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I was wondering, what happens to the covariance matrix of the errors, when I assume that all the errors are stochastically independent?

Is the covariance matrix still: $$\sigma^2I = \begin{bmatrix}\sigma^2 & ... & 0\\. & . &.\\ 0 & ... & \sigma^2\end{bmatrix}$$

I really appreciate your answer!!!

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What do you mean 'still'? The covariance matrix is

$$ \boldsymbol{\Sigma}_u= \begin{bmatrix} E(u_1^2)&E(u_1u_2) & \cdots & E(u_1u_n)\\ E(u_1u_2)&E(u_2^2) & \cdots & E(u_2u_n)\\ \vdots & \vdots & \ddots & \vdots\\ E(u_1u_n)&E(u_2u_n) & \cdots & E(u_n^2)\\ \end{bmatrix} $$ so under the assumptions that $$ E(u_i)=0 \, \forall \, i\\ E(u_iu_j)=0\, \forall \, i\neq j\\ E(u_i^2)=\sigma^2 \, \forall \, i $$ we can see that $$ \boldsymbol{\Sigma}_u= \begin{bmatrix} \sigma^2&0 & \cdots & 0\\ 0&\sigma^2& \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0&0 & \cdots & \sigma^2\\ \end{bmatrix} =\sigma^2 \boldsymbol{I}_n $$ as you say.

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    $\begingroup$ His question might go a bit deeper assuming he means that maybe the assumptions do longer hold. In any case he'll have to specify. $\endgroup$ – IMA Apr 26 '13 at 7:27
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    $\begingroup$ Note the requirement of the assumption that the error terms all have the same variance; that is, $E(u_i) = 0 \forall i$ and $E(u_i^2) = \sigma^2 \forall i$. This is necessary for saying the covariance matrix has the same $\sigma^2$ value on all the diagonals. $\endgroup$ – Firefeather Apr 26 '13 at 14:55
  • $\begingroup$ Instead of "still" one might even say, "even more so", as uncorrelatedness of the errors is already enough to get $\sigma^2I$, but if independence moreover holds, uncorrelatedness is implied. $\endgroup$ – Christoph Hanck Feb 24 '15 at 12:34

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