<p>If you want to explain the fraction of variance explained, use <span class="math-container">$R^2$</span>. If you want to compare <span class="math-container">$R^2$</span>'s but adjust (<em>very weakly</em>) for model size, use adjusted <span class="math-container">$R^2$</span>'s. If you want to compare these models, use an <span class="math-container">$F$</span>-test since Model 1 is nested in Model 2.</p>

CC BY-SA 4.0
2020-08-12T00:31:22.690Z
2020-08-12T00:31:22.690Z
kurtosis
<p>I want to compare coefficients of determination (i.e., <span class="math-container">$R^2$</span>) of the following two models. <span class="math-container">$fe_{t+h}$</span> is a forecast error in <span class="math-container">$t+h$</span> and <span class="math-container">$\mathbf{\varepsilon}_t$</span> is a shock of interest occurring between <span class="math-container">$t$</span> and <span class="math-container">$t+h$</span>. 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.</p>
<p><strong>Model 1</strong>
<span class="math-container">\begin{align*}
fe_{t+h} = \beta_0 \mathbf{\varepsilon}_{t} + ... + \beta_h\mathbf{\varepsilon}_{t+h} + u_t
\end{align*}</span></p>
<p><strong>Model 2</strong>
<span class="math-container">\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*}</span></p>
<p>where <span class="math-container">$F(z_t)$</span> is a smooth logistic function with <span class="math-container">$F(z_t)\in [0,1]$</span> which weights the coefficients <span class="math-container">$\beta_0, ..., \beta_h$</span> between regimes 1 and 2.
In my model at hand, <span class="math-container">$h$</span> is equal to 15.</p>
<p>Is it ok to use the non-adjusted <span class="math-container">$R^2$</span>? I am aware that typically, <span class="math-container">$R^2$</span> will rise with the number of additional explanatory variables. However, in my case I have no additional variables but only weighted coefficients.</p>

<econometrics>
Coefficient of Determination with Weighted Coefficients
chopschoc
# Coefficient of Determination with Weighted Coefficients

(Asked by: username_0 on 2020-08-11)

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.

**Tags:** econometrics


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## 1 answers

### Answer by username_1 on 2020-08-12. Score: 1 | Accepted answer

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.


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