# Hayashi Econometrics Seemingly Unrelated Regressions (SUR) Eq 4.5.13'-15'

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.

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

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

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

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

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

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

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

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

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 $$P_ZZ=Z$$, $$P_XZ=Z+P_ZDLD'Z-P_ZD'LDZ-DLD'Z+DLD'Z=Z.$$