I would like to check my understanding of the Chow Forecast test.The test is used to perform a test on the following models: $y_1=\beta X+\epsilon_1$ and $y_2=\beta X+ \nu + \epsilon_2$, with $H_0:\nu=0$ and $H_1: \nu\neq 0$.

Now I wonder whether that means that the following holds true: \begin{align*} F&=\frac{(e_R'e_R-e_1'e_1)/g}{e_1'e_1/(n-k)}\\ &=\frac{\left(\sum\limits_{i=1}^{n+g} (y_i-x_i'\beta)^2-\sum\limits_{i=1}^n (y_i-x_i'\beta)^2 \right)/g}{\sum\limits_{i=1}^n (y_i-x_i'\beta)/(n-k)}\\ &=\frac{\sum\limits_{i=n}^{n+g} (y_i-x_i'\beta)^2/g}{\sum\limits_{i=1}^n (y_i-x_i'\beta)/(n-k)} \end{align*}

Are these true, and if not, why not?

  • $\begingroup$ Never mind. Question can be deleted. $\endgroup$
    – abc
    Feb 2, 2014 at 14:29
  • $\begingroup$ If you know the answer, please post it here; you can help other people that may stumble upon a similar problem. And earn some reputation. $\endgroup$
    – user88
    Feb 2, 2014 at 15:26

1 Answer 1


If the $\beta$ in both regressions is the same then the numerator should say $\frac{\displaystyle\sum_{i=n+1}^{n+g}(y_i-x_i\beta)^2}{g}$ but I don't think they are. So they don't cancel out when you subtract one from the other.


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.