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I've been given a problem asking that I show that, in a normal error regression model, $$E(SSR)=σ^2+β'X'(I_n-(1/n) J_n)Xβ$$ I am new to applying matrix algebra to statistics, but I do know that the quadratic form of SSR is $$SSR=Y'[H-(1/n)J]Y$$ And after a few videos (and reading cross-validated) I understand why $$E(Y'AY)=tr(AΣ)+(μ'Aμ).$$ My attempted work so far (please correct me if I'm wrong) is that $μ=E(Y)=Xβ$, so my understanding is that I can rewrite $μ'=Y'=β'X'$ and $μ=Y=Xβ$, which gets me to $$E(Y'AY)=trace([H-(1/n)J]Σ)+β'X'[H-(1/n)J]Xβ$$ Just based on the pattern, I could hazard a guess that $trace([H-(1/n)J]Σ)=σ^2$, but I can't figure out why. Moreover, I don't see how I can make $β'X'[H-(1/n)J]Xβ$ become $β'X'(I_n-(1/n) J_n)Xβ$. Perhaps all of this is trivial, but I'm new to it and it was not really explained prior to the question being asked.

EDIT: I've presented this exactly as it was presented to me. I believe $J_n$ refers to an $n$ x $n$ matrix filled with 1's. I would assume that $n$ is the same for all $J$ and $H$ mentioned, but there was no additional detail provided.

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  • $\begingroup$ You should define $J$, $J_n$, $H$ $\endgroup$ – Christoph Hanck Oct 12 at 10:12
  • $\begingroup$ I presented the required solution as it was given to me. I believe $J$ and $J_N$ refer to a matrix of 1's and an $n$x$n$ matrix of 1's, respectively. I would assume that $n$ is the same for the entire problem, but it was not specified. $\endgroup$ – AJV Oct 12 at 12:20
  • $\begingroup$ I find it difficult to start here. $H$ might, in common notation, be the hat matrix $X(X'X)^{-1}X'$, but then a common way to write $SSR$ is $Y'(I_H)Y$, which makes me wonder how the above result comes about. So we will need more context, I believe $\endgroup$ – Christoph Hanck Oct 13 at 5:02

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