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I am self-studying some econometrics/linear regression model. In the text (Greene 2018), it is assumed that $\text{E}[\mathbf{\varepsilon}\mathbf{\varepsilon}'\mid \mathbf{X}]=\sigma^2\mathbf{I}$. Then the book states that, by using the variance decomposition formula, we find $$ \text{Var}[\mathbf{\varepsilon}] = \text{E}[\text{Var}[\mathbf{\varepsilon}\mid\mathbf{X}]] + \text{Var}[\text{E}[\mathbf{\varepsilon}\mid\mathbf{X}]] = \sigma^2\mathbf{I}. $$

Here, $\mathbf{\varepsilon}$ is an $n\times1$ column vector of disturbances and $\mathbf{X}$ is the $n\times K$ data matrix.

I cannot understand how equality is derived. Could someone please explain it for me? Thanks a lot in advance.

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Presumably it was assumed that $\operatorname E(\varepsilon\mid \mathbf X) = 0$ regardless of the value of $\mathbf X.$ Consequently the variance of that expression is $0$ and so the entire second term in the sum is $0.$

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  • $\begingroup$ Thank you for your answer. I have a follow-up question: Yes, it was assumed that $\text{E}[\mathbf{\epsilon}|\mathbf{X}]=\mathbf{0}$, so I can see that $\text{Var}[\mathbf{0}]=\mathbf{0}$. But why is the first term $\sigma^2\mathbf{I}$? $\endgroup$
    – Beerus
    Commented Dec 2 at 2:10
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    $\begingroup$ Because $\text{E}[\mathbf{\varepsilon}\mathbf{\varepsilon}'\mid\mathbf{X}] = Var[\varepsilon \mid \mathbf{X}] \implies \text{E} [Var[\varepsilon \mid \mathbf{X}]] = \text{E}[ \sigma^2 \mathbf{I} ] = \sigma^2 \mathbf{I}$ $\endgroup$
    – Stats
    Commented Dec 2 at 3:19
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    $\begingroup$ @Beerus : The vector $\varepsilon$ is $n\times1$ so $\varepsilon'$ is $1\times n$ and so $\varepsilon\varepsilon'$ is $n\times n.$ The $i,j$ entry in $\varepsilon\varepsilon'$ is the product of the $i,1$ entry in $\varepsilon$ and the $1,j$ entry in $\varepsilon'.$ And that is the product of the $i\text{th}$ and $j\text{th}$ residuals. Since the residuals have expectation $0,$ that is the covariance between those two residuals. The covariance is $0$ if $i\ne j$ and is $\sigma^2$ if $i=j,$ i.e. the covariance between a random variable and itself is the variance. So all of$\,\ldots\qquad$ $\endgroup$
    – Michael Hardy
    Commented Dec 2 at 17:46
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    $\begingroup$ $\ldots\,$the diagonal entries are $\sigma^2$ and all of the off-diagonal entries are $0. \qquad$ $\endgroup$
    – Michael Hardy
    Commented Dec 2 at 17:47
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    $\begingroup$ @MichaelHardy I really appreciate your help! $\endgroup$
    – Beerus
    Commented Dec 2 at 20:57

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