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Econometrics I: Instrumental Variables Matrix
Algebra
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
 
Motivation
So far, we only considered the case of one endogenous
variable and one instrument.
Now we will consider the general case.
We will also talk about specification and overidentification
tests
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
Before we move, let’s do a recap on OLS
The OLS assumptions are (focus on the large sample case):
1 Linearity: yi = x0
ib + ei, where b is a K ⇥ 1 vector with the
b coefficient.
2 Full rank: E[xix0
i] = Q, a positive definite K ⇥ K matrix
3 Exogeneity: E[eixi] = 0
4 {yi, xi}N
i=1 is a sequence of i.i.d. observations.
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
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Before we move, let’s do a recap on OLS
What happens if assumption 3 fails?
Assume that E[eixi] = g 6= 0
We can show that all coefficients might be inconsistent.
Not only the ones associated to variables that are correlated
with the error!
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Instrumental Variables
Now we relax the assumption that E[eixi] = 0.
To estimate the model, we need an additional set of
variables, Z, that satisfy the exogeneity and the relevance
conditions.
The IV assumptions are:
1 Linearity: yi = x0
ib + ei, where b is a K ⇥ 1 vector with the
b coefficient.
2 Full rank: E[zix0
i] = Qzx, has rank K (relevance)
3 Exogeneity: E[eizi] = 0 (exogeneity)
4 {yi, xi, zi}N
i=1 is a sequence of independent observations.
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
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The set of instruments zi will include the xi variables that
are not correlated with ei + other variables.
Lets partition xi in:
xi1, the set of K1 variables not correlated with the error
term.
xi2, the set of K2 variables correlated with the error term.
Let L be the number of variables of zi.
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
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Identification
b is identified if we can write it as an explicit function of population
moments of random variables that we observe.
Let’s consider again the first stage and reduced form regressions...
First stage:
xi = G0zi + ui (1)
where we can define G such that E[ziu0
i ] = 0.
Therefore, G = (E[ziz0i ])
�1E[zix0i ] ) G is identified.
Reduced form:
yi = x0i b + ei = z0il + vi (2)
where l = Gb and vi = ei + u0
i b. IV assumptions ) E[zivi] = 0.
Therefore, l = (E[ziz0i ])
�1E[ziyi] ) l is identified.
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Identification
We can identify l and G, and we know that
l = Gb (3)
However, we want to estimate b. Main question: is b
identified?
Eq 3 is a set of L equations and K unknowns.
) We will have a unique solution for b if, and only if,
rank(G) = K.
This condition is equivalent to rank(E[zix0
i]) = K,
A necessary (but not sufficient) condition for the rank
condition is that L � K.
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Instrumental Variable Estimator
Let’s start with the case L = K.
The instrumental variables estimator will be given by:
bIV = (Z0X)�1Z0y
We can show that bIV is consistent and asymptotically
normal.
Assume that var[ziei] = s2E[ziz0i] = s2Qzz (this is
equivalent to homoskedasticity)
Under some technical conditions, we have that
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Two Stage Least Squares
What if we have more instruments (L > K)?
In this case, Z0X will not be a square matrix, so we cannot
find (Z0X)�1
Ideas?
1 One idea would be to choose K variables among the L
variables in Z.
2 Another idea would be to use all variables, considering
linear combinations of the Z variables.
If all instruments are valid, then linear combinations of the
instruments will be valid (you need to check the rank
condition).
This is what 2SLS does.
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Two Stage Least Squares
For a L ⇥ K matrix F, the n ⇥ K matrix eZ = ZF will be a
linear combination of the instruments.
You can use eZ to construct an IV estimator:
eb = (eZ0
X)�1eZ0
y
Question: which F should you choose?
One idea is to consider F = (Z0Z)�1Z0X. In this case:
eb = (eZ0X)�1eZ0y = (X0PZX)�1X0PZy
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
Two Stage Least Squares
For a L ⇥ K matrix F, the n ⇥ K matrix eZ = ZF will be a
linear combination of the instruments.
You can use eZ to construct an IV estimator:
eb = (eZ0
X)�1eZ0
y
Question: which F should you choose?
One idea is to consider F = (Z0Z)�1Z0X. In this case:
eb = (eZ0X)�1eZ0y = (X0PZX)�1X0PZy
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
Two Stage Least Squares
Since PZ is idempotent, note that:
eb = (X0PZPZX)�1X0PZy = (bX0 bX)�1 bX0
y = b2SLS
This is why b2SLS is called 2-stage least squares. This
estimator can be obtained by the following procedure:
1 Regress X on Z. Store the predicted values bX.
2 Regress y on bX.
Notes:
Recall! The standard error you obtain from the second
stage regression would not be valid.
Under homoskedasticity, 2SLS is the most efficient way to
use your instruments.
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
Two Stage Least Squares
Since PZ is idempotent, note that:
eb = (X0PZPZX)�1X0PZy = (bX0 bX)�1 bX0
y = b2SLS
This is why b2SLS is called 2-stage least squares. This
estimator can be obtained by the following procedure:
1 Regress X on Z. Store the predicted values bX.
2 Regress y on bX.
Notes:
Recall! The standard error you obtain from the second
stage regression would not be valid.
Under homoskedasticity, 2SLS is the most efficient way to
use your instruments.
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
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Specification Tests
What if we run 2SLS when the OLS assumptions are valid?
2SLS would still be consistent, but OLS would be consistent
and efficient...
If we believe 2SLS is valid, then we can test whether OLS
would have beenvalid.
Now it makes sense to look at 1
n  xi ûi, where ûi is the
residual from the 2SLS regression.
This is essentially what is done in the Hausman
specification test.
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Specification Tests
What if we run 2SLS when the OLS assumptions are valid?
2SLS would still be consistent, but OLS would be consistent
and efficient...
If we believe 2SLS is valid, then we can test whether OLS
would have been valid.
Now it makes sense to look at 1
n  xi ûi, where ûi is the
residual from the 2SLS regression.
This is essentially what is done in the Hausman
specification test.
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
Hausman Specification Test
Idea:
If E[xiei] = 0, then both OLS and 2SLS are consistent, but
OLS is efficient.
If E[xiei] 6= 0, then both 2SLS is consistent, but OLS is not.
So we can look at the difference: bIV � bOLS
If this is very different from zero, you reject that E[xiei] = 0
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Hausman Specification Test
As usual, we need to define what is very different from zero.
Since bIV and bOLS are asymptotically normal, bIV � bOLS
is asymptotically normal as well. Therefore, under the null:
H = (bIV � bOLS)
0[var(bIV � bOLS)]
�1(bIV � bOLS)
d! c2
J
where J is the number of endogenous variables.
In general, var(bIV � bOLS) will not have full rank, so we should
use a generalized inverse (we won’t go into details on that).
The key result in the Hausman test is that, given that OLS
is efficient under the null (if we have homoskedasticity):
var(bIV � bOLS) = var(bIV)� var(bOLS)
This is a general result.
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
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Overidentification Tests
The IV estimator is based on the orthogonality conditions:
E[ziei] = 0
There are two possible cases:
1 L = K: The model is just-identified.
In this case, the IV estimator would set 1
n  zi ûi = 0
We cannot test if the orthogonality assumption holds...
2 L > K: The model is over-identified.
It would not be the case that 1
n  zi ûi = 0 by construction.
But if all instruments are valid, then 1
n  zi ûi should be
close to zero for all moments.
We can test that...
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
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Overidentification Tests
Let’s test! We need a test statistic... Let:
m =
1
n
n
Â
i=1
zi ûi =
1
n
n
Â
i=1
zi(yi � x0
ibIV)
As usual, m will be asymptotically normal, so we can look
at:
W = m0 [var(m)]�1 m d! c2
L�K
Where we can estimate var(m) using:
\var(m) =
1
n2
n
Â
i=1
(zi ûi)(zi ûi)
0 =
1
n2
n
Â
i=1
û2
i ziz0i
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
Overidentification Tests - Notes
1 The degrees of freedom of the Chi-square distribution is
only L � K.
This happens because we can only detect the failure of
L � K moments equations.
2 If the overidentification test rejects, it does not tell you
which instrument is invalid.
3 Intuition of the test:
If you have more than one instrument, you can construct
different IV estimators;
Under the assumption that all instruments are valid, all of
them should converge to the same place;
Implicitly, the overidentification test compares these
different IV estimators.
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
Overidentification Tests - Notes
If the treatment effect is heterogeneous, then you might
reject the null even if all instruments are valid.
We will only see that in Microeconometrics...
Bruno Ferman Econometrics I: Instrumental Variables Matrix Algebra
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References
Greene, chapter 8
Hansen, chapter 11
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