In a SURE framework, if all X are the same in all regressions I was under the impression that there is no efficiency gain. Recently an assistant professor told me that the beta coefficients would be the same as OLS, but the standard errors would decrease due to the SURE framework even though all the X are the same.

Looking at the derivation in Greene's 7th edition, section 10.2.2, I believe I am correct.

Can anyone clarify further? Does SURE give an efficiency improvement if all X are the same in all regressions?

  • 1
    $\begingroup$ As far as I know, you are correct. I think chapter 3 or 4 in luktepohl's text might give a proof of your claim. $\endgroup$
    – mark leeds
    Jul 8 at 13:01

Assume that for each observation $i = 1,\ldots, N$, we have $M$ equations: $$ y_{i,j} = x_{i,j}\beta_j + \varepsilon_{i,j} $$ Where $i = 1,\ldots, N$ enumerates individuals and and $j = 1,\ldots, M$ enumerates the equations. here $x$ is of size $1 \times k_j$ and $\beta_j$ is of size $k_j \times 1$ and $k_j$ is the number of covariates for regression $j$. Stacking over all $i = 1,\ldots N$, we get $M$ equations: $$ y_j = X_j \beta_j + \varepsilon_j $$ where now $X_j$ is of size $N \times k_j$. For simplicity, assume that $X_j$ are non-stochastic. Next, assume that for all $i = 1,\ldots, N$ and $j = 1,\ldots, M$: $$ \begin{align*} &\mathbb{E}(\varepsilon_{i,j}) = 0,\\ &\mathbb{E}(\varepsilon_{i,j}^2) = \sigma_{jj} \end{align*} $$ For the covariance between equations, let for all $i = 1,\ldots, N$ and $j,\ell = 1,\ldots, M$: $$ \mathbb{E}(\varepsilon_{i,j} \varepsilon_{i,\ell}) = \sigma_{j,\ell} $$ while for all $j,\ell = 1,\ldots, M$ and $i,i' = 1,\ldots, N$ with $i \ne i'$: $$ \mathbb{E}(\varepsilon_{i,j}, \varepsilon_{i',k}) = 0 $$ This means that errors for the same individual might be correlated across equations, while errors for different individuals are uncorrelated.

This can be expressed more compactly as: $$ cov(\varepsilon_j, \varepsilon_{\ell}) = \sigma_{j,\ell}I_N $$ Now, let us stack the various equations, one on top of the other: $$ y = Z\beta + \varepsilon, $$ where: $$ y = \begin{bmatrix} y_1\\y_2\\ \vdots\\y_M\end{bmatrix}, \varepsilon = \begin{bmatrix} \varepsilon_1 \\ \vdots \\ \varepsilon_M \end{bmatrix},\\ Z = \begin{bmatrix} X_1 & 0 & \ldots & 0\\ 0 & X_2 & \ldots & 0,\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \ldots & X_M \end{bmatrix}, \beta = \begin{bmatrix} \beta_1\\ \vdots \\ \beta_M\end{bmatrix} $$ The variance-covariance matrix of $\varepsilon$ takes the form: $$ \mathbb{E}(\varepsilon \varepsilon') = V = \begin{bmatrix} \sigma_{11} I_N & \sigma_{12}I_N & \ldots & \sigma_{1M} I_N\\ \sigma_{21} I_N & \sigma_{22} I_N & \ldots & \sigma_{2N} I_N\\ \ldots & \ldots & \ddots & \vdots\\ \sigma_{M1} I_N & \ldots & \ldots & \sigma_{MM}I_N \end{bmatrix} = \Sigma \otimes I_N $$ where $\otimes$ is the Kronecker product and: $$ \Sigma = \begin{bmatrix}\sigma_{11} & \sigma_{12} & \ldots & \sigma_{1M}\\ \sigma_{21} & \sigma_{22} & \ldots & \sigma_{2M}\\ \vdots & \vdots & \ddots & \vdots\\ \sigma_{M1} & \sigma_{M2} & \ldots & \sigma_{MM} \end{bmatrix} $$ $\Sigma$ gives the variance covariance matrix of the errors for a fixed individual.

For the Kronecker product, we have the rules: $(A \otimes B)^{-1} = A^{-1} \otimes B^{-1}$ and $(A \otimes B)(C \otimes D) = AC \otimes BD$ and $(A \otimes B)' = A' \otimes B'$ .

Let $\hat \Sigma$ be the estimate of $\Sigma$ based on an initial OLS estimation of $y_j$ on $X_j$ and let $\hat V = \hat \Sigma \otimes I_N$. Then the feasible GLS estimator is given by: $$ \begin{align*} \hat \beta &= (Z' \hat V^{-1} Z)^{-1} Z' \hat V^{-1} y,\\ &=(Z'(\hat \Sigma \otimes I_N)^{-1}Z)^{-1}Z'(\hat \Sigma \otimes I_N)^{-1}y,\\ &= (Z'(\hat \Sigma^{-1}\otimes I_N)Z)^{-1}Z'(\hat \Sigma^{-1}\otimes I_N)y,\\ &= \beta + (Z'(\hat \Sigma^{-1}\otimes I_n)Z)^{-1}Z'y \end{align*} $$

Now, let us assume that all the $X_i$ are identical, say $X$, then $Z = I_M \otimes X$ and we can further simplify: $$ \begin{align*} \hat \beta &= (Z'(\hat \Sigma^{-1}\otimes I_N)Z)^{-1}Z'(\hat \Sigma^{-1}\otimes I_N)y,\\ &= ((I_M \otimes X)'(\hat \Sigma^{-1}\otimes I_N)(I_M \otimes X))^{-1}(I_M \otimes X)'(\hat \Sigma^{-1}\otimes I_N)y,\\ &= ((I_M \hat \Sigma^{-1}\otimes X'I_N)(I_M \otimes X))^{-1}(I_M \hat \Sigma^{-1} \otimes X' I_N)y,\\ &= (\hat \Sigma^{-1} \otimes X'X)^{-1}(\hat \Sigma^{-1}\otimes X')y,\\ &= (\hat \Sigma \otimes (X'X)^{-1})(\hat \Sigma^{-1}\otimes X')y,\\ &= (\hat \Sigma\hat \Sigma^{-1} \otimes (X'X)^{-1}X')y\\ &= (I_M \otimes (X'X)^{-1} X')y \end{align*} $$ Notice that $\hat \Sigma$ disappeared from this equation. The last one equation can be written in the following way: $$ \hat \beta = \begin{bmatrix} (X'X)^{-1}X'y_1\\ (X'X)^{-1} X'y_2\\ \vdots\\ (X'X)^{-1}X' y_1 \end{bmatrix} = \beta + \begin{bmatrix}(X'X)^{-1}X'\varepsilon_1,\\ (X'X)^{-1}X'\varepsilon_2\\\vdots \\ (X'X)^{-1}X'\varepsilon_M\end{bmatrix} $$ So the feasible GLS estimates are identical to the OLS estimates from an equation by equation estimation. Notice that this also means that the residuals $\hat \varepsilon_j$ will be identical to the residuals from an OLS estimation.

Now to estimate the variance covariance matrix, we take the product $(\hat \beta - \beta)(\hat \beta - \beta)'$ which gives a matrix with entries: $$ \begin{align*} (\hat \beta_{j} - \beta_j)(\hat \beta_j - \beta_j)' &= [(X'X)^{-1}X' \varepsilon_j][(X'X)^{-1}X'\varepsilon_j]',\\ &= (X'X)^{-1}X'\varepsilon_j \varepsilon_j'X(X'X)^{-1} \end{align*} $$ Then for equation $j$, we have the variance covariance matix: $$ V(\hat \beta_j) = \mathbb{E}((\hat \beta_j - \beta_j)(\hat \beta_j - \beta_j)) = \sigma_{jj}\left(X'X\right)^{-1}, $$

As $\sigma_{jj}$ is not known, it is usually estimated by $\hat \sigma_{jj} = \frac{1}{N}\sum_i \hat \varepsilon_{i,j}^2$ where $\hat \varepsilon_{i,j}$ are the residuals of the feasible GLS estimator. However, in this case, these will be identical to the residuals of an OLS estimator (as the estimators $\hat \beta$ are identical). As such, the estimates of the variances of $\hat \beta$ for the SUR will be identical to the variance estimates of the OLS estimates (equation by equation).

  • 2
    $\begingroup$ Another thorough answer from you – a joy to read! $\endgroup$ Jul 10 at 12:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.