 ρ 2,2 ρ 2,K ρ 2,K ρ 2,K ρ 2,K ρ 2, ρ 2, ρ 2, ρ 2, ρ 2,2 ρ 2,2 ρ 2,2 ρ 2,K ρ 2,K ρ 2,K ρ 2,K ρ K, ρ K, ρ K, ρ K, ρ K,2 ρ K,2 ρ K,2 ρ K,2 ρ K,K ρ K,K ρ K,K ρ K, ρ K, ρ K, ρ K, ρ K,2 ρ K,2 ρ K,2 ρ K,2 ρ K,K ρ K,K ρ K,K ρ K, ρ K, ρ K, ρ K, ρ K,2 ρ K,2 ρ K,2 ρ K,2 ρ K,K ρ K,K ρ K,K ρ K, ρ K, ρ K, ρ K, ρ K,2 ρ K,2 ρ K,2 ρ K,2 ρ K,K ρ K,K ρ K,K (3) where the K submatrices on the diagonal are of dimensions N N, N 2 N 2, N K N K To sample from N(0, Σ X ) using the standard Cholesky factorization requires O(N 2 ) calculations per sample However, we can exploit the block structure presented in Σ X to reduce the cost of sampling Below, we propose two methods: the implied factor model and a quasi-factor model The former approach reduces the problem to a factor model that is easy to implement, but it only works with some special cases The latter method needs more effort to implement but can be applied to any cases where Σ X has block structure and is positive semidefinite Since the correlations are usually specified by user, it may happen that Σ X is not positive semidefinite (as a result, the sampling problem is not well defined) We present efficient methods to detect and fix such problems 3 Implied Factor Model For Some Special Cases As we mentioned in the introduction, one advantage of factor models is that sampling can be carried out efficiently (with O(N) calculations) Therefore, it is desirable to convert (if possible) a multivariate Gaussian distribution to a factor model We note that, in Andersen et al (2003), the authors propose an iteration procedure to find a In the identical case, ie when i = j and m = n, we have, of course ρ(x (i) m, X (j) n ) = 4 wwwstandardandpoorscom

5 factor model approximation of a general Gaussian distribution In this paper, we focus on a correlation matrix that has a block structure and derive a factor model (or a quasi-factor model) that exactly replicates the original distribution Moreover, compared to the procedure of Andersen et al (2003), our approach is much faster because we do not need any iteration and because most of the matrix operations are performed on much smaller group level matrices For Gaussian distributions with block structure, we show in this section that, for some special cases, we can derive a factor model that induces the same distribution We define the K K group level correlation matrix 2 R = (ρ i,j ), where ρ i,j is as defined in Equation(2) for i =,, K; j =, 2,, K Suppose that R is positive semidefinite Then we can find a square root G of R, ie: G G T = R (4) by using the Cholesky factorization Now we consider the following factor model For each X m (i) random variable from group i, let: as the m th X (i) m = where ɛ j, j =, 2,, K and all the η m (i) that the following result is true K j= G(i, j) ɛ j + ρ i,i η (i) m (5) are independent standard normal random variables It is easy to check Proposition If R is positive-semidefinite, then the factor model defined in equation (5) induces the correlation matrix Σ X Hence we have converted the original model (specified by correlation matrix) to a factor model, with K common factors Notice that, for each group, the "factor loadings" of every obligor in that group are all the same It is easy to see that a random sample from this factor model needs O(N + K 2 ) calculations Notice that, in order for this method to work, we have to assume that R is positive-semidefinite, which might not be true even when the original "full" matrix Σ X is positive definite The following case shows such an example Consider the following correlation matrix with block structure Σ X = The matrix is positive definite because each diagonal entry is greater than the sum of all other entries in the same row Now we consider the corresponding group level "correlation matrix" R = The determinant of the matrix = , indicating that this matrix is not positive-semidefinite 2 Notice that, in general, ρi,i, because it is the intra group correlation, not the correlation of an entity with itself, so R is not really a correlation matrix in the usual sense 5

6 On the other hand, the correlation matrix Σ X itself is not always positive-semidefinite when it is defined using group level correlations 3 Therefore, it is important to check whether Σ X is positive-semidefinite to make sure the model is well defined An ordinary method involves calculating all the eigenvalues to see whether they are all non-negative, however, such a method need O(N 3 ) calculations, which is expensive when N is large When Σ X has the block structure, this task can be performed more efficiently using the much smaller R matrix with O(K 3 ) calculations, as the following result indicates Proposition 2 Σ X is positive-semidefinite if and only if the K K matrix R + N ( ρ, ) N 2 ( ρ 2,2 ) N 3 ( ρ 3,3 ) N K ( ρ K,K ) N K ( ρ K,K ) (6) is positive-semidefinite We will prove this conclusion after Proposition 5 This result can also be used by analysts when they determine the correlation structure For example, if they want to make sure that the correlation numbers they specify always lead to a well defined distribution (ie Σ X is positive-semidefinite) regardless of the number of obligors in each group, then they should make sure that R is positive-semidefinite because when the N k, the matrix defined in (6) converges to R 4 A Quasi-Factor Model For General Cases The factor model approach described in the previous section only works if the group level "correlation matrix" R is positive-semidefinite In this section, we study the more general cases where we only require that the Σ X matrix is well defined (ie positive-semidefinite) The method we adopt here can be summarized as follows: We derive an eigenvector decomposition of Σ X and exploit patterns shown in the decomposition to get an efficient random sampling First, we present some simple results that lead to the decomposition For each k =, 2,, K, let V k be the subspace of R N such that its elements are of the following form: where ˆx = (0, 0,, 0, 0, 0,, 0,, 0, 0,, 0 }{{}}{{}}{{}, x, x 2,, x Nk, 0, 0,, 0,, 0, 0,, 0 }{{}}{{}}{{} )T N N 2 N k N k N k+ N K x + x x Nk = 0 Let V 0 be the subspace of R N such that its elements are of the following form: ˆx= (x, x,, x }{{}, x 2, x 2,, x 2 }{{},, x K, x K,, x K }{{} )T (7) N N 2 N K (8) 3 If ΣX is derived, for example, empirically from a complete set of historical data, then it is positive-semidefinite However, in practice, Σ X is usually defined directly and is not always positive-semidefinite More details are given in the following sections in regard to this 6 wwwstandardandpoorscom

7 Let x = (x, x 2,, x k ) T, we define the mapping R K R N where ˆx is as in (8) The following lemma can be easily proved P (x) = ˆx ˆx 2 Lemma All the subspace V k, k = 0,,, K are perpendicular to each other, and they span R N, ie: Each V k is an invariant subspace of Σ X, ie: R N = V 0 V V K Σ X V k V k and for each k 0, V k consists of eigenvectors of Σ X with eigenvalue ( ρ k,k ), ie, for each v V k, we have Σ X v = ( ρ k,k )v Now we consider the following two different cases: V k, k 0 and V 0 For the case V k, k 0, to get an orthonormal basis of V k, we define a matrix F m of dimension m (m ) for any integer m > F m = It is straightforward to check that (m )m (m )m (m )m (m )m m where I m,m is the identity matrix of dimension (m ) (m ) For any k 0, we define (m )m (9) F T mf m = I m,m (0) U k = 0 N N k 0 N2 N k 0 Nk N k F Nk 0 Nk k+ N k 0 NK N k where 0 i j is just the zero matrix of dimension i j It is straight forward to check that each column v of U k is an unit vector (whose L 2 -norm is ) in V k and all the columns of U k are orthogonal to each other So by definition, 7

8 the columns of U K form an orthonormal basis of the subspace V k Note that each column of V k is an eigenvector of Σ X with respect to the eigenvalue ρ k,k, but these eigenvectors only depends on the size N k, and have nothing to do with any of the group correlations (ρ i,j ) Next, we discuss an efficient method to derive the eigenvectors from V 0 Formally, we are looking for vectors from v V 0 that are eigenvectors of Σ X, ie, Σ X v = λv () where v is of the form of (8) and λ is some real number For that purpose, we define B = + (N )ρ, N 2 ρ,2 N 3 ρ,3 N K ρ,k N ρ 2, + (N 2 )ρ 2,2 N 3 ρ 2,3 N K ρ 2,K N ρ 3, N 2 ρ 3,2 + (N 3 )ρ 3,3 N K ρ 3,K N ρ K, N 2 ρ K,2 N K ρ K,3 + (N K )ρ K,K It is straightforward to check that () holds if and only if B(x, x 2,, x K ) T = λ(x, x 2,, x K ) T (2) which means that we can reduce the problem of () from N dimensional space to K dimensional space We restate this result in the following Lemma 2 x R K is an eigenvector of B with respect to an eigenvalue λ if and only if P (x) is an eigenvector of Σ X with respect to eigenvalue λ If the eigenvalues of B are all different, then the K eigenvectors (P (x k )) of Σ X, mapped from the K eigenvectors (x k ) (k =, 2,, K) of B, are automatically orthogonal to each other 4 On the other hand, if for some eigenvalue λ k of B, there are multiple eigenvectors v, v 2,, v m, then P (v ), P (v 2 ),, P (v m ) may not be orthogonal to each other But we can use the Gram-Schmidt process to orthogonalize P (v ), P (v 2 ),, P (v m ) We have shown that the problem of decomposing Σ X (with a cost of O(N 3 )) can be reduced to the problem of decomposing B (with a cost of O(K 3 )) and we summarize this in the following result Proposition 3 Σ X has the following eigenvector decomposition: Σ X = UDU T where U is an orthogonal matrix and D is a diagonal matrix Furthermore F N U = G 0 F N2 0 0 (3) F NK where each F Nk is as defined in (9) and G is an N K matrix, each column of G equals P (x) with x being an eigenvector of B The first K entries on the main diagonal of D are the eigenvalues of B; Beginning from the K + entry on the main diagonal, we have the following eigenvalues: ρ, with multiplicity of N, ρ 2,2 with multiplicity of N 2,, ρ K,K with multiplicity N K 4 This is a known result of linear algebra The proof is actually very simple: Suppose ΣX v = λ v and Σ X v 2 = λ 2 v 2, where λ λ 2, then λ < v, v 2 >=< Σ X v, v 2 >=< v, Σ T X v 2 >=< v, Σ X v 2 >= λ 2 < v, v 2 >, hence < v, v 2 >= 0 8 wwwstandardandpoorscom

9 Once we have the decomposition as presented in Proposition 3, a sample can be obtained by calculating: X = U Dɛ (4) where ɛ is an N vector of independent standard normal random variables Note that in (4), the first K factors (ɛ,, ɛ K ) affect all variables, the next N factors (ɛ K+,, ɛ K+N ) affect only variables in group, then the next N 2 2 factors affect only variables in group 2, and so on Such a structure bears similarities to a factor model, the difference is that, in factor models we have one "idiosyncratic" factor for each obligor, but in (4), we have N k "idiosyncratic" factors for each group k Because of such similarities, we call the model in (4) a quasi-factor model So Proposition 3 indicates that a model with block correlation matrix can be converted to a quasi-factor model The following result is just a simple observation, but we state it as a proposition due to its importance Proposition 4 A sample from the quasi-factor model (4) (and therefore any model with block correlation structure) can be obtained with O(K 2 ) + O(N) steps Proof First, since D is a diagonal matrix, Dɛ can be calculated in O(N) steps In (3), the G matrix has the same rows for each group, ie, the first N rows are all the same, the next N 2 rows are all the same,, so the multiplication of G with the first K numbers of Dɛ can be carried out by O(K K) calculations Now, for each group k, we need to calculate F Nk ɛ k, where ɛ k is the sub-vector of ɛ corresponding to the sub matrix F Nk in (3) By the definition of F Nk in (9), any two adjacent rows of F Nk are all the same except for (at most) 2 entries Therefore, given F Nk (i, :)ɛ k (where F Nk (i, :) is the i th row of F Nk ), we only need a fixed amount of steps to get F Nk (i, :)ɛ k Therefore, F Nk ɛ k can be carried out in O(N k ) steps So the total amount of steps needed to get one sample from the quasi-factor model (4) is O(K 2 ) + O(N) Note that the cost of getting one random sample using the quasi-factor model is in the same order of that of the implied factor model of section 3 (of course, when R is positive-semidefinite) It is natural to ask whether the sampling time can be further improved This seems to be hopeless if we want to exactly replicate Σ X using (quasi-) factor model On the other hand, in practice, it might suffice just to approximate Σ X using fewer number of common factors As we mentioned earlier, Andersen et al (2003) proposed an approximation method which converts the general whole matrix Σ X to a factor model As we noted earlier, this approach might be slow when the portfolio is large It is interesting to see whether we can combine the technique developed here and that of Andersen et al (2003) to come up with an efficient approximation procedure when Σ X has block structure This is a research that we are currently undertaking By Lemma 2, we can always find a set of real eigenvectors of B that span R K On the other hand, B is not symmetric, and our experience shows that ordinary numerical routines often produce complex eigenvectors So it is interesting to see whether we can transform the problem of (2) into another problem which only involves a symmetric matrix That can be done by the following: Proposition 5 Define a symmetric matrix + (N )ρ, N N 2 ρ,2 N N 3 ρ,3 N2 N ρ 2, + (N 2 )ρ 2,2 N2 N 3 ρ 2,3 B = N3 N ρ 3, N3 N 2 ρ 3,2 + (N 3 )ρ 3,3 NK N ρ K, NK N 2 ρ K,2 NK N 3 ρ K,3 + (N K )ρ K,K N N K ρ,k N2 N K ρ 2,K N3 N K ρ 3,K and let M = diag( N, N 2,, N K ), then for x R K, Bx = λx if and only if B[Mx] = λ[mx] Proof It is straightforward to check that B = M BM, so Bx = λx if and only if M BMx = λx if and only if BMx = λmx 9

10 So we do not have to solve problem (2) directly, instead, we can solve the following: By = λy and we get the solution of (2) by taking x = M y As a by-product of Proposition 3 and 5, we can now prove Proposition 2 in the previous section Proof of Proposition 2 By Proposition 3, Σ X is positive-semidefinite if and only if the eigenvalues of B are all non-negative, which in turn is true if and only if B is positive-semidefinite, by Proposition 5 On the other hand, the matrix defined in (6) equals N N NK B N N NK Hence Σ X is positive-semidefinite if and only if the matrix of (6) is positive-semidefinite So far, we have assumed that the model with block structure (3) is well-defined, ie, Σ X is positive-semidefinite But in practice, it is possible that the user-defined Σ X might not be positive-semidefinite For the remaining part of this section, we consider the practical problem of "fixing" Σ X when it is not positive-semidefinite Methods of "fixing" a general correlation can be found in Rebonato and Jäckel (2000), here we focusing on more efficient methods when correlation matrix has a block structure First, we present a common method for "fixing" Σ X when it is not positive definite 5 Consider any eigenvector decomposition 6 where Ũ is an orthogonal matrix (ie ŨŨT = I) and D is a diagonal matrix: D = Σ X = Ũ DŨ T (5) λ λ λ λ N (6) Each of λ i is an eigenvalue of Σ X, and Ũ(:, i), the i th column of Ũ is the corresponding eigenvector, ie: Σ X Ũ(:, i) = λ i Ũ(:, i) If Σ X is specified through user-defined ρ i,j, it is possible that Σ X is not positive definite, ie, one or more of the eigenvalues λ i might be negative We can fix this problem by setting all the negative λ i s in (4) to 0 to get a fixed D and use (5) (with a fixed D matrix) to get the new positive-semidefinite ΣX Of course, to use such a method to fix Σ X, we need to find the decomposition of (5), which in general induces a cost of O(N 3 ) calculations Our question now is: can we achieve that efficiently? The answer is yes, because the specific decomposition as described in Proposition 3 can be carried out in O(K 3 ) calculations, instead of O(N 3 ) The only technical detail left is: will this "fixing" procedure using the specific decomposition of Proposition 3 lead to the same positive-semidefinite Σ X, as induced by any decomposition of (5)? For that we have the following result 5 This is the spectral decomposition method introduced in Rebonato and Jäckel (2000) Other methods for "fixing" can also be found in the same paper In this research, we focus on developing efficient spectral decomposition method, but the general idea of reducing the complexity of matrix operations can also be applied to other "fixing" methods when the correlation matrix has block structure 6 In general, the eigenvector decomposition is not unique, so we use Ũ instead of U to differentiate an arbitrary eigenvector decomposition and that of proposition 3 0 wwwstandardandpoorscom

11 Proposition 6 The resulting positive-semidefinite Σ X of the aforementioned fixing procedure does not depend on the choice of Ũ Proof Σ X = Ũ DŨ T Suppose there is a negative eigenvalue λ < 0, and without loss of generality, let s assume λ = λ 2 = = λ m = λ < 0, and no other eigenvalues equal λ We fix this eigenvalue λ by setting it to 0, and we end up with a new diagonal matrix D new, and as a result, a new Σ X = Ũ D new Ũ T It is simple to check that where Ũ(:, : m) is the first m columns of Ũ Assume we have another decomposition Σ X = Σ X λũ(:, : m)ũ(:, : m)t Σ X = ÛDÛ T The columns Û(:, ), Û(:, 2),, Û(:, m) of Û span the subspace of RN that consists of all λ eigenvectors of Σ X, the same space as that spanned by Ũ(:, ), Ũ(:, 2),, Ũ(:, m), in other words, each of the two sets is an orthonormal basis of the same subspace Therefore Û(:, : m) = Ũ(:, : m)a where A is a m m orthogonal matrix So the fixing procedure using Ũ results in the following covariance matrix: Σ X λû(:, : m)û(:, : m)t = Σ X λũ(:, : m)aat Ũ(:, : m) T = Σ X λũ(:, : m)im mũ(:, : m)t = Σ X λũ(:, : m)ũ(:, : m)t So the fixing procedure is independent of the choice of U So far, we have only "fixed" one negative eigenvalue, and we can repeat this procedure and fix all negative eigenvalues to get a positive definite Σ X, which does not depend on the choice of Ũ 5 Performance Of Our Method In Real Life Problems We have done some performance tests by comparing Monte Carlo simulations using both standard Cholesky factorization and quasi-factor model methods All the tests are carried out on Intel Due CPU 233 GHz 95 GB Ram systems The programming language we use is C++, together with the numerical package provided by Numerical Algorithms Group (NAG) Tests are performed for two large loan portfolios, whose size and group information is presented in table The group-specific correlations, ie the ρ i,j s are determined by assigning intra-group (i = j) and inter-group (i j) correlations to each pair of groups according to their regions and sectors Some statistics of the correlations are presented in table 2 To prepare for the Monte Carlo simulation using the standard method, we perform a Cholesky factorization using Numerical Recipe algorithm If the correlation matrix is not positive-semidefinite, this factorization will fail, and we will perform the "fixing" procedure as described in previous sections The eigenvalue/vector computations are carried out using NAG s nag_real_symm_eigenvalues (f02aac) function After the "fixing" procedure, we perform the Cholesky factorization again Similarly, to prepare for the Monte Carlo simulation using the quasifactor model we proposed, we perform eigenvalue/vector computations (using nag_real_symm_eigenvalues) of

12 Table Portfolios Portfolio ID # of Obligors # of Groups # of obligors in each group min max mean Table 2 Statistics of group correlations Portfolio ID intra-group inter-group min max mean min max mean the matrix in Proposition 5 and fix it if there are negative eigenvalues Table 3 shows the time needed for the preparation Note that the original correlation matrix of portfolio is not positive-semidefinite and needs "fixing"; the correlation matrix of portfolio 2 is positive-semidefinite We see that the time needed to prepare for the quasi-factor model is much less The time needed for performing Monte Carlo simulations is presented in table 4 Again the calculation time is greatly reduced when the quasi-factor model approach is used 2 wwwstandardandpoorscom

13 Table 3 Time needed to prepare for simulation Portfolio ID standard quasi-factor fixing time Cholesky fact 3 hours 2 hours < minute 2 5 hours < minute Table 4 Simulation time comparison number of trials standard quasi-factor Portfolio 0,000 5 minutes 3 minutes 00,000 hour 40 minutes 24 minutes 500,000 8 hours 0 minutes hour 20 minutes Portfolio 2 0,000 hour 7 minutes 3 minutes 00,000 5 hours 30 minutes 28 minutes 500,000 day 2 hours 2 hours 28 minutes,000,000 4 days 4 hours 28 minutes 2,000,000 8 days 8 hours 30 minutes 6 Acknowledgments We would like to thank Bill Morokoff, Craig Friedman, Jayson Rome and other colleagues of the Quantitative Analytics Research Group for helpful suggestions and discussions References L Andersen, J Sidenius, and S Basu All your hedges in one basket Risk, (Noverr ber):67 72, 2003 Fitch Ratings The fitch default vector model-user manual Fitch Ratings Report, 2005 A McNeil, R Frey, and P Embrechts Quantitative Risk Management Princeton University Press, 2005 R Rebonato and P Jäckel The most general methodology to create a valid correlation matrix for risk management and option pricing purpose The Journal of Risk, 2:7 28, 2000 Standard & Poor s CDO evaluator system version 4 user guide Standard & Poor s Structure Finance Group,

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