Solutions to Review Question

by Qiang Gao, updated at Mar 13, 2017

Review Question 1.1.4 (Normally distributed ramdom sample)

Consider a random sample on consumption and disposable income, $( CON_i, YD_i )$, $( i = 1, 2, \ldots, n )$. Suppose the joint distribution of $( CON_i, YD_i )$ (which is the same across $i$ because of the random sample assumption) is normal. Clearly, Assumption 1.3 is satisfied; the rank of $\mathbf{X}$ would be less than $K$ only by pure accident. Show that the other assumptions, Assumptions 1.1, 1.2, and 1.4, are satisfied. Hint: if two random variables, $y$ and $x$, are jointly normally distributed, then the conditional expectation is linear in $x$, i.e.,

$$\mathrm{E} ( y | x ) = \beta_1 + \beta_2 x,$$

and the conditional variance, $\mathrm{Var} ( y | x )$, does not depend on $x$. Here, the fact that the distribution is the same across $i$ is important; if the distribution differed across $i$, $\beta_1$ and $\beta_2$ could vary across $i$.

Solution (with flaw)

(1) We define the error term as

$$\varepsilon_i = CON_i - \beta_1 - \beta_2 YD_i,$$

then Assumption 1.1 is satisfied.

(2) Because

$$\begin{align} \mathrm{E} (\varepsilon_i | \mathbf{X}) & = \mathrm{E} ( CON_i - \beta_1 - \beta_2 YD_i | \mathbf{X} ) \qquad \text{(definition of $\varepsilon_i$)} \\ & = \mathrm{E} ( CON_i | YD_i ) - \beta_1 - \beta_2 YD_i \qquad \text{(linearity of conditional expectations)} \\ & = \beta_1 + \beta_2 \mathit{YD}_i - \beta_1 - \beta_2 \mathit{YD}_i \qquad \text{(hint)} \\ & = 0, \end{align}$$

Assumptoin 1.2 holds.

(3) To prove Assumption 1.4,

$$\begin{align} \mathrm{E} ( \varepsilon_i^2 | \mathbf{X} ) & = \mathrm{Var} ( \varepsilon_i | \mathbf{X} ) + \mathrm{E} ( \varepsilon_i | \mathbf{X} )^2 \qquad \text{(definition of $\mathrm{Var} (\cdot)$)} \\ & = \mathrm{Var} ( \varepsilon_i | \mathbf{X} ) \qquad \text{(Assumption 1.2)} \\ & = \mathrm{Var} ( CON_i - \beta_1 - \beta_2 YD_i | YD_i ) \qquad \text{(definition of $\varepsilon_i$)} \\ & = \mathrm{Var} ( CON_i | YD_i ) \qquad \text{($\mathrm{Var} (ax + b) = a^2 \mathrm{Var} (x)$ )} \\ & = \sigma^2 > 0, \qquad \text{(hint)} \end{align}$$

and

$$\begin{align} \mathrm{E} ( \varepsilon_i \varepsilon_j | \mathbf{X} ) & = \mathrm{E} ( \varepsilon_i | \mathbf{x}_i ) \mathrm{E} ( \varepsilon_j | \mathbf{x}_j ) \qquad \text{(Review Question 1.1.2)} \\ & = 0. \qquad \text{(Assumption 1.2)} \end{align}$$

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