so I am trying to see whether there is a breakpoint in my data set at a certain time. Basically I have 200 data points and am wondering if there is a breakpoint at t=100, ie whether the slopes of the two regression lines change significantly. Le the slope of line 1= b1 and the slope of line 2= b2.

I want to see whether $b1-b2=0$ using a t test, and so I think that would reduce to evaluating the statistic $t= (b1-b2)/($std error$(b1-b2))$ The problem I have is that I don't know how to compute that standard error since b1 and b2 might be correlated. Furthermore I am not sure what degrees of freedom the corresponding t distribution would have? Finally, how do I even see that the statistic I have written down does follow a t distribution in the first place?

Thanks in advance for anyone who can help me with any of these questions!


You are comparing two models: the first fits a common slope, the second fits two slopes. So fit them both and then compare the fit of the two models. If the second has an improvement in fit you needed two slopes. You do not say how you are handling the junction between the two parts of the dataset, whether you are constraining the two lines to meet there but if you are also allowing two separate intercepts then the procedure I outline should work just as well. If you want to read further then in some fields these are known as broken stick models which may help you. Note also that you are assuming you know the breakpoint, if you estimate it things get more complicated.

We assume we know the breakpoint $c$. We define two basis functions. \begin{equation} B_l(x) = \begin{cases} c - x & \text{if } x < c \\ 0 & \text{otherwise} \end{cases} \end{equation} and \begin{equation} B_r(x) = \begin{cases} x - c& \text{if } x > c \\ 0 & \text{otherwise} \end{cases} \end{equation}

We now fit \begin{equation} y = \beta_0 + \beta_1B_l(x) + \beta_2 B_r(x) + \epsilon \end{equation} using standard methods.

Note that the coefficients are for the new variables here and the way they are coded means that a negative value shown for $B_l(x)$ corresponds to a positive slope.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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