Coefficient of Determination with Weighted Coefficients

Question

I want to compare coefficients of determination (i.e., $R^2$) of the following two models. $fe_{t+h}$ is a forecast error in $t+h$ and $\mathbf{\varepsilon}_t$ is a shock of interest occurring between $t$ and $t+h$. The idea is to interpret the coefficient of determination as the share of forecast error variance (FEVD) which can be explained by my shock of interest.

Model 1 \begin{align*} fe_{t+h} = \beta_0 \mathbf{\varepsilon}_{t} + ... + \beta_h\mathbf{\varepsilon}_{t+h} + u_t \end{align*}

Model 2 \begin{align*} fe_{t+h} = F(z_t) \left(\beta_0^I \mathbf{\varepsilon}_{t} + ... + \beta_h^I \mathbf{\varepsilon}_{t+h}\right) + \left( 1-F(z_t)\right) \left(\beta_0^{II}\mathbf{\varepsilon}_{t} + ... + \beta_h^{II} \mathbf{\varepsilon}_{t+h}\right) + u_t \end{align*}

where $F(z_t)$ is a smooth logistic function with $F(z_t)\in [0,1]$ which weights the coefficients $\beta_0, ..., \beta_h$ between regimes 1 and 2. In my model at hand, $h$ is equal to 15.

Is it ok to use the non-adjusted $R^2$? I am aware that typically, $R^2$ will rise with the number of additional explanatory variables. However, in my case I have no additional variables but only weighted coefficients.

Answer 1

If you want to explain the fraction of variance explained, use $R^2$. If you want to compare $R^2$'s but adjust (very weakly) for model size, use adjusted $R^2$'s. If you want to compare these models, use an $F$-test since Model 1 is nested in Model 2.