This question emerged from reading Wooldridge's Econometric Analysis of Cross Section and Panel Data, second edition, section 14.4.3, where the asymptotic distribution of the GMM (Generalized Method of Moment) estimator with efficient instruments is derived.

I have an issue which looks like a simple algebra problem, but for some reason I am missing something and I am not able to solve it. Sorry in advance : it takes a bit of notation and derivation to explain the issue at stake with enough clarity.

We have a moment condition

$$E(r(w_i, \theta_0) | x_i ) = 0,$$

from which we derive

$$ E ( Z(x_i) r(w_i, \theta_0) | x_i ) = 0,$$

for any $Z$ function of $x_i$ only. Then by the LIE we have the unconditional moment condition

$$ E ( Z(x_i) r(w_i, \theta_0) ) = 0,$$

from which we can derive the GMM estimator $\hat{\theta}_Z$ . Wooldridge tells us that the most efficient choice for $Z$ is

$$ Z^*(x_i) \equiv \big[\underbrace{Var\big(r(w_i,\theta_0 | x_i)\big)}_{\equiv \Omega_0(x_i)}\big]^{-1} \underbrace{E(\nabla_{\theta} r(w_i,\theta_0) | x_i)}_{\equiv R_0(x_i)}$$

and then derives the asymptotic distribution of $\hat{\theta}_{Z^*}$

$$ \sqrt{N} (\hat{\theta}_{Z^*} - \theta_0) \rightarrow^d N ( 0, [G_0 \Lambda_0^{-1} G_0 ]^{-1}),$$


$$G_0 \equiv E\big(R_0 (x_i)' \Omega_0^{-1}(x_i) R_0 (x_i) \big) $$


$$\Lambda_0 = E\big(R_0 (x_i)' \Omega_0^{-1}(x_i) r(w_i, \theta_0)r(w_i, \theta_0)' \Omega_0^{-1}(x_i) R_0 (x_i) \big).$$

Now my problem is with the next claim

$$ E\big(R_0 (x_i)' \Omega_0^{-1}(x_i) r(w_i, \theta_0)r(w_i, \theta_0)' \Omega_0^{-1}(x_i) R_0 (x_i) \big) = E\big(R_0 (x_i)' \Omega_0^{-1}(x_i) R_0 (x_i) \big),$$

which in turn implies that $G_0 = \Lambda_0$.

For the last claim to work, we seem to need

$$ r(w_i, \theta_0)r(w_i, \theta_0)' = \Omega_0 (x_i). (1)$$

I understand that because $ E ( r(w_i, \theta_0)|x_i ) = 0$, we have $E[r(w_i, \theta_0)r(w_i, \theta_0)'|x_i] = \Omega_0$, but I do not see how this leads to (1). Or is it something completely different? What am I missing?

Thanks in advance for your help.


By the Law of Iterated Expectations (careful with parentheses)

$$\Lambda_0 = E\big[R_0 (x_i)' \Omega_0^{-1}(x_i) r(w_i, \theta_0)r(w_i, \theta_0)' \Omega_0^{-1}(x_i) R_0 (x_i) \big] $$

$$=E\Big(E\big[R_0 (x_i)' \Omega_0^{-1}(x_i) r(w_i, \theta_0)r(w_i, \theta_0)' \Omega_0^{-1}(x_i) R_0 (x_i)\mid x_i \big] \Big)$$

Then "take out what's known"

$$=E\Big(R_0 (x_i)' \Omega_0^{-1}(x_i) \cdot E\big[r(w_i, \theta_0)r(w_i, \theta_0)' \mid x_i \big]\cdot \Omega_0^{-1}(x_i) R_0 (x_i)\Big)$$

and since as you too write, $$ \Omega_0 = E\Big[r(w_i,\theta_0 )r(w_i,\theta_0 )'\mid x_i\Big]$$

substitute to get the result.

  • $\begingroup$ Use the LIE, as usual uh...;) Many thanks @AlecosPapadopoulus. $\endgroup$ – Martin Van der Linden Dec 5 '14 at 14:30
  • $\begingroup$ Yeah, it is almost like a magic wand. $\endgroup$ – Alecos Papadopoulos Dec 5 '14 at 14:31

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