When Adding A Predictor Raises Adjusted R-Squared

Why the Partial F Statistic Appears in Adjusted R-Squared

In ordinary least squares, adding a predictor to a model cannot increase its residual sum of squares when the models are fit to the same observations and the larger model contains the smaller one. That guarantees that ordinary $R^2$ cannot fall. Adjusted $R^2$, however, also accounts for the residual degrees of freedom used by the larger model. The decrease in residual error must be large enough to offset that adjustment.

For one added predictor, the comparison has a particularly useful form: when the partial F statistic is defined, adjusted $R^2$ increases exactly when that statistic is greater than 1. This is an algebraic criterion for the change in adjusted $R^2$, not a claim that the predictor is statistically significant at a conventional test level. The distinction is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Definition: Consider two ordinary least-squares regression models fit to the same $n$ observations, both with an intercept. The reduced model has $k$ predictors; the full model adds one predictor, so it has $k+1$. The models are nested if the reduced model can be obtained from the full model by setting the coefficient of the added predictor to zero. Write $SSE_R$ and $SSE_F$ for their residual sums of squares, and let $\nu=n-k-2$ be the full model's residual degrees of freedom. The partial F statistic for the added predictor is $$ F_{\mathrm{partial}}=\frac{SSE_R-SSE_F}{SSE_F/\nu}, $$ provided $SSE_F>0$ and $\nu>0$.

The formula assumes the added predictor contributes one new, estimable model parameter: the design matrix rank increases by one. Because both models use the same observations and the models are nested, $SSE_R\geq SSE_F$. Thus the statistic is nonnegative when defined. For its usual null-distribution interpretation as an F statistic, the classical linear-model assumptions are required: errors are independent, normally distributed with mean zero and common variance, conditional on the predictors. The adjusted-$R^2$ comparison below is algebraic and does not require normality.

The condition $SSE_F>0$ matters: if the full model fits perfectly, the denominator of this finite partial F statistic is zero. That boundary case requires a separate comparison rather than blindly applying the statistic.

Deriving the Increase Criterion

With an intercept, define $SST=\sum_{i=1}^n(y_i-\bar y)^2$. For a model with $p$ predictors and positive residual degrees of freedom, adjusted $R^2$ is

R^2_{\mathrm{adj}}=1-\frac{SSE/(n-p-1)}{SST/(n-1)}.

This comparison requires $SST>0$, meaning the response is not constant, so the total mean square in the denominator is positive. For the reduced model, $p=k$, giving $n-k-1=\nu+1$ residual degrees of freedom. For the full model, $p=k+1$, giving $\nu=n-k-2$. We also require both residual degrees of freedom to be positive.

Write the adjusted values.
The common factor $SST/(n-1)$ is positive, so the comparison reduces to comparing the two residual mean squares: $$ R^2_{\mathrm{adj},F}=1-\frac{SSE_F/\nu}{SST/(n-1)},\qquad R^2_{\mathrm{adj},R}=1-\frac{SSE_R/(\nu+1)}{SST/(n-1)}. $$

Set the condition for an increase.
The full model has higher adjusted $R^2$ exactly when its residual mean square is smaller: $$ R^2_{\mathrm{adj},F}>R^2_{\mathrm{adj},R} \quad\Longleftrightarrow\quad \frac{SSE_F}{\nu}<\frac{SSE_R}{\nu+1}. $$

Rearrange the inequality.
Since $\nu$ and $\nu+1$ are positive, multiplying by their product preserves the inequality: $$ (\nu+1)SSE_F<\nu SSE_R \quad\Longleftrightarrow\quad SSE_R-SSE_F>\frac{SSE_F}{\nu}. $$

Recognize the partial F statistic.
When $SSE_F>0$, divide the last inequality by $SSE_F/\nu>0$. The result is $$ R^2_{\mathrm{adj},F}>R^2_{\mathrm{adj},R} \quad\Longleftrightarrow\quad F_{\mathrm{partial}}>1. $$

Theorem: For nested least-squares models fit to the same observations, with an intercept, one estimable predictor added, positive residual degrees of freedom, and $SST>0$, if $SSE_F>0$, adjusted $R^2$ increases exactly when $F_{\mathrm{partial}}>1$. It is unchanged when $F_{\mathrm{partial}}=1$ and decreases when $F_{\mathrm{partial}}<1$.

The threshold is 1 because the comparison is between residual mean squares: the predictor must reduce the residual sum of squares enough that the full model’s residual mean square is smaller. A partial F statistic above 1 says precisely that. It is not the same as comparing the statistic with a larger critical value from a significance test.

Applying the Criterion to Nested Models

Worked Example: A Small Error Reduction Is Not Enough

A hypothetical insurer models annual claim counts for 24 policy groups. A reduced model with $k=2$ predictors has $SSE_R=500$. Adding one predictor gives $SSE_F=480$. The response has $SST=600$. Calculate the partial F statistic and both adjusted $R^2$ values, then decide whether adjusted $R^2$ rises.

Method. Use the partial F statistic for the nested-model comparison, then verify the conclusion directly from adjusted $R^2$. The common mistake to avoid is concluding that adjusted $R^2$ must rise just because the full model has smaller $SSE$.

The full model has three predictors, so its residual degrees of freedom are $\nu=24-3-1=20$. The reduced model has $24-2-1=21$ residual degrees of freedom. The partial F statistic is $$ F_{\mathrm{partial}}=\frac{500-480}{480/20} =\frac{20}{24} =0.8333. $$

Since $0.8333<1$, the criterion says adjusted $R^2$ decreases. Check by calculation. The total mean square is $600/23$. For the reduced model, $$ R^2_{\mathrm{adj},R} =1-\frac{500/21}{600/23} =1-\frac{11500}{12600} =0.0873016\ldots. $$ For the full model, $$ R^2_{\mathrm{adj},F} =1-\frac{480/20}{600/23} =1-\frac{552}{600} =0.0800000. $$

Rounded to four decimal places, the values are $0.0873$ and $0.0800$, respectively. Adjusted $R^2$ falls, as predicted. The smaller $SSE$ does improve ordinary $R^2$, but not enough to offset the added predictor’s degrees-of-freedom cost.

Worked Example: A Predictor That Raises Adjusted R-Squared

A hypothetical portfolio study uses $n=30$ observations. A reduced model with $k=3$ predictors has $SSE_R=400$, while the full model adds one predictor and has $SSE_F=360$. Given $SST=900$, calculate the partial F statistic and adjusted $R^2$ values.

Method. Find the full model’s residual degrees of freedom, compare the partial F statistic with 1, and check both adjusted values. This avoids treating every drop in $SSE$ as sufficient evidence of an adjusted-$R^2$ increase.

The full model has four predictors, so $\nu=30-4-1=25$; the reduced model has $26$ residual degrees of freedom. The partial F statistic is $$ F_{\mathrm{partial}} =\frac{400-360}{360/25} =\frac{40}{14.4} =2.7778. $$

Because $2.7778>1$, adjusted $R^2$ increases. The total mean square is $900/29$. Directly, $$ R^2_{\mathrm{adj},R} =1-\frac{400/26}{900/29} =1-\frac{11600}{23400} =0.5042735\ldots, $$ and $$ R^2_{\mathrm{adj},F} =1-\frac{360/25}{900/29} =1-\frac{10440}{22500} =0.5360000. $$

Rounded to four decimal places, the adjusted values are $0.5043$ and $0.5360$. The full model’s residual mean square is $14.4$, below the reduced model’s $400/26$, approximately $15.3846$, which is the direct reason adjusted $R^2$ rises.

Worked Example: The Boundary Between an Increase and a Decrease

A reduced model with one predictor is fit to $n=18$ observations and has $SSE_R=320$. Adding a predictor gives $SSE_F=300$, and $SST=500$. Determine whether adjusted $R^2$ changes.

Method. Use the partial F statistic and verify the adjusted values. The exam trap here is to assume that the criterion must say “increase” or “decrease”; equality is possible.

The full model has two predictors and $\nu=18-2-1=15$ residual degrees of freedom. The reduced model has $16$. Thus $$ F_{\mathrm{partial}}=\frac{320-300}{300/15} =\frac{20}{20} =1. $$

The total mean square is $500/17$. The reduced adjusted value is $$ R^2_{\mathrm{adj},R} =1-\frac{320/16}{500/17} =1-\frac{340}{500} =0.32. $$ The full adjusted value is $$ R^2_{\mathrm{adj},F} =1-\frac{300/15}{500/17} =1-\frac{340}{500} =0.32. $$

The two values are equal: the added predictor reduces $SSE$, but by exactly the amount needed to leave the residual mean square unchanged. No rounding is needed for the final adjusted-$R^2$ values.

When the Full Model Fits Perfectly

If $SSE_F=0$, the finite partial F statistic in the definition is undefined because its denominator is zero. Do not divide by zero or claim that a finite statistic exceeds 1. The adjusted-$R^2$ comparison itself can still be made when $SST>0$ and both models have positive residual degrees of freedom. The full model then has adjusted $R^2=1$. Since residual sums of squares are nonnegative, the reduced model has adjusted $R^2\leq 1$: it is lower if $SSE_R>0$, and equal if $SSE_R=0$ as well.

Worked Example: Zero Residual Error in the Full Model

A reduced model with one predictor is fit to $n=12$ observations and has $SSE_R=16$. Adding a predictor gives $SSE_F=0$, and $SST=100$. Compare adjusted $R^2$, and state whether a finite partial F statistic can be used.

Method. Calculate the adjusted values directly, since the partial F statistic has a zero denominator. This avoids incorrectly applying the finite-statistic criterion in a case where it is undefined.

The reduced model has $12-1-1=10$ residual degrees of freedom, and the full model has $12-2-1=9$. The total mean square is $100/11$. The reduced adjusted value is $$ R^2_{\mathrm{adj},R} =1-\frac{16/10}{100/11} =1-\frac{176}{1000} =0.824. $$ The full model’s residual mean square is zero, so $$ R^2_{\mathrm{adj},F} =1-\frac{0/9}{100/11} =1. $$

Adjusted $R^2$ increases from $0.824$ to $1.000$, with no rounding required. The partial F statistic is undefined because $SSE_F/9=0$. The direct adjusted-$R^2$ calculation gives the conclusion without assigning an artificial finite value to that statistic.

Common Mistakes and Practical Interpretation

Using the raw $R^2$ rule. The full model’s $SSE$ cannot be larger for nested least-squares models fit to the same data, but adjusted $R^2$ can fall. Check the residual mean squares or the partial F statistic.
Using the wrong degrees of freedom. If the reduced model has $k$ predictors and an intercept, its residual degrees of freedom are $n-k-1$; after adding one predictor, they are $n-k-2$. The denominator of the partial F statistic uses the full model’s degrees of freedom.
Treating $F_{\mathrm{partial}}>1$ as a significance conclusion. It only establishes that adjusted $R^2$ increases. A hypothesis test compares the statistic with a critical value determined by the test level and degrees of freedom.
Ignoring the comparison conditions. The models must use the same response and observations and be nested. The algebra also requires $SST>0$, positive residual degrees of freedom, and—when using the finite partial F criterion—$SSE_F>0$.

The rule helps explain the trade-off encoded by adjusted $R^2$, but it is a model-comparison calculation, not by itself a guarantee that a selected model will predict well. Interpret it in the context of the modeling objective and the data used for fitting.

Check Your Understanding

For each nested-model comparison, use the full model’s residual degrees of freedom. Give adjusted-$R^2$ results to four decimal places where a numerical value is requested.

A reduced model with $k=2$ predictors is fit to $n=20$ observations and has $SSE_R=180$. The full model has $SSE_F=168$. Calculate the partial F statistic and state whether adjusted $R^2$ rises. (Answer: $\nu=16$, $F_{\mathrm{partial}}=1.1429$; adjusted $R^2$ increases.)
A reduced model with $k=4$ predictors is fit to $n=25$ observations; $SSE_R=250$ and $SSE_F=240$. Determine the change in adjusted $R^2$ using the partial F criterion. (Answer: $\nu=19$, $F_{\mathrm{partial}}=0.7917$; adjusted $R^2$ decreases.)
A reduced model with one predictor is fit to $n=16$ observations; $SSE_R=140$ and the full model has $SSE_F=130$. What is the partial F statistic, and what happens to adjusted $R^2$? (Answer: $\nu=13$, $F_{\mathrm{partial}}=1.0000$; adjusted $R^2$ is unchanged.)
A reduced model with $k=2$ predictors is fit to $n=40$ observations, with $SSE_R=600$, $SSE_F=540$, and $SST=1200$. Calculate the partial F statistic and both adjusted $R^2$ values. (Answer: $\nu=36$, $F_{\mathrm{partial}}=4.0000$; reduced $R^2_{\mathrm{adj}}=0.4730$, full $R^2_{\mathrm{adj}}=0.5125$, so adjusted $R^2$ increases.)
For a valid nested comparison, suppose $SST>0$, both residual degrees of freedom are positive, and $SSE_F=0$. Is the finite partial F statistic defined, and what can be concluded about adjusted $R^2$? (Answer: The finite partial F statistic is undefined; full adjusted $R^2=1$, so it is higher if $SSE_R>0$ and equal to 1 if $SSE_R=0$.)

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