Solutions to Review Question

by Qiang Gao, updated at Mar 13, 2017

Review Question 1.1.6 (An exercise in conditional and unconditional expectations)

Show that Assumptions 1.2 and 1.4 imply

$$\begin{align} \mathrm{Var} ( \varepsilon_i ) & = \sigma^2 \qquad (i = 1, 2, \ldots, n) \\ \text{and} \qquad \mathrm{Cov} ( \varepsilon_i, \varepsilon_j ) & = 0 \qquad (i \neq j; i, j = 1, 2, \ldots, n). \tag{$\ast$} \end{align}$$

Solution

Strict exogeneity implies $\mathrm{E} (\varepsilon_i) = 0$. So $(\ast)$ is equivalent to

$$\begin{align} \mathrm{E} ( \varepsilon_i^2 ) & = \sigma^2 \qquad (i = 1, 2, \ldots, n) \\ \text{and} \qquad \mathrm{E} ( \varepsilon_i \varepsilon_j ) & = 0 \qquad (i \neq j; i, j = 1, 2, \ldots, n). \end{align}$$

(1) For $i = 1, 2, \ldots, n$,

$$\begin{align} \mathrm{E} (\varepsilon_i^2) & = \mathrm{E} [\mathrm{E} (\varepsilon_i^2 | \mathbf{X})] \qquad \text{(law of total expectations)} \\ & = \sigma^2. \qquad \text{(Assumption 1.4)} \end{align}$$

(2) For $i \neq j; i, j = 1, 2, \ldots, n$,

$$\begin{align} \mathrm{E} (\varepsilon_i \varepsilon_j) & = \mathrm{E} [ \mathrm{E} (\varepsilon_i \varepsilon_j | \mathbf{X}) ] \qquad \text{(law of total expectations)} \\ & = 0. \qquad \text{(Assumption 1.4)} \end{align}$$

Copyright ©2017 by Qiang Gao