Download simultaneous equations

SIMULTANEOUS EQUATIONS (beautiful algebra for IDENTIFICATION, etc)
t = 1, 2,…, T j=1,.. M. The STRUCTURE means behavioral economic equations.


Y  XB  U Y 1  XB 1  U 1 Y  X  B 1  V
Y is TM and goes with Γ, which is M M in structure.
X is TK and goes with B, which is KM. Error Matrix U is TM in structure.
or Y=XΠ +V is the reduced form where the errors are uncorrelated with X.
•
Imposter model where one simply post-multiplies the structure by a nonsingular
matrix F
• YF +XBF =UF. The reduced form is still the same: FF1 cancels out as identity
mtx. YFF11 +XBFF11 =UFF11
Y=XΠ +V is still the reduced form. Identification means avoiding such imposter models.
Instrumental Variable Estimation jth str’l equation y j  Y j  j  X j j   j  Z j  j   j
Later on we will make j=1 for convenience and without loss of generality.
Wj is T (Mj + Kj) matrix such that p lim


1
W j' Z j   WZ is finite non-singular
T
More importantly we assume that the matrix E(VWj)=0. (DMCK text eq (8.20) seq)


 j,IV  Wj' Z j

1
Wj' y j is consistent estimator of j parameters of the structure.
with asymptotic variance (DMCK text eq(8.12)) given by:
 W j' Z j 

p lim 
 T 
T


 jj
1
1
 W j' Z j 
 where the consistent estimate of jj is by
W Wj 
 T 



'
j

'





 Yj  Z j  j, IV   Yj  Z j  j, IV 
 Re sidSS  

 (DMCK (8.19))


T
T


df  T  M j  K j is sometimes used as the denominator, but asymptotically
T  T  M j  K ;  , that is, it does not matter asymptotically even if the denominator is
shown to be T.
Identification Finding the unique coefficients of the structure from those of the reduced
form.
STR: Y   X B  U unknowns are  and B and elements of the Covariance matrix of
M M
kM
U, which will be symmetric.
Reduced Form: Y  X  V estimates  are available K*M known numbers
K M
and we want to estimate 3 matrices:
B and   . Even if we ignore error
( KM )
M  M M M
covariance matrix across M equations, and note that we need not estimate the coefficient
of the variable on the left side. When we regress consumption on income, coefficient of
consumption is always 1. We do not need to estimate it. So M coefficients are known to
be 1 for each of the M jointly dependent variable.
Little algebra shows that we have M(M – 1)Too many unknowns. We want to get Γ and
B matrices but got Π only. This means in generally identification is impossible, without
some assumptions and restrictions. (Recall rainfall absent in demand equation and
income absent in supply equation needed for identification of demand/supply). Here we
focus on the formal and more general characterization of identification problem.
Focus only on the first equation of the system of M equations, without loss of generality.
Let us take Y=X +V and write it as 3 sets of columns. For the left side the 3 sets of
columns are
(i) the left hand side variable y1, [for example if the first equation has consumption on left
side, y1=consumption]. This is only one column.
(ii) the included endogenous variables in the first equation is collected in columns Y1 ,
[for example if the first equation has income on the right side, Y1=income] This has M1
columns.
(iii) the excluded endogenous variable in Y*1. This has M1* columns.
This describes the left side partition. Each term on the right side also must have same
kind of columns, namely 1, M1 and M1* by matrix algebra rules.
Now what about rows? The distinction between included and excluded exogenous
variables must be reflected by the choice of two rows.
X= [X1 X*1], where the number of columns in X1 is K1 and number of columns in X1* is
K1*.
This is a 1 2 partition so  will have to be 21 partition to remain conformable. Thus
X will have to be partitioned into two sets of rows where the first set has K1 rows and
second set has K1* rows and 3 sets of columns (namely 1, M1 and M1*)
y
1
 
Y1 Y  X 1
LHS
*
1
RHS
M1
1
* 


11
11
11

  v V V * 
X
*
 21  21  21   11 M11 M11* 
 1

*
1

Excluded
Included
Excluded
Included=1, Included RHS=M1 , Excluded Endo=M1*
Included Exogenous = K1, Excluded Exogenous=K1*
We will see below that the most important matrix is 21 whose rows are K1* and
columns are M1. We will see that this 21 matrix will have to be invertible to have
identification (uniquely going from reduced form to the structure).
(If the first equation is demand equation, rainfall is absent means K1*=1)
How to get the coefficient vectors and matrices of interest from the validly computable Π
matrix? By definition we can write
   B 1
   B
second equation comes from the first by post-multiplicaiton by Γ
 .1  .1
The third equation comes from the second by focusing on nonzero components of Γ and
B matrices, since the zero components are not to be estimated.
Spell out the definitional relation Π Γ = B in greater detail we have two blocks of
equations where the coefficients of the first structural equation are now denoted by
subscript 1. Now Γ is MM matrix but the first equation is M1 vector, which is further
split into 1 to be the left side of the equation, 1 to be the coefficients associated with the
“included endogenous variable” and zero as the coefficient for the excluded (absent)
endog. variables in the first equation. Similarly the matrix B of the original structure has
k rows and M columns. Of the M columns only the first is relevant and again only the
coefficient vector 1 for the “included exogenous variables” needs to be estimated. All
the excluded (absent) exogenous variables must have coefficient zero.
Recall that Y is TM and goes with Γ, which is M M in structure.
X is TK and goes with B, which is KM. Error Matrix U is TM in structure.
Π is KM and contains the readily estimated coefficients of reduced form. Going from
reduced form to the structure involves only matrices of coefficients (not T data points)
Π Γ = B represents KM matrices on both sides. The conformable partitioning of K
rows refers to K1 for included exogenous and K1* excluded exogenous variables.
These K rows are now viewed as 2 rows with 2 blocks of K1 and K1* rows respectively.
Assuming M=3 for convenience, Π Γ = B relation for first equation has 21 blocks on
both sides as:
 11

 21
 11
 21
( 23)
 1 
*

 11
 
  1    1 
* 
0
 21 
 0  ( 21)

and errors v 1
( 31)
V1
 1 
V   1   u 1
 0 
*
1

Now do the formal row-column multiplication of the two blocks yielding 2 equations
given below. Note the stuff with zero coefficients need not be written out.
Equation 1: Thus 11  11 1  1 Eq.2 is:  21   21  1 and errors: v1  V1  1  u 1
Get  1 by solving the 2nd equation  21   21  1 . Remember that entire 23 matrix Π is
known from the valid estimation of the reduced form and we want to get 1 and 1 vectors
for the first structural equation, without loss of generality. We will solve the second block
for 1 first, plug the answer for the second block into the first block and finally get 1.
 211  21   1
We have to be able to solve this to get the 1 of interest. This
solution requires ability to invert the matrix, which must have full rank.


Rank Condition  21  Min K1* , M 1 is necessary and sufficient.
Rank must be at least m1 to get the inverse or K1*  M 1 (order condition, necessary but
not sufficient)
Over identified 1st equation if K1*  M 1
more variables are absent
Just identified 1st equation if K1*  M 1
just equal
Under identified 1st equation if K1*  M 1
Lack of identification (no good)
If the first equation is demand equation, rainfall is absent means K1*=1.
M1=1 also because we have price as the included endogenous variable on RHS. This
means that the demand equation is just identified K1*=1=M1.
Structure is: YΓ +XB =U. It has M equations. Rearrange it so that i-th equation is the
first equation in the structure (without loss of generality) and focus on that equation. So
we are inly looking at the first column of Γ and first column of B with coefficient vectors
i and i respectively. There is no point in showing excluded endogenous or exogeneous
variables whose coefficients are assumed to be zero. Then we write the coefficients of
interest as:
If ith eq is: y i  Yi  i  X i  i  u i , Zi  Yi
 
X i  ,  i   i  , i = 1, 2, …, M
 i 
ILS: Indirect least sq, X=[X : X*] is T(Ki+Ki*)  ILS   X Z i  X yi applicable only if
^
1
system is “just identified”. Ki*=Mi is needed to define XZi properly. Number of
columns in Zi is Mi +Ki.(included endo+included exo) Another way to get ILS is to
estimate  211  21   1 first. Substitute 1 in 11  11 1  1 to get 1.
The Keynesian macro model with simple consumption function and identity is eligible to
be estimated by indirect least squares. This was the only method before others developed.
RECURSIVE system (Wold) simplest to estimate, just use OLS, no problem
If  is upper triangular, we have a recursive system. This implies rather peculiar chain of
causation and not realistic for demand supply case, for example. y1=f(x), y2=f(y1,x)
y3=f(y1,y2,x), y4=f(y1,y2,y3,x). The endogenous variables appear on RHS in a pattern so
that fitted OLS y1 from first equation can be substituted on the right side of second
equation, and so on. Recursion means that by the time second equation comes around, the
y1 is pre-determined and not correlated with the errors of second equation. There is a
sequence.