Having consulted a number of sources, I still can't find a complete proof that Regression Sum of Squares ($SS_{regression}$) and ($SS_{residual}$) are independent random variables. I'll be doubly pleased with a proof in matrix form. If it is too involved to be typed up here, I am happy to check it out myself if people can direct me to a book that contains the proof.

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    $\begingroup$ stats.stackexchange.com/questions/117406/… should be helpful $\endgroup$ Commented Feb 19, 2021 at 6:49
  • $\begingroup$ I see. Is the idea that $SS_{regression}$ is a function of $\hat{y}$ and $SS_{residual}$ a function of $\hat{\epsilon}$ and thus conclude $SS_{residual}$ is independent of $SS_{regression}$ once we can show that $\hat{y}$ and $\hat{\epsilon}$ are uncorrelated, given the assumption of marginally normal errors, with constant variance? $\endgroup$ Commented Feb 19, 2021 at 13:45

1 Answer 1


Assume $y \sim \operatorname{Normal}(\beta X, \Sigma)$ with constant diagonal covariance $\operatorname{Cov}(y)=\Sigma=\sigma^2 \mathbb I$ and mean $\bar y=\mu$. Using the hat matrix $\mathbb H$ we have that:

$$\hat y =\mathbb H y$$


$$\epsilon=y-\hat y=(\mathbb I -\mathbb H)y$$

Then (because the hat matrix is idempotent $\mathbb H^2 = \mathbb H$) $$\begin{cases} SSR = (\hat y - \mu)^T(\hat y - \mu)= y^T \mathbb Hy \color{red}{+n\mu^2-2\mu\mathbf 1^T\hat y}= y^T \mathbb Hy \color{red}{-n\mu^2}\\ SSE = \epsilon ^T \epsilon = y^T (\mathbb I -\mathbb H)y \end{cases}$$

$$\operatorname{Cov}(SSR,SSE) = \operatorname{Cov}(y^T \mathbb Hy,y^T (\mathbb I -\mathbb H)y)$$

Using that $\operatorname{Cov}(x^TAx,x^TBx)=4\mu^TA\Sigma B\mu + 2\operatorname{tr}(A\Sigma B\Sigma)$ (see Prove that $\mathrm{Cov}(x^TAx,x^TBx) = 2 \mathrm{Tr}(A \Sigma B \Sigma) + 4 \mu^TA \Sigma B \mu$)

$$\operatorname{Cov}(SSR,SSE) =4\mu^T\mathbb H\Sigma (\mathbb I -\mathbb H)\mu + 2\operatorname{tr}(\mathbb H\Sigma (\mathbb I -\mathbb H)\Sigma)\\ $$

Since $\Sigma=\sigma^2 \mathbb I$ is constant diagonal:

$$\operatorname{Cov}(SSR,SSE) =4\sigma^2\mu^T\mathbb H (\mathbb I -\mathbb H)\mu + 2\sigma^4\operatorname{tr}(\mathbb H (\mathbb I -\mathbb H))\\ =4\sigma^2\mu^T (\mathbb H -\mathbb H^2)\mu +2\sigma^4\operatorname{tr}(\mathbb H -\mathbb H^2)\\ =0$$

  • $\begingroup$ Very elegant proof. I like it but I am not clear on something and not convinced for the case when x is non-random. As I see it, key to this proof is the assumption that $y$~$N(0,\sum)$. Given that the dependent variable in our sample is what it is and its population average is unknown, what does it mean to assume that the random vector y is multivariate normally distributed with mean 0 and constant variance. Don't we make assumptions only about the distribution of the error term and derive the distribution of the random vector y? $\endgroup$ Commented Feb 19, 2021 at 16:21
  • $\begingroup$ When x is non-random, the random vector y will, by design, not have mean 0. In fact the $y_i$ will not all have the same mean. Would you disagree? $\endgroup$ Commented Feb 19, 2021 at 16:21
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    $\begingroup$ Hi, nice post. Just a quick comment. That $Hy$ and $(I-H)y$ are uncorrelated follows immediately since $H$ and $I-H$ are orthogonal. Due to joint normality, we know $Hy$ and $(I-H)y$ are independent. After that, we know that any functions of those are independent, including their sums of squares, giving the result. $\endgroup$
    – user257566
    Commented Apr 9, 2021 at 21:25
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    $\begingroup$ I see, thanks for explaining. FWIW, your post didn't prove that the sums of squares were independent, just that they were uncorrelated. Am I misreading? $\endgroup$
    – user257566
    Commented Apr 9, 2021 at 21:34
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    $\begingroup$ @user257566 No, you got it right. Now that you mentioned it, I had it in my mind that uncorrelated Chi-square variables are independent, though I'm not so sure now. I'll have to check though their relationship to squared Normals probably helps. Perhaps it brings back to your initial assessment. $\endgroup$
    – Firebug
    Commented Apr 10, 2021 at 1:28

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