Estimating unknown matrix by OLS Rolando Morales1 Ciess-Econométrica (Bolivia) Universidad Mayor de San Andrés (Bolivia)
Summary Let xij to be the elements of a matrix X with known margins. We suppose that these variables are underlying to the determination of the margins of matrix X. We suggest several strategies to evaluate the xij‟s according to the availability of additional information beside the margins of X. The additional information could be: i) A reference matrix close to X, ii) Exogenous variables that can explain the variation of the xij variables, iii) Linear restrictions including sign restrictions. We show that the estimation of X could be done by Singular Ordinary Least Square. Our approach is based on matrix theory with an extensive use of the Penrose pseudo inverse. To illustrate this method, we study the incidence of the foreign market in the structure of Bolivian Gross Domestic Product. Keywords: OLS, pseudo inverse, JEL: c10, c19, c61, c67.
1
[email protected]
This work was supported by funds from the Swiss Program for Research on Global Issues for Development (r4d program) under the thematic research module "Employment in the context of sustainable development" and the research project “Trade and Labor Market Outcomes in Developing Countries”. The Swiss Program for Research on Global Issues for Development is being implemented jointly by the Swiss Agency for Development and Cooperation (SDC) and the Swiss National Science Foundation (SNSF). The views expressed here are the authors‟ and do not necessarily reflect those of the SDC or those of SNSF. All error are our responsibility.
1 Introduction Let X to be a matrix whose elements xij are unknown. The xij are not observable, we call it subjacent variables. Thus the table X contains the subjacent structure of the relationship between rows and columns of this matrix. We suppose that the sums of its elements on each row and the sums of its elements in each column are known. The methodology that we suggest is developed by steps, each one having its own importance. The first stage is based on the ignorance hypothesis, that is to say that there is no additional information other than the margins of X matrix. In a second stage, It is assumed that there is a reference or support matrix which is close to the unknown matrix but has different margins. In a third stage, it is introduced the general problem where the unknown variables xij are dependent on a set of exogenous variables. In addition, we analyze extensions to this problem adding linear constraints and sign restrictions. This methodology covers problems such as the analysis of underlying structures in economy, the evaluation of migratory and transport flows, the estimation of transition probability, input-output tables updating, social accounting matrices and the design of causal models where pairs of endogenous variables have in common a same subset of exogenous variables. In addition, it is possible to apply this methodology in a similar way to that used the Leontief model based on the input-output matrix. Papers with objectives analogous to the present, but using different approach have been done by Stone (1963), Bacharach (1970), Chilton-Poet (1973), Ninjkamp (1975), Batty and March (1976), Willekens (1979), Van Der Ploeg (1982), among others. However, as far as we know this methodology is new.
2 Mathematical notation
Let X to be a matrix with r-rows and m-columns to be evaluate; let be wi =1,2,… r the row sums and wr+j , j=1,2,..m the column sums. We will suppose that we are interesting on evaluate n matrix of this kind, eventually, n=1. The row and column sums will put in a vector w‟=(w1‟,w2‟). For convenience, the unknown variables xijt will be put in a column vector xt, t=1,..n, beginning with the lines of each Xt matrix. We define H1 and H2 matrix in the following way: : H1 = Ir * f ‟m
H2 = f ‟r * Im
Where the symbol * identify the Kronecker product. Matrix I is the identity matrix and fs is a vector where all its elements are equal to one. The diacritical mark „ identifies the transposition of a matrix.
The matrix H1 has r lines and rm columns; the matrix H2 has m lines and rm columns. We have that H1xt=w1t and H2xt=w2t. The vector w1t contains the rows sums of X and the vector w2t contains the columns sums of X. Matrix H1 y H2 will be sub matrix of a new matrix HO =(H1 \ H2) with (r+m) rows and rmcolumns. The rank of H0 is (r+m-1). With this matrix, the rows and columns constraint on X can be writen as follows: Hoxt = wt for t = 1,2,..n
The set of n-restrictions of this kind can be written: Hx=w With H=In*H0. The vectors x, w contains the n vectors xt and wt, t=1,…n.
Two useful matrix are H0+, the Penrose pseudo inverse of H0 and the matrix P0 defined as P0=Irm-H0+H0. For the n-observations, we will define the following matrix: H+=In*H0+ P=In*P0=Inrm-H+H H0+ has rm rows and (r+m) columns. The rank of H0+ is the same of H0, that is (r+m-1). P is a square matrix with rm rows and columns. The rank of P is [rm-(r+m-1)].
Let EA to be a vector space generated by the columns of a matrix A. Therefore, if x belong to EA, there is a vector b, that x=Ab. In the next section, we will develop an analytical expression for H+ with n=1. The extension for n>1 is directly.
3 A pseudo inverse for H Let be; g‟ = (f‟r | - f‟m) G = Im+r – (1/m+r) gg ‟ Hg= ([1/m]H‟1 | [1/r] H‟2 – [1/rm] frmf‟m) G is square matrix with (m+r) rows and columns. Matrix Hg has rm rows and (r+m) columns; the first r columns of Hg are given by (1/m)H1‟, while the next m columns are generate by [1/r] H‟2 – [1/rm] frmf‟m. It is possible to demonstrate (Morales 2015) that the analytical form of the Penrose pseudo inverse of H is given by: H+=HgG
For all vector w verifying g‟w=0 (implying Gw=w), we have that the linear applications H+w and Hgw can be written as: x+ij = wi /m+wr+j /r – wo /rm Notice that x+ij is the mean of row i more the mean of column j less the global mean. If each wi is equal to mμ and each wmj is equal to rμ, where μ is any number, we have that x+=μ. Verifying the following axioms (Boulluion-Odel 1971), the Hg matrix is a pseudo inverse of H: HHgH = H HgH = (HgH)‟ And verifying the following axioms, H+ is a Penrose pseudo-inverse (Rao 1973) of H: A1.- HH+H = GH = H A2.- H+HH+ = H+G = H+ A3.- HH+ = G (Symmetric and idempotent) A4.- H+H = HgGH = HgH (Symmetric and idempotent) The HH+=G matrix is a orthogonal projector (symmetric and idempotent matrix) on EH, and H+H is a orthogonal projector on the vector space generate by the columns of H+. The matrix P=(Irm-H+H) is a orthogonal projector on the complementary vector space of EH+.. This means that if x belong to EH+ and y belong to EP, then x and y are orthogonal. Other useful results are: H+HP=PH+H=0 and Pfrm=0, Px+=0 if x+=H+w for any vector w.
4 Matrix estimation in the absence of additional information In the absence of additional information other than the margins of the matrix X, we can accept the hypothesis of ignorance, that is to say that there is no sufficient reasons to believe that some xij are larger than others, which is equivalent to assume that the vector x (containing elements xij) is close to a multiple of the vector frm containing only ones. Let be x0 = μfrm this vector with some scalar μ. In this vector all its components are equal. The solution to the following problem provides estimates for the xij: Min(x) ||x-x0||2 With x0=μfrm under Hx=w
The general solution to equation Hx=w is given by x=x++Pz with someone vector z. The ignorance hypothesis imply that the solution x to our problem should have minimum norm among all the solutions of the equation Hx=w. Given that x+ and Pz are orthogonal, we have that: ||x||2 = ||x+|||2 + ||Pz||2, thus, x+ is the solution having the smallest norm among all possible solutions to Hx=w.
5 Normality hypothesis We will assume that vectors wt, t=1 to n, are normally distributed with expected value equal to μt and covariance matrix Фt. Given that g‟wt=0, the rank of matrix Фt. is less than m+r. Vector x+t=H0+wt are also normal with expected value equal to H0+μt and covariance matrix Ωt=H0+ ФtH0+‟.The rank of matrix Ωt is less to m+r. For n-observations, we have E(w)=μ, Cov(w)=Ф.
Therefore, x+ is a normal vector with E(x+)=H+μ and Cov(x+)=H+ФH+‟. Let Ω be the covariance matrix of vector x. The rank of this matrix is less than n(m+r). A particular case of matrix Ω is given when Ф is block diagonal. That is to say that vector wt and vector w t‟, if t≠t‟, are independent in probability. In this case, Ω is also block diagonal. If Ф is block diagonal and all blocks are equal to Ф0, we can write Ф=(In* Ф0) and we have that Ω=(In* Ω0) with Ω0=H+ Ф0H+‟. This case is done when vectors wt (and vectors xt) are issues from a stationary process. If the matrix Ω was known, we can considerer to obtain generalized ordinary least square estimators (Greene 2000, Chapter 11) for the matrix X. Given that, in general matrix Ω is not known, the developments in the next sections are focuses on the traditional O.L.S. estimators. Notice that least square estimators are unbiased, consistent and are asymptotically normally distributed. However, they are no longer efficient and the usual inference procedures based on the F and t distributions will no longer be appropriate (Greene 2000, p457-458).
6
Estimation of the X matrix with exogenous variables
In this section, we will explain the estimation of the matrix X with two kind of exogenous information. The first one is a reference matrix X0 close to matrix X but with different margins. In the second case, it is supposed that there is a set of exogenous variables that can explain the xij variable through linear equations. Formally, both problems can be expressed as the search of a vector x which is the solution to the following optimization problem:
Min(x) ||x-x0|| Hx=w x0 ϵ EA
Where x0 is a fixed point, or alternatively, is some point belonging to the vector space generated by the column vectors of a known matrix A. Matrix A has n observations of q linearly independent vectors susceptible to explain the variations of the vector x. However, as we will show in the appendix, to solve this problem, matrix A needs to be expanded to qrm columns
6.1 When a support or reference matrix exists We suppose that there is a support matrix X0 for X, that is to say that the unknown matrix X should be close to X0. If x belong to vector space EH={x / Hx=w}, x can be written as x=x++Pz for any vector z. Let x to be the optimal solution to this problem. In the particular case that x0 belong to EH, we have that x=x0. Now we will study the case in which x0 doesn‟t belong to EH . There is a vector that x=x++P . Adding and subtracting x from the objective function, we obtain ||x-x0||2 = (x-x)-(x0- x) 2 = (x- x0)||2 + x-x 2 -2(x-x)‟(x0- x) Replacing x by (x+ +P ) and (x-x) by P(z- ), the last expression become: ||x-x0||2 = (x- x0||2 + P(z- ) 2 -2(z- )‟P(x0- x) Given that Px+=0, we have Px=P and this expression can be written as ||x-x0||2 = (x- x0||2 + P(z- ) 2 -2(z- )‟P(x0- )
Given that x is supposed to be the solution to minimization problem, we have that: (||P(z- ) 2 -2(z- )‟P(x0- )) ≥ 0 Thus, the objective function takes its minimum value when this expression is equal to cero, that is to say when the vector (x0- ) belong to kernel of P. In this case, a vector v exist such as (x0- )=H+Hv. Therefore, = x0-H+Hv. Replacing en x=x++P , we have x=x++Px0 . This is the solution to our problem. Notice that P(z- ) 2 vanishes. The vector x+ is orthogonal to vector Px0. Therefore, it is possible to decompose the norm of x between the margin effect x+ and the exogenous information x0.
6.2 Estimation of the X matrix by linear regression Suppose that x0 belong to a vector space EA, for which the column vectors of the A matrix form a generator set, then there is some vector b in Rqrm such that x0 = Ab. Each row of the matrix A is associated with one of the cells of the matrix X, then for each observation, A has nrm rows. The matrix A has n rows and qrm columns.
We will study two problems, called primal and dual, to estimate the matrix X with exogenous variables,
The primal problem aims to estimate vector b solving the following twofold-minimization problem: Min(b){Min(x) ||x-x0)||2} Con Hx=w x0=Ab
Taking into account the previous results, the problem of estimate X with exogenous variables is equivalent to the following minimization problem: Min(b)||x-Ab)||2. With x= x++Px0 = x++PAb Thus, we search to identify vector b that minimize x-Ab)||2 under the restriction Hx=w and knowing that x= x++Px0 = x++PAb. Therefore, the objective function of this problem can be written as: ||x-Ab)||2=||x++PAb-Ab||2=||x+-H+HAb||2 Let be A=H+HA. Note that A is the orthogonal projection of A on EH+. Our minimization problem becomes: Min(b)||x+-Ab
2
Thus, this is a problem of Ordinary Least Square where the vector space EA is the projection space. If x+ belongs to EA, vector b is a solution of the equation Ab = x+. In the general case, x+ doesn‟t belong to EA ( i.e rank (A x) will not be equal to the rank of A, therefore the system will inconsistent). This problem is solved by least squares approximation (Rao 1973). That is to say that b is a solution to the normal equation system: (A‟A) b =A‟x+ Matrix C=(A‟A) has qrm rows and columns and it rank is q(r+m-1), then the normal equations has multiple solutions that can be written as:
b=(A‟A)+A‟x+ = A+x+ + (I- A+A)z
With z an arbitrary vector.
However, by the Projection Theorem (easily demonstrable here), the projection of (unknown) vector x in EA is unique (Luenberger 1997) and it is given by: x= Ab= AA+x+ . To make easy the resolution of the dual problem, that we will explain later, we develop a new expression for vector b.
Let be D=HA. We have that: A+=(H+HA)+=(H+D)+ =D+H A+A=D+D x+=H+w. D+HH+=(HA)+HH+=A+H+HH+=A+H+=(HA)+=D+, Replacing these expressions in the solution b, we have that b=D+w + (I-D+D)z
The dual problem aims to find a vector b that the vector HAb be near to observed w. That is to say that we look for to identify a vector b solution to the following problem: Min(b)||w-HAb||2 Replacing HA by D, we have: Min(b)||w-Db||2
A banal situation is given when there is a solution for the equation w=Db, implying that w belongs to vector space generate by the columns of matrix D, that is to say that rango(D|w)=rango(D). Otherwise, this equation system is inconsistent. The rank of D is q(r+m-1), while D has n(r+m) rows and rmq columns, therefore the O.L.S solution is not unique. A general solution to the normal equation system can be written as: b=D+w + (I-D+D)z
Therefore, the solution to dual problem is the same that the solution to primal problem. The margins of the solution x=Ab are the best approximations to observed margins w. Inspired in the regression models, we can decompose the norm of x+ in the following way:
Origen
SS
SS
Margins conditioned to knowledge of A
SSM/A ||x+-Ab
Interaction of margins and matrix A
SSMA
Margins (Total)
SSM
Ab
2
2
||x+||2
Following the ordinary Least Square framework, this decomposition let us to define the multiple coefficient correlation R2: R2= 1-||x+-Ab 2/ ||x+||2
In addition, the solution of the dual problem provides the following decomposition of the norm of w:
Origen
SS
SS
Margins conditioned to knowledge of A
SSM/A
Interaction of margins and matrix A
SSMA
Margins (Total)
SSM
w-
2
= x
2
||w||2
Thus, we can define the multiple correlation index in the following way: Rw2 =1- w-
2
/||w||2
7 Estimation of X with exogenous variables and lineal restrictions
Now, for the evaluation of matrix X we have beside its margins, exogenous variables and linear restrictions on the xij. Like as the previous cases, we aim to search a vector x that is the nearest to a fix point x0 or to some point in a vector space EA generated by the columns vectors of a matrix A. Therefore the evaluation of X is based on finding a solution to following problem: Min (x – xo)‟(x – xo) HX = w
(4)
Rx = d
(5)
“xo en EA” The first relationship Hx = w means that the sums of rows and columns of unknown matrices Xt, t = 1,2,... n are known. The second relationship Rxt = d defines a set of linear constraints. It is assumed that the R matrix has k rows and nmr columns. The known vector d has k components. In the case where there is no restrictions across different periods, the matrix R contains in its main diagonal rectangular blocks and if they are all equal to a Ro matrix, the R matrix can be
written as R = In * Ro. If for each period t, Roxt = do, the vector d can be written in the form d = fn * do. In the third relationship "xo in EA", the vector xo is a reference vector. Alternatively, x0 is any vector in a vector space EA. If the columns vector of a matrix A generate a vector space EA, for all xo in EA, a vector b exists such that xo = Ab where A is a matrix with nrm rows and qrm columns and it is rang is qrm
Rank A=qrm Rank R=k Rank (RPR‟)=k
( k<=mr-(m-r-1))
Rank RA=kmin(r,m)
The matrix F has (n+r+k) rows y nrm columns. The matrix RA has nk rows and qrm columns.
Let EF to be the vector space generated by the column vector of F, Thus, we aim to determine the solution vector x searching the minimum of the distance between the reference vector x0 and some vector in EF or, alternatively, as the minimum of the distance between one vector in EF and other vector in EA. To include both cases in the same conceptual framework, when x0 is a fix point we will say that EA has only a point.
To make easier what we will develop further down, we will introduce the following notation: Table 1. Definition of a set of particular matrix and its ranks (n=1) Matrix or vector H
Rows r+m
Columns rm
Rank r+m-1
H+
rm
r+m
r+m-1
rm
rm
r+m-1
rm
rm
rm-(r+m-1)
k
k
k
K=R'(RPR') R PK
rm rm
rm rm
k k
B=Irm-PK
rm
rm
mr-k
BP F'=(H'|R')
rm rm
rm r+m+k
mr-(m+r-1)-k (m+r-1)+k
F+=(BH+|PR'(RPR')-1)
r+m+k
rm
(m+r-1)+k
F+F=Irm-BP
rm
rm
(m+r-1)+k
+
HH +
P=Irm-H H RPR’ -1
+
+
x =H w
rm
1
h=PR’(RPR’) d
rm
1
+
rm
1
+
z=h+Bx
All the previous matrix can be generalized for n>1 applying the Kronecker product on the base of the identity matrix In. However, for n>1 it is necessary to specify rows and column sum and the rank of the following matrix.
Table 2. For n>1, some matrix and its rank
Matrix A D=HA RA
A =F+FA
rows nrm n(r+m) nk nrm nrm
colums qrm qrm qrm qrm qrm
rank qrm (m+r-1)q km (m+r-1)q (m+r-1)q+km
Notice that H+H, P, BP y PKP are symmetric and idempotent matrix, thus they are orthogonal projectors. If k=mr-(m+r-1). The matrix BP is equal to cero and F+F=Inrm
7.1 Analysis of the restrictions
Let us to analyze the following equation system:: Fx = c Con F‟ = (H‟, R‟) y c‟ = (w‟, d‟ ).
We will exclude the banal case in which a vector x exists such that Fx = c, which implies that the vector c belongs to the vector space generated by the columns of the matrix F, that it is to say that rank(F|c) = rank(F).
In the general case, a vector x which approximates a vector in Fx to vector c in the sense of the Euclidean distance is given by (Rao 1973): x = F+c + (I–F+F)y y is any vector and F+ is the Penrose pseudo inverse of matrix F.
Replacing F+ by (BH+|PR'(RPR')+) and taking into account that F+F is equal to (I-BP), the last expression can be written as: x=BH+w+PR(RPR‟)-1d +BPy Given that H+w=x+ and that h= PR(RPR‟)-1d, we have: x=Bx+ + h +BPy
The three vectors at the right of this relation are orthogonal. To demonstrate that, we can show that that h‟B=0, B‟BP=BP y x+‟BP=0.
For the subsequent development, it is useful to write x as: x=z+BPy
with z=h+Bx+
Notice that vector z depends only on the known vectors w y d. Given that Bx+=x+- PKx+ and PKh=h, an alternative form to write x is as follow: x=x++PK(h-x+)+BPy
In the last expression, the three vectors at the right of equality are orthogonal.
For any general solution x, we have Rx=d. To show that, we only need to show that RPK=R, RBP=0 and Rh=d.
7.2 Development of the general solution
a) There is a support matrix
We begin supposing that x0 is a fixed point and that we should solve the following problem: Min || x-xo||2 (14) Fx = c "xo in EA" The general solution to the restriction Fx = c is given by x=Bx+ + h +BPy. To the extent that the three vectors that appear to the right of this equality are orthogonal and that the first two rely exclusively on H, R, w and d, this problem is equivalent to finding the minimum of the norm of the vector (BPy - xo) in relation to the vector y. Since BP is an orthogonal projector, the solution to this problem is given by y = xo. Then, the optimal solution is given by: x = h+ Bx++ BPxo
Since BP is an orthogonal projector in the vector space EBP , if x0 belongs to EBP, we have (BPx0 - x0) = 0; otherwise e * =(BPx0-x0) ≠0. Since BP = I – F+F, we can write e*=F+Fx0. b) The solution in a vector space
As in the previous section, the general case where EA is a vector space will be analyze in two steps associated to primal and dual problem,
The primal problem is: Min(b){Min(x) || x-xo||2} Fx = c "xo=Ab in EA" Replacing xo by Ab and x by the general solution to constraint Fx=c, that is
x= z + BPy, we have that the objective function can be written as: || x-xo||2 = ||z+BPy-Ab||2 = ||z||2 + ||BPy-Ab||2 + 2z‟(BPy-Ab) Given that z‟BP=0, this expression become || x-xo||2 = ||z+BPy-Ab||2 = ||z||2 + ||BPy-Ab||2 - 2z‟Ab) Vector z doesn‟t depend on y. Given that BP is a projector, ||BPy-Ab||2 takes it minimum value related to y when y=Ab. Replacing y by Ab in the general solution x=z+BPy, we have x=z+BPAb.
To determine b, we will replace x by z+BPAb in the objective function and we obtain: || z + BPAb-Ab||2= || z-F+FAb||2 = z- b||2 With BP= I – F+F = and
= F+FA.
Thus, vector b is a solution to an OLS problem. Matrix
has nrm rows y qrm columns and
it rank is q(rm-1). We have a solution for vector b solving the following normal equation system: ( ‟ )b =
‟
z
The matrix ( ‟ ) has qrm rows y columns and it rank is q(rm-1), thus it doesn‟t have inverse. The general solution to these equations is given by: b=( ‟ )+ ‟z +(I- ( ‟ )+ ‟)y
Where y is any vector and ( ‟ )+ is the Penrose pseudo inverse of ( ‟ ). As the Projection Theorem says, the optimal x solution for this problem, x= b, is unique for all possible b solutions, in particular for b=( ‟ )+ ‟z =
+
z
The dual problem is to find a vector b such that FAb is the nearest vector to vector c in the sense of the Euclidean distance, i.e., that is to say to find a solution to the following problem: Min(b) ||c-FAb||2 That is to find a vector b, such that the vector x = FAb is the best approximation for to have Hx = w and Rx = d simultaneously. We will show that if a vector b is a solution to the primal problem, is also a solution to this problem. In effect: We have Fh=(0\d), FB=(H\0) y Hx+=w. Thus Fz=F(h+FBx+)=c and b=
+
z=(F+FA)+z = A+F+Fz.
Therefore, the vector b can be written as: b = A+F+c = (FA)+c
This is the solution OLS to problem Min(b)||c-FAb||.
7.3 The triangular decomposition of the norm
From the classical triangular decomposition of the norm in the OLS framework, we have:
Origen Regression Error Total
Given that z- b 2=||h+Bx+-x
2
SS b 2 z- b ||z||2
and P x=h, we have
2
z- b 2=||B(x+-x) 2. In addition
||z||2=||h||2+||Bx+||2. From these results and from the triangular decomposition of the norm of z, we have: Origen Regression Error Total
SS b 2 ||B(x+-x) |2 ||z||2
The norm of the error gives us the distance between x+ and x using the B‟B metrics. The multiple correlation coefficient associated to this regression is: R2=1- z- b 2/||z||2 The norm of vector z can be written as ||z||2= x+B(x+-x) 2. Thus, the R2 is an indicator of the proximity of vector x to vector x+. Thus, the final form of this coefficient is: R2= 1-||B(x+-x) |2/ x+B(x+-x)
2
8 Evaluation of X with sign restrictions The problem of evaluation of X with exogenous variables and linear restrictions subject to the constraint that all of its components are non-negative deserves a special attention.
Formally we will try to solve the following problem:
Min (x - xo) '(x – xo)/2 HX = w Rx = d x≥0 "xo in EA"
Let L (x) to be the function of Lagrange associated with this problem: L (x) = 1/2 (x - xo)'(x – xo) + t'(Hx = w) + s'(Rx – d)
Let v to be the vector of partial derivatives of L(x): v = (x - xo) + H‟t + R‟s
Pre-multiplying this expression by BP on both sides of the equality gives:
BPv = BP (x-xo) Remind that BPH‟=0 and BPR‟=0. Taking into account that BP=I-F+F, from the last expression we obtain:; x=BPv + F+Fx + BPx0 Notice that F+Fx= F+c=(BH+ PR‟(RPR‟)-1)(w\d) =Bx++ h. Therefore: x= BPv + (h+Bx+ + BPxo)
If xo is a fixed point or if x0 belong to vector space EA, as we have shown in the previous section, the vector x = h+ Bx++ BPxo is the solution to the problem without the restriction of sign: x = BPv+ x There are nrm equations and 2nrm unknown in this equation. Because F+FBP=0, premultiplying this relation by F+F, we obtain that F+Fx=F+Fx. Thus, if μ is a vector belonging to the kernel of F+F, we have that x= μ + x. Vector μ should be chosen in the kernel of F+F verifying μ ≥- x and among all vectors verifying this condition, the vector having a smallest norm. Notice that μ 2 = v‟BPv and that BP is symmetric and idempotent,
9 Final comments
The presented methodology covers a broad field of application; as geometric approximation method allows you to generate information; as causal model is an alternative for to understand some structures.
10 Appendix. An example: The trade impact on the economy 10.1 The problem From the equality between the global supply and global demand we have: GDP+M=GC+PC+I+E Where: GDP: Gross Domestic Product; M: Imports; GC: Government Consumption
PC: Private Consumption; I: Gross investment; E: Exports We aim to identify the trade price impact on each of the variables composing the global supply and the global demand for Bolivia and to estimate the demand multipliers. Table 3. Global Offer and Global demand GDP
Imports
Margins
Government Consumption
x11
x12
w1
Private Consumption
x21
x22
w2
Investment
x31
x32
w3
Exports
x41
x42
w4
Margins
w5
w6
w0
The demand multipliers are defining as follow: Let S to be a vector with the row sums of X and Λ=diag(S). Let T to be T=Λ-1X. Matrix T‟ contains the demand multipliers. To avoid the inertia effect, we will transform Table 3 in first differences divided by GDP of the previous year. Thus GDP becomes its growth rate. We will estimate the xij given that margins of X are known by a linear model taking trade index as exogenous variables.. Let px, pm, tcr be the export, import prices and tcr the real exchange rate. We assume that these variables have an impact on the structure of X.
10.2 Notation Thus our main hypothesis is that the variables xij are lineal functions of px,pm, tcr, we can writte: xijt= μij + αijpxt + βijpmt + γijtcrt Years are identified by the letter t. We have 34 observations from 1981 to 2014. For any year t, this model can be writen as: xijt=v‟tbij
With v‟t=(1, pxt, pmt, tcrt) y bij=(μ, αij ,βij, γij )‟ The matrix Xt will be written as a column vector xt, beginning by its rows. Vector xt has rm (r=4, m=2) elements. In its matrix form, this model can be written as:
x11 x12 x21 x22 x31 x32 x41 x42
v' v' v' =
v' v' v' v' v'
b11 b12 b21 b22 b31 b32 b41 b42
Therefore, for any t, we can write; xt=Atb with At= Irm*v‟t Where * identifies the Kronecker product. For all periods, this model can be written as:
xt=1 xt=2 .. .. xt=n
At=1 At=2 = .. .. At=n
b
Or, in compact form: x=Ab Notice that vector x has rmxn components, matrix A has rmxn rows and rmxk columns, where k is the number of exogenous variables (k=4)
10.3 The OLS estimation to explain the xij
Table 4 shows OLS estimators for the variables xij. As we have said in section 5, least square estimators are unbiased, consistent and are asymptotically normally distributed. However, they are no longer efficient and the usual inference procedures based on the standard error, F and t distributions will no longer be appropriate (Greene 2000, p457-458). Notice that OLS estimators for the margins are equal to the sum of the parameters in the same row or in the same column. This provides an interesting linear decomposition of this estimates. Table 4. OLS parameter estimates for variables xij in X Variables Constante Government px Consumption pm tcr R2 Constante Private px Consumption pm tcr R2 Constante px Investment pm tcr R2 Constante px Exports pm tcr R2 Constante px Margins pm tcr R2
GDP 0.4417 0.2024 -0.2979 -1.7614 0.6012 1.4424 0.7079 -0.9666 0.5706 0.3402 0.6702 0.1511 -0.2086 -3.4689 0.2340 1.1002 0.5996 -0.8480 1.2401 0.4255 3.6544 1.6611 -2.3212 -3.4196 0.4086
Imports -0.0910 -0.1114 0.1509 -0.7905 0.1898 0.9097 0.3941 -0.5177 1.5415 0.2377 0.1375 -0.1627 0.2403 -2.4979 0.1618 0.5675 0.2858 -0.3991 2.2111 0.2049 1.5237 0.4059 -0.5256 0.4642 0.0293
Marginsl 0.3506 0.0910 -0.1470 -2.5519 0.5527 2.3521 1.1021 -1.4843 2.1121 0.2716 0.8078 -0.0116 0.0316 -5.9668 0.1785 1.6677 0.8855 -1.2472 3.4512 0.3116 5.1782 2.0670 -2.8468 -2.9554 0.2875
10.4 Demand multipliers
Demand multipliers shows great time variations. Table 5 and Table 6 shows means of these estimates by periods based on the estimation of matrix X without exogenous variables.
Table 5. Demand multipliers for the GDP, 1981-2014 Government
Private
Consumption Consumption Años 1981-1990 -1.77 3.53 1991-2000 1.32 0.63 2001-2010 1.27 0.64 2011-2014 0.84 0.57
Investment
Exports
0.72 -4.61 0.57 0.46
0.49 0.52 0.60 0.62
Table 6. Demand multipliers for Imports Government
Private
Consumption Consumption Años 1981-1990 2.77 -2.53 1991-2000 -0.32 0.37 2001-2010 -0.27 0.36 2011-2014 0.16 0.43
Investment
Exports
0.28 5.61 0.43 0.54
0.51 0.48 0.40 0.38
Finally, Table 7 shows the demand multipliers estimated with exogenous variables for 2014. Table 7. Demand multipliers for GDP and Imports, 2014 Año 2014 GDP IMPORTS
Government Private Consumption Consumption
0.54 0.46
0.51 0.49
Investment
Exports
0.51 0.49
0.51 0.49
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