2
$\begingroup$

Evaluate the determinant $\det \Omega $ and find the inverse matrix $\Omega^{-1}$ of:

$$\Omega = \begin{bmatrix} \beta_1^2(1+\theta_1^2) & \beta_1 \beta_2 & ... & \beta_1 \beta_{k-1} & \beta_1 \beta_k \\ \beta_2 \beta_1 & \beta_2^2(1+ \theta_2^2) & ... & \beta_2 \beta_{k-1} & \beta_2 \beta_k \\ ... & ... & ... & ... & ... \\ \beta_k \beta_1 & \beta_k \beta_2 & ... & \beta_k \beta_{k-1} & \beta_k^2(1+\theta_k^2) \end{bmatrix}$$

Application

I got this problem when attempting to understand the methodology of the Worldwide Governance Indicators where the authors specify a log-likelihood function of three unknown parameters to solve the maximisation at page 97-99 here Governance Matters VII: Aggregate and Individual Governance Indicators, 1996-2007 (page 97-99).

$\endgroup$

2 Answers 2

7
$\begingroup$

Your matrix is diagonal plus rank 1. Use the Sherman-Morrison formula and the matrix determinant lemma.

$\endgroup$
0
$\begingroup$

$\Omega = \begin{bmatrix} \beta_1^2(1+\theta_1^2) & \beta_1 \beta_2 & ... & \beta_1 \beta_{k-1} & \beta_1 \beta_k \\ \beta_2 \beta_1 & \beta_2^2(1+ \theta_2^2) & ... & \beta_2 \beta_{k-1} & \beta_2 \beta_k \\ ... & ... & ... & ... & ... \\ \beta_k \beta_1 & \beta_k \beta_2 & ... & \beta_k \beta_{k-1} & \beta_k^2(1+\theta_k^2) \end{bmatrix}= \begin{bmatrix} \beta_1^2\theta_1^2 & & & \\ & \beta_2^2\theta_2^2 & & \\ & & \ddots & \\ & & & \beta_k^2\theta_k^2\\ \end{bmatrix} + \begin{bmatrix} \beta_1 \\ \beta_2 \\ \vdots \\ \beta_k \\ \end{bmatrix} \cdot \begin{bmatrix} \beta_1 \: \beta_2 \: \ldots \:\beta_k \end{bmatrix} $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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