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I'm trying to understand the derivation of a ML-estimator and more specifically the rewriting of the covariance matrix $\Sigma$. In this rewriting, a lemma is used to show that: $$ \tag{1} \Omega=\sigma^2_{c}\boldsymbol{1}\boldsymbol{1'} + \sigma^2_{\varepsilon}I_T=(\sigma^2_{\varepsilon}+T\sigma^2_{c})\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}'+\sigma^2_{\varepsilon}(I_T- \boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}')$$ $$\Omega^{-1}=\frac{1}{\sigma^2_{\varepsilon}+T\sigma^2_{c}}\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}'+\frac{1}{\sigma^2_{\varepsilon}}(I_t-\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}')$$ $$|\Omega|=(\sigma^2_{\varepsilon}+T\sigma^2_{c})\sigma^{{2(T-1)}}_{\varepsilon}$$ Here, $\boldsymbol{1}$ is a T-vector of ones.

The Lemma states:

For $\lambda_1 > 0$ and $\lambda_2 > 0$ we define the covariance $\Sigma$ by: $$\Sigma = \lambda_1 P + \lambda_2 Q ,$$ so that $$x'\Sigma x = \lambda_1 T \bar{x}^2+\lambda_2\sum\nolimits_{i=1}^T (x_i-\bar{x})^2$$ Then $$\Sigma^{-1} = \lambda_1^{-1} P + \lambda_2^{-1} Q ,$$ so that $$x'\Sigma^{-1} x = \lambda_1^{-1} T \bar{x}^2+\lambda_2^{-1}\sum_{i=1}^T (x_i-\bar{x})^2 .$$ The eigenvalues of $\Sigma$ are $\lambda_1$ with multiplicity one, and $\lambda_2$ with multiplicity $T-1$, so that $$ |\Sigma| = \lambda_1\lambda_2^{T-1} .$$

Can anyone explain the second equality in (1)? I.e. this equality $$\sigma^2_{c} + \sigma^2_{\varepsilon}I_T=(\sigma^2_{\varepsilon}+T\sigma^2_{c})\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}'+\sigma^2_{\varepsilon}(I_T- \boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}')$$

Please let me know if the question is off topic.

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You can simply multiply $\Omega^{-1}$ as given by the author times $\Omega$ and see that the product is $I$. This confirms that the formula for $\Omega^{-1}$ is correct.

The key here is that

$(I-1(1'1)^{-1}1')(1(1'1)^{-1}1')=1(1'1)^{-1}1'-1(1'1)^{-1}1'=0$

so the cross terms vanish.

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  • $\begingroup$ Thanks for your answer. Actually I ment the second equality in the first line in (1). Do you understand what happens here? Sorry for not being clear here! $\endgroup$
    – Sunv
    Mar 30, 2014 at 17:50
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    $\begingroup$ You aren't being clear about which equality you're trying to understand. My answer gets to why $\Omega^{-1}$ is what it is. The second line of (1) reads $\Omega^{-1}=...$. Are you referring to some other equation? $\endgroup$
    – Brian Borchers
    Mar 31, 2014 at 12:19
  • $\begingroup$ Sorry for the confusion. I made it clear in the buttom of the question (the original question has now been edited). $\endgroup$
    – Sunv
    Mar 31, 2014 at 16:34
  • $\begingroup$ $(1'1)^{-1}$ is simply the scalar $1/T$. Thus $T\sigma_{c}^{2}1(1'1)^{-1}1'=\sigma_{c}^{2}11'$. $\endgroup$
    – Brian Borchers
    Mar 31, 2014 at 21:51

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