GMM

Question 1

Moment conditions:

Answer 1

Equations that relate functions of the data and the unknown parameters to zero

Question 2

When do we use GMM?

Answer 2

When the model is non-linear in β and when L≥K, there are more equations than unknown parameters. (L=vector of moments, K=variables)

Question 3

GMM assumptions:

Answer 3

1: The observations of yi, xi and the instrumental variables zi, need to be random samples from the population
2: β is the vector of unknown coefficients (Kx1)
3: m(yi,xi,zi,β)=mi(β) is the vector of moment functions (Lx1)
4: mi(β) is continuously differentiable in β; ∂mi(β)/∂β’ is the matrix of derivatives of the moment and is LxK with rank K
5: There exist a sequence of possible random LxL symmetric positive definite matrices Wn, such that Wn–>W (in probability)
6: E(mi(β))≠0

Question 4

Under the assumptions of linearity, relevance, exogeneity and random sampling and L>K, we obtain the form of the GMM:

Answer 4

βGMM^=(X’ZWnZ’X)^-1X’ZWnZ’y

Question 5

Properties of GMM estimator:

Answer 5

1: βGMM^ is consistent
2: βGMM^ is asymptotically normal
3: βGMM^

Question 6

βGMM^–>β0 (in probability) implies:

Answer 6

Ḡ(βGMM)–>Ḡ(β0) (in probability), where Ḡ(β0)=E(∂mi(β0)/∂β0’)

Question 7

Optimal weighting matrix:

Answer 7

W=(E(mi(β0)mi(β0)’)^-1

Question 8

To minimize the asymptotic variance of βGMM^, we choose:
And we get βGMM^=:

Answer 8

S^-1=(E(ziεizi’εi))^-1, because it makes βGMM^ asymptotically efficient.
βGMM^=(X’ZSn^-1Z’X)^-1X’ZSn^-1Z’y

Question 9

What if L=K?
And what if Z=X?

Answer 9

Then X’Z is invertible and we get βGMM^=βMM^=βIV^
Then L=K, X’X is invertible and βGMM^=βMM^=βOLS^

Question 10

dim(xi)=dim(zi)
dim(xi)>dim(zi)
dim(xi)<dim(xi)

Answer 10

exact identification
underidentification
overidentification

Question 11

What is a dynamic model?

Answer 11

A dynamic model contains data that is collected over time. xt and yt are variables at time t and y_(t-k), k>0 are lags of the dependent variable. Assumptions for linear model: E(εt)=0, Var(εt)=σ^2, corr(εt,εs)=0 for t≠s.

Question 12

When can we use y_(t-1-k)?

Answer 12

In the IV model, we can use zt=y_(t-k) as an instrument and also y_(t-1-k) if we have autocorrelation of order 1 (error term is a function of the lagged observation of the error term)

Question 13

When can we guarantee the existence of the GMM estimator?

Answer 13

1: If Θ is compact & the objective function Qn(θ) is continuous in θ
2: If Θ is convex & Qn(θ) is convex everywhere and strictly convex in a neighbourhood of θ0, then a unique maximum exists

Question 14

When is Qn(θ) uniquely minimized at true parameter value θ0? (Global identification)

Answer 14

If Q(θ)>Q(θ0) for θ≠θ0. Wn is positive definite and the solutiosn of the moment conditions are unique.

Question 15

What do we need to check for local identification?

Answer 15

G=E(∂g(wi,θ0)/∂θ’)=E(∂g(wi,θ0)/∂θ1’, … , ∂g(wi,θ0)/∂θp’) has full rank p (p is number of parameters (check if det≠0)

Question 16

When is GMM consistent?

Answer 16

1: Parameter space Θ is convex
2: Qn(θ) is convex
3: g(wi,θ) and Qn(θ) are continuous in θ for any wi
4: Wn–>W (in probability), W is positive definite
5: E(g(wi,θ0))=0 and E(g(wi,θ))≠0 for θ≠θ0
6: wi is independent

Question 17

When is GMM asymptotically normal?

Answer 17

CLT has to hold, so:
1: wi is i.i.d.
2: E(g(wi,θ0))=0 exists
3: S=Var(g(wi,θ0))=E(g(wi,θ0)g(wi,θ0)’) exists and is positive definite

Question 18

What happens with the choice of W if we have exact identification? (p=k, p=number of estimated coefficients, k=number of moment conditions or used instruments)

Answer 18

The choice of W is irrelevant as G and W are square invertible matrices, so W drops out: V=G^-1S(G’)^-1

Question 19

What is the asymptotic variance-covariance matrix of the GMM in a linear regression and what is its distribution?

Answer 19

V=(G’WG)^-1G’WSWG(G’WG)^-1 and √n(θn^-θ0)–>N(0,V) (in distribution)

Question 20

What happens with V if we have overidentification? (p<k, p=number of estimated coefficients, k=number of moment conditions or used instruments)

Answer 20

V=(G’WG)^-1G’WSWG(G’WG)^-1=(G’S^-1G)^-1 +DD’
D=(G’WG)^-1G’WS^(1/2)-(G’S^-1G)^-1S^(-1/2)
Note: when DD’≥0, optimal weighting matrix W=S^-1, then D=0 and V=(G’S^-1G)^-1

Question 21

Two-step GMM estimator when S=Var(g(wi,θ0)) is unknown:

Answer 21

Step 1: Use an initial weighting matrix for the first GMM estimation to obtain θn^, like Wn=Ik (identity) or Wn=(Z’Z)^-1 (2SLS)
Step 2: Estimate Sn^=(1/n )Σg(wi,θn^)g(wi,θn^)’. Perform the second GMM estimation with Wn=(Sn^)^-1

Question 22

Hansen test to test if E(g(wi,θ0))=0 is valid:

Answer 22

H0: E(g(wi,θ0))=0 vs H1: E(g(wi,θ0))≠0
Jn=n((1/n)Σg(wi,θn^))’(Sn^)^-1((1/n)Σg(wi,θn^))~χ^2(k-p)
and reject H0 if Jn≥χ^2(1-α, k-p).
Interpretation: if we reject H0, at least one instrument is invalid

Question 23

Wald test (GMM case):

Answer 23

H0: Rθ0-q=0 vs. H1: Rθ0-q≠0
Wn=n(Rθn^-q)’(RVn^R’)^-1(Rθn^-q)–>χ^2J
Reject H0 if Wn≥χ^2(1-α, J)
Interpretation: if we reject H0, at least one of the coefficients differs significantly from their hypothesis value

Question 24

Likelihood-ratio test (GMM case):

Answer 24

H0: Rθ0-q=0 vs. H1: Rθ0-q≠0
θn~ is the restricted estimator, θn^ is the unrestricted estimator
LRn=n(Qn(θn~-Qn(θn^))–>χ^2J
Reject H0 if LRn≥χ^2(1-α, J)
Interpretation: if we reject H0, at least one of the coefficients differs significantly from their hypothesis value

Question 25

Hausman-test for GMM:

Answer 25

H0: E(εi|xi)=0 (θLS^ consistent, θMM^ consistent) vs. H1: E(εi|xi)≠0 (θLS^ inconsistent, θMM^ consistent) Hn=(θMM^-θLS^)'(Var(θMM^)-Var(θLS^))^-1(θMM^-θLS^)~χ^2_p Reject H0 when Hn≥χ^2_(1-α, p). (p=number of compared parameters) Interpretation: if we reject H0, βOLS^ is inconsistent and the certain variable is endogenous)