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According to 4.5.13

$\hat{A}_{mh}=\left(\frac{1}{n}\sum_{i=1}^{n}z_{im}x_{i}^{\prime}\right)\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{i}^{\prime}\right)^{-1}\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}z_{ih}^{\prime}\right)$

I have no idea how $x_i$ "disappears" and the above equation becomes

$\hat{A}_{mh}=\frac{1}{n}\sum_{i=1}^{n}z_{im}z_{ih}^{\prime}$

The dimension of $\frac{1}{n}\sum_{i=1}^{n}z_{im}x_{i}^{\prime}$ is $L_m\times \sum L_m$

The dimension of $\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{i}^{\prime}\right)^{-1}$ is $\sum L_m\times \sum L_m$

The dimension of $\frac{1}{n}\sum_{i=1}^{n}x_{i}z_{ih}^{\prime}$ is $\sum L_m\times L_h$

many thanks.

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2 Answers 2

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The paragraph just below the equations provides the proof.

Let $D_{m}=\left[e_{\sum_{\ell=1}^{m-1}L_{\ell}+1},e_{\sum_{\ell=1}^{m-1}L_{\ell}+1},\ldots,e_{\sum_{\ell=1}^{m}L_{\ell}}\right]$

$D_{h}=\left[e_{\sum_{\ell=1}^{h-1}L_{\ell}+1},e_{\sum_{\ell=1}^{h-1}L_{\ell}+1},\ldots,e_{\sum_{\ell=1}^{h}L_{\ell}}\right]$

where $e_{j}$ is dim-$\sum_{m=1}^ML_m$ vector where the $j$th element is 1 and all other elements are 0.

Then we have $z_{im}=D_{m}^{\prime}x_{i}$ and $Z_{ih}=D_{h}^{\prime}x_{i}$

Thus the estimator for $\delta$ is

$\left(\frac{1}{n}\sum_{i=1}^{n}z_{im}x_{i}^{\prime}\right)\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{i}^{\prime}\right)^{-1}\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}z_{ih}^{\prime}\right)$

=$\left(\frac{1}{n}\sum_{i=1}^{n}D_{m}^{\prime}x_{i}x_{i}^{\prime}\right)\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{i}^{\prime}\right)^{-1}\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{i}^{\prime}D_{h}\right)$

=$D_{m}^{\prime}\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{i}^{\prime}\right)\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{i}^{\prime}\right)^{-1}\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{i}^{\prime}\right)D_{h}$

=$D_{m}^{\prime}\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{i}^{\prime}\right)D_{h}$

=$\frac{1}{n}\sum_{i=1}^{n}\left(D_{m}^{\prime}x_{i}\right)x_{i}^{\prime}D_{h}$

=$\frac{1}{n}\sum_{i=1}^{n}\left(D_{m}^{\prime}x_{i}\right)\left(D_{h}^{\prime}x_{i}\right)^{\prime}$

=$\frac{1}{n}\sum_{i=1}^{n}z_{im}z_{ih}^{\prime}$

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SUR describes a situation in which the regressors $z_{ih}$ are a subset of the instruments $x_i$. In matrix notation and without the factors of $n$ which are unnecessary for the argument, your first display can be written as $$ Z'X(X'X)^{-1}X'Z. $$ Here, $P_X:=X(X'X)^{-1}X'$ is the projection matrix on $X$. We have $P_XZ=Z$ as we project $Z$ on (among other things) itself, and the fitted values of a regression on itself (plus possibly something else) are just itself. (Hence, the argument is also not specifically related to SUR.)

Geometric explanations would go along lines of saying that $Z$ is already in the column space of $X$.

Think of explaining weight by height, age, gender, other things...and weight - no better way to predict someone's weight than by knowing his/her weight. Clearly not an "interesting" regression from a practical point of view, but that's the computational step here.

Showing this with matrix algebra is a little ugly (at least how I do it here):

Let $X:=(Z:D)$, where $D$ are the "other" variables. We then want to show that $P_XZ=Z$. Using partitioned inverses, we have that, letting $$ L:=(D'M_ZD)^{-1} $$ with the residual maker matrix $M_Z:=I-Z(Z'Z)^{-1}Z'$, $$ \begin{align*}P_X&=(Z:D)\begin{pmatrix}Z'Z&Z'D\\D'X&D'D\end{pmatrix}^{-1}\begin{pmatrix}Z'\\D'\end{pmatrix}\\ &=(Z:D)\begin{pmatrix}(Z'Z)^{-1}+(Z'Z)^{-1}Z'DLD'Z(Z'Z)^{-1}&-(Z'Z)^{-1}Z'DL\\-LD'Z(Z'Z)^{-1}&L\end{pmatrix}\begin{pmatrix}Z'\\D'\end{pmatrix} \end{align*} $$ Multiplying out gives $$P_X=P_Z+P_ZDLD'P_Z-P_ZD'LD-DLD'P_Z+DLD'.$$ Hence, since $P_ZZ=Z$, $$ P_XZ=Z+P_ZDLD'Z-P_ZD'LDZ-DLD'Z+DLD'Z=Z. $$

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