# Chow test

The Chow test is a statistical and econometric test of whether the coefficients in two linear regressions on different data sets are equal. The Chow test was invented by economist Gregory Chow in 1960. In econometrics, the Chow test is most commonly used in time series analysis to test for the presence of a structural break. In program evaluation, the Chow test is often used to determine whether the independent variables have different impacts on different subgroups of the population.

structural break program evaluation

At ${\displaystyle x=1.7}$ there is a structural break, regression on the subintervals ${\displaystyle [0,1.7]}$ and ${\displaystyle [1.7,4]}$ delivers a better modelling than the combined regression(dashed) over the whole interval.

Comparison of 2 different programs (red, green) existing in a common data set, separate regressions for both programs deliver a better modelling than a combined regression (black).

Suppose that we model our data as

${\displaystyle y_{t}=a+bx_{1t}+cx_{2t}+\varepsilon .\,}$

If we split our data into two groups, then we have

${\displaystyle y_{t}=a_{1}+b_{1}x_{1t}+c_{1}x_{2t}+\varepsilon .\,}$

and

${\displaystyle y_{t}=a_{2}+b_{2}x_{1t}+c_{2}x_{2t}+\varepsilon .\,}$

The null hypothesis of the Chow test asserts that ${\displaystyle a_{1}=a_{2}}$, ${\displaystyle b_{1}=b_{2}}$, and ${\displaystyle c_{1}=c_{2}}$, and there is the assumption that the model errors ${\displaystyle \varepsilon }$ are independent and identically distributed from a normal distribution with unknown variance.

Let ${\displaystyle S_{C}}$ be the sum of squared residuals from the combined data, ${\displaystyle S_{1}}$ be the sum of squared residuals from the first group, and ${\displaystyle S_{2}}$ be the sum of squared residuals from the second group. ${\displaystyle N_{1}}$ and ${\displaystyle N_{2}}$ are the number of observations in each group and ${\displaystyle k}$ is the total number of parameters (in this case, 3). Then the Chow test statistic is

${\displaystyle {\frac {(S_{C}-(S_{1}+S_{2}))/(k)}{(S_{1}+S_{2})/(N_{1}+N_{2}-2k)}}.}$

## References

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