## 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.