<p>We have
<span class="math-container">\begin{align}
\frac1n \sum_{i=1}^n Z_i \tilde{e}_i^2 Z_i' &amp;= \frac1n \sum_{i=1}^n Z_i \Big[e_i^2 - 2e_i X_i' (\tilde\beta - \beta) + (\tilde\beta - \beta)' X_i X_i' (\tilde\beta - \beta)\Big] Z_i' \\ &amp;= A_1 - 2A_2 + A_3.
\end{align}</span>
because <span class="math-container">$\tilde{e}_i = y_i - X_i'\tilde\beta = (X_i'\beta + e_i) - X_i'\tilde\beta = e_i - X_i' (\tilde\beta - \beta).$</span></p>
<p>First,
<span class="math-container">$$A_1 = \frac1n \sum_{i=1}^n Z_i e_i^2 Z_i' \to_p E[Z_i e_i^2 Z_i'] = \Omega \text{ by LLN}.$$</span>
Next,
<span class="math-container">$$A_2 = \frac1n \sum_{i=1}^n Z_i e_i X_i' (\tilde\beta - \beta) Z_i',$$</span>
the <span class="math-container">$(j,k)$</span> element of which is
<span class="math-container">$$A_{2,jk} = \frac1n \sum_{i=1}^n z_{ij} e_i X_i' (\tilde\beta - \beta) z_{ik} = \left( \frac1n \sum_{i=1}^n z_{ik} z_{ij} e_i X_i' \right) (\tilde\beta - \beta).$$</span>
The first term (in the brackets) is <span class="math-container">$O_p(1)$</span>; it can be <span class="math-container">$o_p(1)$</span> if <span class="math-container">$X_i$</span> is exogenous but it does not matter. The second term is <span class="math-container">$o_p(1)$</span> because <span class="math-container">$\tilde\beta$</span> is consistent. Thus, <span class="math-container">$A_2 = o_p(1)$</span>. Finally,
<span class="math-container">$$A_3 = \frac1n \sum_{i=1}^n Z_i (\tilde\beta - \beta)' X_iX_i' (\tilde\beta - \beta) Z_i',$$</span>
the <span class="math-container">$(j,k)$</span> element of which is
<span class="math-container">\begin{align}
A_{3,jk} &amp;= \frac1n \sum_{i=1}^n z_{ij} (\tilde\beta - \beta) X_i X_i' (\tilde\beta - \beta) z_{ik}\\
&amp;= (\tilde\beta - \beta)' \left[ \frac1n \sum_{i=1}^n X_i z_{ij} z_{ik} X_i' \right] (\tilde\beta - \beta)\\
&amp;= o_p(1) O_p(1) o_p(1) = o_p(1),
\end{align}</span>
and thus <span class="math-container">$A_3 = o_p(1)$</span>.</p>
<p>We have shown that <span class="math-container">$\hat{W}^{-1} \to_p \Omega$</span>. Under the assumption that <span class="math-container">$\Omega$</span> is invertible, <span class="math-container">$\hat{W} \to_p \Omega^{-1}$</span>.</p>

CC BY-SA 4.0
2021-07-19T12:34:08.533Z
2021-07-19T12:39:59.003Z
2021-07-19T12:39:59.003Z
chan1142
<blockquote>
<p>(Hansen Exercise 13.3) Take the model <span class="math-container">$Y = X'\beta + e$</span> with <span class="math-container">$\mathbb{E}[Ze] = 0 $</span>. Let <span class="math-container">$\tilde e_i = Y_i - X_i'\tilde \beta$</span> where <span class="math-container">$\tilde\beta$</span> is consistent for <span class="math-container">$\beta$</span> (e.g. a GMM estimator with some weight matrix). An estimator of the optimal GMM weigh  matrix is
<span class="math-container">$$\hat W  = \left( \frac1n \sum_{i=1}^n Z_i Z_i' \tilde e_i^2 \right)^{-1}.$$</span>
Show that <span class="math-container">$\hat W \to_p \Omega^{-1}$</span> where <span class="math-container">$\Omega = \mathbb{E}[ZZ'e^2]$</span>.</p>
</blockquote>
<p>Since <span class="math-container">$\tilde \beta$</span> is consistent, <span class="math-container">$\tilde e_i$</span> also converges in probability to <span class="math-container">$e_i$</span> for all <span class="math-container">$i$</span>. Also, by LLN,
<span class="math-container">$$\frac1n \sum_{i=1}^n Z_i Z_i' \tilde e_i^2 \to E[Z_iZ_i' \tilde e_i^2].$$</span>
I think I need to combine these two results in some way to get the final result. Can you give some help?</p>

CC BY-SA 4.0
2021-07-18T21:31:19.750Z
At the level of a single random term, $\tilde e_i$ will not converge in probability to $e_i$. Averaging is necessary (LLN).
Bertrand
CC BY-SA 4.0
2021-07-19T10:13:51.287Z
Note $\tilde{e}_i = e_i - X_i' (\tilde{\beta}-\beta)$ so $\tilde{e}_i^2 = e_i^2 - 2e_i x_i' (\tilde\beta - \beta) + (\tilde\beta - \beta)' X_i X_i' (\tilde\beta - \beta)$. Multiply $Z_iZ_i'$, add up, and then you will see that the second term and the third term converge in probability to zero due to the consistency of $\tilde\beta$. The first term converges in probability to $E[Z_i Z_i' e_i^2] = \Omega$ by LLN. Note that it's not $E[Z_iZ_i' \tilde{e}_i^2]$.
chan1142
CC BY-SA 4.0
2021-07-19T10:49:50.657Z
@Bertrand Well, $\tilde{e}_{i}$ does converge in probability to $e_{i}$, it's just from consistency of $\tilde{\beta}$ and slutsky lemma. But this method won't work because the infinite sum of little op is not necessary a little op,  $\tilde{e}_{i}=e_{i}+o_{p}\left(1\right)$ doesn't yield $\frac{1}{n}\sum_{i=1}^{n}Z_{i}Z_{i}^{\top}\tilde{e}_{i}=\frac{1}{n}\sum_{i=1}^{n}Z_{i}Z_{i}^{\top}e_{i}+o_{p}\left(1\right)$.
Q9y5
CC BY-SA 4.0
2021-07-19T12:10:35.987Z
In my previous comment, "add up" is suppose to be "average".
chan1142
CC BY-SA 4.0
2021-07-20T09:51:51.113Z
Thanks for your clarification.
Bertrand
<econometrics>
Show that the estimator of GMM weigh matrix is consistent
shk910
# Show that the estimator of GMM weigh matrix is consistent

(Asked by: username_0 on 2021-07-18)

> (Hansen Exercise 13.3) Take the model $Y = X'\beta + e$ with $\mathbb{E}[Ze] = 0 $. Let $\tilde e\_i = Y\_i - X\_i'\tilde \beta$ where $\tilde\beta$ is consistent for $\beta$ (e.g. a GMM estimator with some weight matrix). An estimator of the optimal GMM weigh matrix is
> $$\hat W = \left( \frac1n \sum\_{i=1}^n Z\_i Z\_i' \tilde e\_i^2 \right)^{-1}.$$
> Show that $\hat W \to\_p \Omega^{-1}$ where $\Omega = \mathbb{E}[ZZ'e^2]$.

Since $\tilde \beta$ is consistent, $\tilde e\_i$ also converges in probability to $e\_i$ for all $i$. Also, by LLN,
$$\frac1n \sum\_{i=1}^n Z\_i Z\_i' \tilde e\_i^2 \to E[Z\_iZ\_i' \tilde e\_i^2].$$
I think I need to combine these two results in some way to get the final result. Can you give some help?

**Tags:** econometrics
> **Comments:**
> **username_1**: At the level of a single random term, $\tilde e_i$ will not converge in probability to $e_i$. Averaging is necessary (LLN).  
> **username_2**: Note $\tilde{e}_i = e_i - X_i' (\tilde{\beta}-\beta)$ so $\tilde{e}_i^2 = e_i^2 - 2e_i x_i' (\tilde\beta - \beta) + (\tilde\beta - \beta)' X_i X_i' (\tilde\beta - \beta)$. Multiply $Z_iZ_i'$, add up, and then you will see that the second term and the third term converge in probability to zero due to the consistency of $\tilde\beta$. The first term converges in probability to $E[Z_i Z_i' e_i^2] = \Omega$ by LLN. Note that it's not $E[Z_iZ_i' \tilde{e}_i^2]$.  
> **username_3**: @Bertrand Well, $\tilde{e}_{i}$ does converge in probability to $e_{i}$, it's just from consistency of $\tilde{\beta}$ and slutsky lemma. But this method won't work because the infinite sum of little op is not necessary a little op,  $\tilde{e}_{i}=e_{i}+o_{p}\left(1\right)$ doesn't yield $\frac{1}{n}\sum_{i=1}^{n}Z_{i}Z_{i}^{\top}\tilde{e}_{i}=\frac{1}{n}\sum_{i=1}^{n}Z_{i}Z_{i}^{\top}e_{i}+o_{p}\left(1\right)$.  
> **username_2**: In my previous comment, "add up" is suppose to be "average".  
> **username_1**: Thanks for your clarification.  

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## 1 answers

### Answer by username_2 on 2021-07-19. Score: 1 | Accepted answer

We have
\begin{align}
\frac1n \sum\_{i=1}^n Z\_i \tilde{e}\_i^2 Z\_i' &= \frac1n \sum\_{i=1}^n Z\_i \Big[e\_i^2 - 2e\_i X\_i' (\tilde\beta - \beta) + (\tilde\beta - \beta)' X\_i X\_i' (\tilde\beta - \beta)\Big] Z\_i' \\ &= A\_1 - 2A\_2 + A\_3.
\end{align}
because $\tilde{e}\_i = y\_i - X\_i'\tilde\beta = (X\_i'\beta + e\_i) - X\_i'\tilde\beta = e\_i - X\_i' (\tilde\beta - \beta).$

First,
$$A\_1 = \frac1n \sum\_{i=1}^n Z\_i e\_i^2 Z\_i' \to\_p E[Z\_i e\_i^2 Z\_i'] = \Omega \text{ by LLN}.$$
Next,
$$A\_2 = \frac1n \sum\_{i=1}^n Z\_i e\_i X\_i' (\tilde\beta - \beta) Z\_i',$$
the $(j,k)$ element of which is
$$A\_{2,jk} = \frac1n \sum\_{i=1}^n z\_{ij} e\_i X\_i' (\tilde\beta - \beta) z\_{ik} = \left( \frac1n \sum\_{i=1}^n z\_{ik} z\_{ij} e\_i X\_i' \right) (\tilde\beta - \beta).$$
The first term (in the brackets) is $O\_p(1)$; it can be $o\_p(1)$ if $X\_i$ is exogenous but it does not matter. The second term is $o\_p(1)$ because $\tilde\beta$ is consistent. Thus, $A\_2 = o\_p(1)$. Finally,
$$A\_3 = \frac1n \sum\_{i=1}^n Z\_i (\tilde\beta - \beta)' X\_iX\_i' (\tilde\beta - \beta) Z\_i',$$
the $(j,k)$ element of which is
\begin{align}
A\_{3,jk} &= \frac1n \sum\_{i=1}^n z\_{ij} (\tilde\beta - \beta) X\_i X\_i' (\tilde\beta - \beta) z\_{ik}\\
&= (\tilde\beta - \beta)' \left[ \frac1n \sum\_{i=1}^n X\_i z\_{ij} z\_{ik} X\_i' \right] (\tilde\beta - \beta)\\
&= o\_p(1) O\_p(1) o\_p(1) = o\_p(1),
\end{align}
and thus $A\_3 = o\_p(1)$.

We have shown that $\hat{W}^{-1} \to\_p \Omega$. Under the assumption that $\Omega$ is invertible, $\hat{W} \to\_p \Omega^{-1}$.


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