A bivariate regression is fitted to 20 sample observations on y and X. I know the following: $X'X=\begin{pmatrix} 20 & 10\\ 10 & 30 \\ \end{pmatrix}$, $X'y=\begin{pmatrix} 30\\ 40 \\ \end{pmatrix}$, $y'y=75$.

I received that $\beta=\begin{pmatrix} 1.4\\ 1.3\\ \end{pmatrix}$.

Then a new observation was obtained: $X=2$, $Y=4$. I should perform a Chow test of parameter constancy. In fact I know the formula for the test: $F=\frac{(RSS_{pooled}-RSS_{1}-RSS_{2})/(k+1)}{(RSS_{1}-RSS_{2})/(n-2k-2)}$. Unfortunately, I have no idea how to calculate the RSS for any of these regressions. Help is needed.

  • $\begingroup$ Do you know what RSS is? $\endgroup$
    – cc7768
    Commented Nov 4, 2015 at 17:32
  • $\begingroup$ @cc7768 yes. RSS=sum of (y-y_estimated)^2 $\endgroup$
    – Pichen'ka
    Commented Nov 4, 2015 at 17:54

1 Answer 1


This should help you figure out how to compute RSS for the different models.

Let's begin with what we have:

We know that

  • $y$ is an $n \times 1$ vector that has observations of the endogenous variable
  • $x$ is an $n \times 2$ matrix of observations of the exogenous variables
  • $\beta$ is a $2 \times 1$ matrix of coefficients

Unfortunately (or fortunately depending on how you look at it), we don't know what $x$ and $y$ look like in this equation. Instead we are given $(x' x)$, $(x' y)$, and $(y' y)$.

Using your comment about what RSS actually is -- Namely, that

$$\text{RSS} = (y_{\text{observed}} - y_{\text{estimated}})' (y_{\text{observed}} - y_{\text{estimated}})$$

First notice that $y_{\text{estimated}}$ is simply $x \beta$. Then replacing this in the previous formula gives us

\begin{align*} \text{RSS} &= (y_{\text{observed}} - y_{\text{estimated}})' (y_{\text{observed}} - y_{\text{estimated}}) \\ &= (y_{\text{observed}} - x \beta)' (y_{\text{observed}} - x \beta) \\ &= y'y - 2y'x\beta + x \beta \beta' x' \end{align*}

Now this is almost what we want. We want things to ultimately be in terms of $(y'y)$, $(x'x)$, $(x' y)$, and $\beta$. The first term is exactly $y'y$, but other terms look like they have pieces we want and we just have to find a way to get them.

Now, notice that each of the elements in our equation above is simply a scalar (aka sizes are such that they are $1 \times 1$). The determinant of a scalar is itself, then by using properties of determinants we can say:

\begin{align*} 2y'x \beta &= \det(2y'x\beta) = \det(2 \beta' x' y) \\ x \beta \beta' x' &= \det(x \beta \beta' x') = \det(\beta' x' x \beta) \end{align*}

I will leave the linear algebra itself as an exercise.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.