15.2 Non-Nested Model

compare models with different non-nested specifications

15.2.1 Davidson-Mackinnon test

15.2.1.1 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 \(H_0: \gamma = 0\)

  1. 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.

\[ y = \beta_0 + ln(x)\beta_1 + \epsilon \\ y = \beta_0 + x(\beta_1) + x^2\beta_2 + \epsilon \] * 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

15.2.1.2 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 \(\hat{y}\)

  1. 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.