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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.