How to obtain the inverse of the variance covariane matrix of GLS

by user9903833   Last Updated January 11, 2019 11:19 AM

In the standard GLS set up how do you find the inverse of the variance covariance matrix?

$$y _ { i t } = \beta _ { 0 } + x _ { i t } ^ { \prime } \beta + \alpha _ { i } + u _ { i t } \hspace{35pt} u _ { i t } \sim I I D ( 0 , \sigma _ { u } ^ { 2 })\hspace{35pt} \alpha _ { i } \sim I I D ( 0 , \sigma _ { \alpha } ^ { 2 } ) $$

$\alpha _ { i }$ is the random error, $u_{it}$ is the error term, $x _ { i t } $ are the control variables, $\Omega$ is the variance covariance matrix of GLS estimator it is symmetric and positive deļ¬nite. $l _ { T } l _ { T } ^ { \prime }$ is a t x t matrix of ones $ I_{T}$ is the t x t identity matrix. $\sigma _ { \alpha } ^ { 2 }$ is the variance of the random effect and $\sigma _ { u } ^ { 2 }$ is the variance of the error term.

$$V \{ \alpha _ { i } l _ { T } + u _ { i } \} = \Omega = \sigma _ { \alpha } ^ { 2 } l _ { T } l _ { T } ^ { \prime } + \sigma _ { u } ^ { 2 } I _ { T } \hspace{35pt} (a)$$

$$\Omega ^ { - 1 } = \sigma _ { u } ^ { - 2 } [ I _ { T } - \frac { \sigma _ { \alpha } ^ { 2 } } { \sigma _ { u } ^ { 2 } + T \sigma _ { \alpha } ^ { 2 } } l _ { T } l _ { T } ^ { \prime } ] \hspace{35pt} (b)$$

$$\Omega ^ { - 1 } = \sigma _ { u } ^ { - 2 } [ ( I _ { T } - \frac { 1 } { T } l _ { T } l _ { T } ^ { \prime } ) + \psi \frac { 1 } { T } l _ { T } l _ { T } ^ { \prime } ] $$

$$\psi = \frac { \sigma _ { u } ^ { 2 } } { \sigma _ { u } ^ { 2 } + T \sigma _ { \alpha } ^ { 2 } }$$

Does anyone know the algebraic steps required to obtain the inverse of the variance covariance matrix $\Omega ^ {-1}$ from the variance covariance matrix i.e. go from $(a)$ to $(b)$. All of the papers I have looked at skip this step so I suspect its straight forward but my matrix algebra is not great so I can't figure it out.



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