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For a multilinear regression model, I'm trying to find the expected value of the sum of squares of regression (SSR). I have so far,

$$E(SSR) = E(\hat y'\hat y) = E((X\hat\beta)'(X\hat\beta)) = E((X(X'X)^{-1} X'y)'((X(X'X)^{-1} X'y)) = E((Hy)'(Hy)) =$$

And that's the extent of my matrix algebra skills :(

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2 Answers 2

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I rewrote your expression using formulae, and in the process, changed an inverse for a transpose in the middle, that you probably entered by mistake. Please check.

Rewrite your final expression as $$ E(y'H'Hy) $$ Next, note that $H$ is symmetric and idempotent, $H=H'$ and $HH=H$, so that we get $$E(y'Hy)$$ This is a scalar, so equal to its trace, which we may permute cyclically: $$tr[E(Hyy')]$$ To continue, we need assumptions. For simplicity, I take $X$ to be fixed (else, we would reason conditional on $X$, so that we can take out $H$ from the expectation:

$$E(SSR)= tr[HE(yy')]$$ Also, under classical assumptions in the linear model, we would have $$Var(y)=E(yy')=\sigma^2I,$$ so that $$E(SSR)= \sigma^2tr[H]$$ Now, $$ tr(H)=tr(X(X'X)^{-1}X')=tr((X'X)^{-1}X'X)=tr(I)=n-k $$

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I was able to answer this question:

$$E(SS_R) = E(y'[X(X'X)^{-1}X'-1(1'1)^{-1}1']y) = trace([X(X'X)^{-1}X'-1(1'1)^{-1}1']\sigma^2I = E(y)'[X(X'X)^{-1}X'-1(1'1)^{-1}1']E(y) = k\sigma^2 + \beta_R'X_C'X_C\beta_R$$

where $1$ is a $(n x 1)$ vector all of whose elements are 1's, $X_C$ is a matrix of centered regressors values and $\beta_R$ is a matrix of the non-intercept regressors.

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