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Part CCSIR NET December 2025centrality-of-the-chi-square-and-f-distributions-needs-the-null-but-independence-of-the-two-sums-of-squares-never-does

Centrality of the chi square and f distributions needs the null but independence of the two sums of squares never does

Let Y=Xβ+εY=X\beta+\varepsilon be a multiple linear regression model with p regressors and an intercept, where β=(β0,β1,,βp)T\beta=(\beta_{0},\beta_{1},\dots,\beta_{p})^{T} and the random error ε\varepsilon~Nn(0,σ2I),σ>0N_{n}(0,\sigma^{2}I), \sigma>0 and n>p+1. The least squares method provides a unique estimator β\betâ. Let the total sum of squares (corrected), sum of squares due to regression, and sum of squares due to error, based on β\betâ, be denoted YᵀAY, YᵀBY and YᵀCY respectively, so YᵀAY=YᵀBY+YᵀCY. Which of the following statements are always true?

  1. A.YᵀAY/σ2/\sigma^{2} follows a central χ2\chi^{2} distribution with (n−1) degrees of freedom.
  2. B.YᵀBY/σ2/\sigma^{2} follows a central χ2\chi^{2} distribution with p degrees of freedom if β1=β2==βp=0\beta_{1}=\beta_{2}=\cdots=\beta_{p}=0.
  3. C.YᵀBY/YᵀCY follows a central F distribution with (p,n−p) degrees of freedom.
  4. D.YᵀBY and YᵀCY are independently distributed if and only if β1=β2==βp=0\beta_{1}=\beta_{2}=\cdots=\beta_{p}=0.

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.

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50 are analysed free — try those first.

Related counterexample: OLS is the BLUE in every linear model

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

From Linear Models and MultivariateGauss–Markov, regression, ANOVA basics

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