10.2 Non-Nested Model

compare models with different non-nested specifications

10.2.1 Davidson-Mackinnon test

10.2.1.1 Independent Variable

Should the independent variables be logged? (decide between non-nested alternatives)

$$\begin{aligned} y = \beta_0 + x_1\beta_1 + x_2\beta_2 + \epsilon && \text{(level eq)} \\ y = \beta_0 + ln(x_1)\beta_1 + x_2\beta_2 + \epsilon && \text{(log eq)} \end{aligned}$$

1. Obtain predict outcome when estimating the model in log equation $\check{y}$ and then estimate the following auxiliary equation,

$$y = \beta_0 + x_1\beta_1 + x_2\beta_2 + \check{y}\gamma + error$$

and evaluate the t-statistic for the null hypothesis $H_0: \gamma = 0$

2. Obtain predict outcome when estimating the model in the level equation $\hat{y}$, then estimate the following auxiliary equation,

$$y = \beta_0 + ln(x_1)\beta_1 + x_2\beta_2 + \check{y}\gamma + error$$

and evaluate the t-statistic for the null hypothesis $H_0: \gamma = 0$

• If you reject the null in the (1) step but fail to reject the null in the second step, then the log equation is preferred.
• If fail to reject the null in the (1) step but reject the null in the (2) step then, level equation is preferred.
• If reject in both steps, then you have statistical evidence that neither model should be used and should re-evaluate the functional form of your model.
• If fail to reject in both steps, you do not have sufficient evidence to prefer one model over the other. You can compare the $R^2_{adj}$ to choose between the two models.

$$\begin{aligned} y &= \beta_0 + ln(x)\beta_1 + \epsilon \\ y &= \beta_0 + x(\beta_1) + x^2\beta_2 + \epsilon \end{aligned}$$

• Compare which better fits the data
• Compare standard $R^2$ is unfair because the second model is less parsimonious (more parameters to estimate)
• The $R_{adj}^2$ will penalize the second model for being less parsimonious + Only valid when there is no heteroskedasticity (A4 holds)
• Should only compare after a Davidson-Mackinnon test

10.2.1.2 Dependent Variable

$$\begin{aligned} y &= \beta_0 + x_1\beta_1 + \epsilon & \text{level eq} \\ ln(y) &= \beta_0 + x_1\beta_1 + \epsilon & \text{log eq} \\ \end{aligned}$$

• In the level model, regardless of how big y is, x has a constant effect (i.e., one unit change in $x_1$ results in a $\beta_1$ unit change in y)
• In the log model, the larger in y is, the effect of x is stronger (i.e., one unit change in $x_1$ could increase y from 1 to $1+\beta_1$ or from 100 to 100+100x$\beta_1$)
• Cannot compare $R^2$ or $R^2_{adj}$ because the outcomes are complement different, the scaling is different (SST is different)

We need to “un-transform” the $ln(y)$ back to the same scale as y and then compare,

1. Estimate the model in the log equation to obtain the predicted outcome $\hat{ln(y)}$
2. “Un-transform” the predicted outcome

$$\hat{m} = exp(\hat{ln(y)})$$

3. Estimate the following model (without an intercept)

$$y = \alpha\hat{m} + error$$

and obtain predicted outcome $\hat{y}$

4. Then take the square of the correlation between $\hat{y}$ and y as a scaled version of the $R^2$ from the log model that can now compare with the usual $R^2$ in the level model.