Solutions to Review Question

by Qiang Gao, updated at Mar 13, 2017

Review Question 1.1.3 (Combining linearity and strict exogeneity)

Show that Assumptions 1.1 and 1.2 imply

$$\mathrm{E} ( y_i | \mathbf{X} ) = \mathbf{x}_i' \boldsymbol{\beta} \qquad \text{($i = 1, 2, \ldots, n$)} \tag{1.1.20}$$

Conversely, show that this assumption implies that there exist error terms that satisfy those two assumptions.

Solution

(1) Firstly, we prove Assumption 1.1 and 1.2 imply (1.1.20).

$$\begin{align} \mathrm{E} ( y_i | \mathbf{X} ) & = \mathrm{E} ( \mathbf{x}_i' \boldsymbol{\beta} + \varepsilon_i | \mathbf{X} ) \qquad \text{(Assumption 1.1)} \\ & = \mathrm{E} ( \mathbf{x}_i' \boldsymbol{\beta} | \mathbf{X} ) + \mathrm{E} ( \varepsilon_i | \mathbf{X} ) \qquad \text{(linearity of conditional expectatioins)} \\ & = \mathrm{E} ( \mathbf{x}_i' \boldsymbol{\beta} | \mathbf{X} ) \qquad \text{(Assumption 1.2)} \\ & = \mathbf{x}_i' \boldsymbol{\beta} \qquad \text{($\mathbf{x}_i$ is known conditional on $\mathbf{X}$, $\boldsymbol{\beta}$ is constant)} \end{align}$$

(2) Conversely, we prove (1.1.20) implies there exist error terms that satisfy Assumption 1.1 and 1.2.

We define the error term as

$$\varepsilon_i = y_i - \mathbf{x}_i' \boldsymbol{\beta},$$

then Assumption 1.1 is satisfied. To prove assumption 1.2, notice that

$$\begin{align} \mathrm{E} (\varepsilon_i | \mathbf{X}) & = \mathrm{E} ( y_i - \mathbf{x}_i' \boldsymbol{\beta} | \mathbf{X}) \qquad \text{(definition of $\varepsilon_i$)} \\ & = \mathrm{E} ( y_i | \mathbf{X} ) - \mathrm{E} ( \mathbf{x}_i' \boldsymbol{\beta} | \mathbf{X} ) \qquad \text{(linearity of conditional expectations)} \\ & = \mathbf{x}_i' \boldsymbol{\beta} - \mathrm{E} ( \mathbf{x}_i' \boldsymbol{\beta} | \mathbf{X} ) \qquad \text{(1.1.20)} \\ & = \mathbf{x}_i' \boldsymbol{\beta} - \mathbf{x}_i' \boldsymbol{\beta} \qquad \text{($\mathbf{x}_i$ is known conditional on $\mathbf{X}$, $\boldsymbol{\beta}$ is constant)} \\ & = 0. \end{align}$$

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