Least Squares Models: SSR Comparison

We are comparing the Sum of Squared Residuals (SSR) for two models fitted using the least squares method:

• Model 1 (Linear): $y_i = \beta_0 + \beta_1 x_i$. The SSR is denoted by $A = \sum_1^n \left(y_i - (\hat{\beta}_0 + \hat{\beta}_1 x_i)\right)^2$.
• Model 2 (Quadratic): $y_i = \beta_0^* + \beta_1^* x_i + \beta_2^* x_i^2$. The SSR is denoted by $B = \sum_1^n \left(y_i - (\hat{\beta}_0^* + \hat{\beta}_1^* x_i + \hat{\beta}_2^* x_i^2)\right)^2$.

Analysis of $A \ge B$

Model 2 is a more flexible model than Model 1 because it includes an additional predictor term ($x_i^2$). The least squares method aims to minimize the SSR. Adding terms to a regression model can only decrease or maintain the SSR; it cannot increase it. This is because the parameter estimates for the more complex model (Model 2) have more freedom to fit the data. Therefore, the SSR of the simpler model ($A$) must always be greater than or equal to the SSR of the more complex model ($B$).

Conclusion: The statement $A \ge B$ is always true.

Analysis of 'It can happen that $B = 0$ but $A > 0$'

Condition for $B = 0$: $B=0$ means Model 2 fits the data perfectly. This occurs if the observed $y_i$ values lie exactly on the quadratic curve defined by $\hat{\beta}_0^* + \hat{\beta}_1^* x_i + \hat{\beta}_2^* x_i^2$. This is possible if the true relationship is quadratic or linear.

Condition for $A > 0$: $A>0$ means Model 1 (the linear model) does not fit the data perfectly. This happens if the data points do not lie exactly on a straight line.

Scenario: Consider data points that follow a true quadratic relationship, such as $(1,1), (2,4), (3,9)$, where $y_i = x_i^2$. Model 2 can capture this relationship perfectly with $\hat{\beta}_0^*=0, \hat{\beta}_1^*=0, \hat{\beta}_2^*=1$, yielding $B=0$. However, Model 1 (linear fit) cannot perfectly represent this quadratic data, so its SSR, $A$, will be greater than 0.

Conclusion: The statement 'It can happen that $B = 0$ but $A > 0$' is true.

Analysis of 'It can happen that $A = 0$ but $B > 0$'

$A = 0$ implies that Model 1 fits the data perfectly, meaning all data points lie exactly on a straight line ($y_i = \hat{\beta}_0 + \hat{\beta}_1 x_i$). If the data perfectly follows a linear relationship, Model 2, which includes linear terms, can also fit the data perfectly by setting $\hat{\beta}_2^* = 0$. Therefore, if $A=0$, then $B$ must also be $0$. It is impossible for $A=0$ while $B>0$.

Conclusion: The statement 'It can happen that $A = 0$ but $B > 0$' is false.

Final Answer Justification

Based on the analysis, the two correct statements are:

• $A \ge B$ (due to model complexity and SSR properties).
• It can happen that $B = 0$ but $A > 0$ (when the true relationship is quadratic and Model 2 fits perfectly, while Model 1 does not).