# Chow test of parameter constancy

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.

• Do you know what RSS is? – cc7768 Nov 4 '15 at 17:32
• @cc7768 yes. RSS=sum of (y-y_estimated)^2 – Pichen'ka Nov 4 '15 at 17:54

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 $(y' y)$.

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

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.