Solutions to Review Question

by Qiang Gao, updated at Mar 20, 2017

Review Question 1.1.2 (Conditional cross-moment of error terms)

Prove the last equality in (1.1.15),

$$\mathrm{E} (\varepsilon_i \varepsilon_j | \mathbf{X}) = \mathrm{E} (\varepsilon_i | \mathbf{x}_i) \mathrm{E} (\varepsilon_j | \mathbf{x}_j) \qquad \text{(for $i \neq j$)}. \tag{1.1.15}$$

Solution

$$\begin{align} \mathrm{E} ( \varepsilon_i \varepsilon_j | \mathbf{X} ) & = \mathrm{E} [ \mathrm{E} ( \varepsilon_i \varepsilon_j | \mathbf{X}, \varepsilon_j ) | \mathbf{X} ] \qquad \text{(law of iterated expectations)} \\ & = \mathrm{E} [ \varepsilon_j \mathrm{E} ( \varepsilon_i | \mathbf{X}, \varepsilon_j ) | \mathbf{X} ] \qquad \text{($\varepsilon_j$ is constant under condition)} \\ & = \mathrm{E} [ \varepsilon_j \mathrm{E} ( \varepsilon_i | \mathbf{x}_i ) | \mathbf{X} ] \qquad \text{(random sample)} \\ & = \mathrm{E} ( \varepsilon_i | \mathbf{x}_i ) \mathrm{E} ( \varepsilon_j | \mathbf{X}) \qquad \text{($\mathbf{x}_i$ is known conditional on $\mathbf{X}$)} \\ & = \mathrm{E} ( \varepsilon_i | \mathbf{x}_i ) \mathrm{E} ( \varepsilon_j | \mathbf{x}_j) \qquad \text{(random sample)} \end{align}$$

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