# What is the number of parameters when modeling a structural break?

Lets assume that we have a data set with a structural break at some point $x$. This can be tested using the Chow test.

To build a model that takes this break into account, I stimate two linear models for the two subintervals defined by the break point. Each model has $n$ coefficients. Now I want to calculate the AIC.

For this purpose, I need to calculate the total number of parameters of the "pooled" model. My first though was that the number of parameters is $2(n+1)$. However, I suspect that this is not true, because the separate models are estimated on a subset of the data. Am I right? What is the (effective) number of parameters of the "pooled" model?

• Strictly speaking the parameters estimated include the variance parameters. Since you did two regressions with different variance estimates, both of those would be included, in addition to both intercepts and both slopes – Glen_b Oct 6 '16 at 0:32
• I agree. However, as argued in my response one would obtain the same estimates in a single regression with interaction coding and the assumption of homoscedastic errors over the entire sample. In many applications this difference is practically not very relevant, though. But it may be... – Achim Zeileis Oct 6 '16 at 0:42

If the breakpoint is known/given in advance and not estimated from the data, then you have simply twice the number of regression coefficients as in the original model. It does not matter that some regression coefficients are just influenced by a subset of the observations - that is the same as for any other dummy variable or interaction.

Moreover, if the original regression model is given by $X \beta$ with

$$\left( \begin{array}{ccc} x_{1,1} & \dots & x_{1,k} \\ & \vdots & \\ x_{n,1} & \dots & x_{n,k} \end{array} \right) \left( \begin{array}{c} \beta_1 \\ \vdots \\ \beta_k \end{array} \right)$$

then the segmented model can also be written as a single regression in interaction coding:

$$\left( \begin{array}{cccccc} x_{1,1} & \dots & x_{1,k} & 0 & \dots & 0 \\ & \vdots & & & \vdots & \\ x_{n_A,1} & \dots & x_{n_A,k} &0 & \dots & 0 \\ 0 & \dots & 0 & x_{n_A + 1,1} & \dots & x_{n_A + 1,k} \\ & \vdots & & & \vdots & \\ 0 & \dots & 0 & x_{n_A + n_B,1} & \dots & x_{n_A + n_B,k} \end{array} \right) \left( \begin{array}{c} \beta^A_1 \\ \vdots \\ \beta^A_k \\ \beta^B_1 \\ \vdots \\ \beta^B_k \end{array} \right)$$

So there are simply twice as many parameters ($2 \cdot k$ in my notation) as in the original model. A subtle issue is whether you assume the errors to be homoscedastic across the entire sample or just within each segment.

Things are different, of course, if the breakpoint is not known in advance and has to be estimated from the data.

• Thanks. So, if the error variance is not the same for both segments, we have to include the error variance as an additional parameter [ 2(k+1) in your notation]? This is of particular interest, if I am interested in comparing a nested model with a non-nested one. – Julian Dec 28 '16 at 10:27
• Yes, it is either 2 * k + 1 in the case of constant variance and 2 * (k + 1) = 2 * k + 2 if the variance changes across the segments. – Achim Zeileis Dec 28 '16 at 17:41
• Thank you very much. Please allow me a follow up question: Does it follow that in order to determine the number of parameters of an OLS models with segmentwise interaction coding I have to test if the error variance for each segment is the same? This could be problematic if I deal with a lot of segments. – Julian Jan 2 '17 at 8:00
• I wouldn't conduct a formal test but simply use the OLS estimate which is consistent under heteroscedasticity. Inference may have to be adjusted by suitable covariance estimators, though. – Achim Zeileis Jan 2 '17 at 17:56