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Part CCSIR NET December 2024a-saturated-model-has-no-residual-degrees-of-freedom

A saturated model has no residual degrees of freedom

Consider a linear model Yi=β1+β2+⋯+βi+εi,1≤i≤nY_{i} = \beta_{1} + \beta_{2} + \cdots + \beta_{i} + \varepsilon_{i}, 1 \le i \le n, where errors εi′s\varepsilon_{i}'s are uncorrelated with zero mean and finite variance σ2>0\sigma^{2} > 0. Let β\betâi_{i} be the best linear unbiased estimator (BLUE) of βi,i=1,2,…,n\beta_{i}, i = 1, 2, \dots, n. Then, which of the following statements are true?

  1. A.The sum of squares residuals is strictly positive with probability 1.
  2. B.For every βi,1≤i≤n\beta_{i}, 1 \le i \le n, there are infinitely many linear unbiased estimators.
  3. C.Var(βVar(\betâ1)=σ2_{1}) = \sigma^{2}✓ Correct
  4. D.Y3−Y2Y_{3} - Y_{2} is the BLUE of β3\beta_{3}.✓ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests boundary and endpoint.

See pricing

50 are analysed free — try those first.

The trap it tests

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

Drill statements like this

Related counterexample: OLS is the BLUE in every linear model

More on this topic

The chapter behind this: Linear models, Gauss–Markov and ANOVA — free to read

From Linear Models and Multivariate › Gauss–Markov, regression, ANOVA basics

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