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

$$ (1) \hspace{1.4 cm}\Omega=\sigma^2_{c}\boldsymbol{1}\boldsymbol{1'} + \sigma^2_{\varepsilon}I_T=(\sigma^2_{\varepsilon}+T\sigma^2_{c})\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}'+\sigma^2_{\varepsilon}(I_T- \boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}')$$ $$\Omega^{-1}=\frac{1}{\sigma^2_{\varepsilon}+T\sigma^2_{c}}\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}'+\frac{1}{\sigma^2_{\varepsilon}}(I_t-\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}')$$ $$|\Omega|=(\sigma^2_{\varepsilon}+T\sigma^2_{c})\sigma^{{2(T-1)}}_{\varepsilon}$$

The Lemma states: enter image description here

Can anyone explain the second equality in (1)?

  • $\begingroup$ I could imagine that this fits better to stats.stackexchange $\endgroup$ – Ric Mar 31 '14 at 7:49
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    $\begingroup$ Could you clarify the first equation in (1)? There are some brackets missing I suppose? And why the notation $(\mathbb{1}^\prime \mathbb{1})^{-1}$? Isn't that just $1/T$? $\endgroup$ – vanguard2k Mar 31 '14 at 8:37

I agree with vanguard2k's comment: A few more details on the notation would be helpful. But, as far as I can tell, the second equality is a simple expansion.

First, $\mathbf{1}'\mathbf{1} = T$ (assuming the vectors are elements of $\mathbb{R}^T$). The expression $\mathbf{1} (\mathbf{1}'\mathbf{1})^{-1} \mathbf{1}'$ is therefore nothing else than a $T\times T$ matrix with $\frac{1}{T}$ in every element.

I assume that scalar addition to a matrix actually means that a matrix with the scalar as each element is added, i.e. that $\sigma_c^2 + \sigma_{\epsilon}^2 \mathbf{I}_T := \sigma_c^2 \mathbf{1}\mathbf{1}' + \sigma_{\epsilon}^2 \mathbf{I}_T$. Then you can write: \begin{align} \sigma_c^2 + \sigma_{\epsilon}^2 \mathbf{I}_T =&\ \sigma_c^2 T \frac{1}{T} \mathbf{1}\mathbf{1}' + \sigma_{\epsilon}^2 \mathbf{I}_T \\ =&\ T \sigma_c^2 \mathbf{1}(\mathbf{1}'\mathbf{1})^{-1} \mathbf{1}' + \sigma_{\epsilon}^2 \mathbf{I}_T + \sigma_{\epsilon}^2\mathbf{1}(\mathbf{1}'\mathbf{1})^{-1} \mathbf{1} - \sigma_{\epsilon}^2\mathbf{1}(\mathbf{1}'\mathbf{1})^{-1} \mathbf{1}' \\ =&\ (\sigma_{\epsilon}^2 + T \sigma_c^2) \mathbf{1}(\mathbf{1}'\mathbf{1})^{-1} \mathbf{1}' + \sigma_{\epsilon}^2 (\mathbf{I}_T - \mathbf{1}(\mathbf{1}'\mathbf{1})^{-1} \mathbf{1}') \end{align}

  • $\begingroup$ Thank you for your excellent answer! I made everything very clear! $\endgroup$ – Sunv Apr 9 '14 at 21:44

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