Handling the measurement error problem by means of panel data: Moment methods applied on firm data



(33)


When either (C1) holds and t, θ, p are different,
o or (C2) holds and
t p, θ p > τ, then
, E[Vip,(∆om)] = E[yv(∆yM)] E[yv(∆Xa)]β = 0.

(34)


When either (B1), (D1), and (D2) hold and t, p, q are different,
or (B2), (D1), and (D2) hold and
t p, t q >τ, then

. E[(∆Xipq)0ta] = E[(∆X)0Vit] - E[(∆Xipq) Xt]β = 0к,.

(35)


When either (C1), (D1), and (D2) hold and t, p, q are different,
or (C2), (D1), and (D2) hold and
t p, t q >τ, then

E^Vipq ) 0it] = E^Vipq ) Vit] - E^Vipq ) Xit] β = 0

The intercept c needs a comment. When mean stationarity of the latent regres-
sor, (D1), holds, then
E(∆Xipq) = 01K and E(∆Vipq) = 0. If we relax (D1), which
cannot be assumed to hold in many situations due to non-stationarity, we get

E[^ xipq ) βit] = E[^ xipq ) 0Vit] - E[^ xipq ) 0 ] c - E[^ Xipq ) Xit] β = 0 K1,

E[^Vipq)0it] = E[^Vipq)Vit] - E[^Vipq)]c - E[^Vipq)xit]β = 0

Eliminating c by means of E(it) = E(Vit) - c - E(Xit)β = 0 leads to the following
modifications of (34) and (35):

When either (B1) and (D2) hold and t, p, q are different,

(36) or (B2) and (D2) hold and t-p, t-q >τ, then

E[(∆Xipq) 0Oit] = E[(∆Xipq)0(Vit- E(Vit))] - E[(∆Xipq)0(Xit- E(Xit))]β = 0K,

When either (C1) and (D2) hold and t, p, q are different,

(37) or (C2) and (D2) hold and t-p, t-q >τ, then

E[(∆ Vipq ) Oit ] = E[(∆ Vipq )( Vit - E( Vit ))] - E[(∆ Vipq )( Xit - E( Xit ))] β = 0.

The OC’s (32) - (37), corresponding to (19) in the general exposition of the
GMM, will be instrumental in constructing our GMM estimators. Not all of these
OC’s, whose number is substantial even for small
T , are, of course, independent.
Let us examine the relationships between the OC’s in (32) - (33) and between
the OC’s in (34) - (35). Some of these conditions are redundant,
i.e., linearly

15



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