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My question is motivated by the switchpoint in texting frequency example at the end of the first chapter of "Probabilistic Programming and Bayesian Methods for Hackers". In that example, the rate of texting is modeled as a poisson distribution where the rate parameter undergoes a sudden change at a point in time. enter image description here

I was wondering if it would be possible to extend this modeling to linear regression. I have data that I expect to have different slopes in two different regimes. The boundary between those regimes would be the "switchpoint." How would one go about calculating the location of the switchpoint and the slopes before and after it for linear regression using probabilistic programming?

EDIT

I wanted to note that ideally there would be no gap before and after the kink. So while the slope would be discontinuous at the switchpoint,the line itself would be connected at the kink.

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    $\begingroup$ You might try searching for "changepoint" instead of "switchpoint", it's a more commonly used term. The answers to these two questions might help you get started: stats.stackexchange.com/questions/24810/… stats.stackexchange.com/questions/19772/…. I think there are several other good answers to similar questions on the site too! $\endgroup$ – jbowman Aug 19 '16 at 21:04
  • $\begingroup$ It is strange that the line plotting "expected number" of messages received is usually greater than any moving average would suggest. Are the blue and cyan features perhaps plotted on two different axes? $\endgroup$ – whuber Aug 19 '16 at 21:44
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Here's a solution to your problem using BUGS.

First, let's make some data (in R):

set.seed(121)
a1 <- 1
a2 <- 1.5
b1 <- 0
b2 <- -.15

n <- 101
changepoint <- 30

x  <- seq(0,1,len=n)
y1 <- a1 * x[1:changepoint]     + b1
y2 <- a2 * x[(changepoint+1):n] + b2
y  <- c(y1,y2) + rnorm(n,0,.05)

enter image description here

The changepoint is not obvious. Let's just plot where the split is:

enter image description here

The model states that there is one change point, $\tau$, that separates the two linear regimes. I'll assume that both regimes share the same variance (which might not be the case):

$$y_i \sim \mathcal{N}(a_1 x_i + b_1, \sigma^2 ),~ ~ i \leq \tau$$ $$y_i \sim \mathcal{N}(a_2 x_i + b_2, \sigma^2), ~ ~ i \gt \tau$$

$$a_i \sim \mathcal{N}(\alpha_a,\beta_a), ~ i=1,2$$ $$b_i \sim \mathcal{N}(\alpha_b,\beta_b), ~ i=1,2$$

$$\tau \sim \text{DiscreteUniform}(x_{init}, x_{end})$$

I will not define $\alpha, \beta$ as hyperparameters, but will just choose reasonable values looking at the available data.

This model can be written in Bugs like this:

model {
  tau ~ dcat(xs[])        # the changepoint

  a1 ~ dnorm(1,3)
  a2 ~ dnorm(1,3)
  b1 ~ dnorm(0,2)
  b2 ~ dnorm(0,2)

  for(i in 1:N) {
    xs[i]  <- 1/N    # all x_i have equal priori probability to be the changepoint

    # the normal's mean depends where the split is
    mu[i]  <- step(tau-i) * (a1*x[i] + b1) + step(i-tau-1) * (a2*x[i] + b2)

    # using the zero's trick
    phi[i] <- -log( 1/sqrt(2*pi*sigma2) * exp(-0.5*pow(y[i]-mu[i],2)/sigma2) ) + C

    dummy[i] <- 0
    dummy[i] ~ dpois( phi[i] )
  }

  sigma2 ~ dunif(0.001, 2)
  C  <- 100000
  pi <- 3.1416
}

I used the zero's trick to define the likelihood (the standard dnorm was giving me errors).

If you run the model on the previous data, these are the results after 100k iterations:

            mean         sd   MC_error  val2.5pc    median val97.5pc start sample
tau    33.830000 14.4600000 0.71290000 22.000000 29.000000 85.000000 10001 100000
a1      1.050000  0.1318000 0.00688100  0.826400  1.039000  1.345000 10001 100000
a2      1.457000  0.1227000 0.00677100  0.925400  1.483000  1.542000 10001 100000
b1     -0.003780  0.0210100 0.00108500 -0.060560 -0.002420  0.031020 10001 100000
b2     -0.122000  0.1123000 0.00623400 -0.187800 -0.146900  0.366500 10001 100000
sigma2  0.002117  0.0003647 0.00001084  0.001565  0.002065  0.002993 10001 100000

Notice that the true value of $\tau$ for my artificial data is at 30, ie, at the 30th data point. The model proposes the median at the 29th.

The next plot shows the median estimates against the true values (the estimated BUGS values correspond to the stronger lines):

enter image description here

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  • $\begingroup$ So I like this answer, but it still has the issue where it is not constrained to be continuous at the switchpoint. I think I am going to try to adapt this to pymc, with the change that $$y_i \sim \mathcal{N}(a_1 x_i + b_1, \sigma^2 ),~ ~ i \leq \tau$$ $$y_i \sim \mathcal{N}((a_1 + a_2) x_i + b_1, \sigma^2), ~ ~ i \gt \tau$$ This way its more expressed as a change just in the slope and hopefully that will lead to a continuous constraint. $\endgroup$ – Mir Henglin Sep 3 '16 at 21:55
  • $\begingroup$ The second one should not use $b_1$ but a term making them both meet at $x[\tau]$ $$y_i \sim \mathcal{N}((a_1 + a_2) x_i + (b_1 - a_2 x[\tau]), \sigma^2), ~ ~ i \gt \tau$$ $\endgroup$ – jpneto Sep 4 '16 at 10:08
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The strucchange package in R is there to do just this. Here is a minimal working example presuming you have already done install.packages('strucchange'):

library(`strucchange`)
set.seed(1)
x=1:250
y=c(0.01*x[1:100],1.5-0.02*(x[101:250]-101))
ynoise=y+rnorm(250,0,0.2)
ans=breakpoints(ynoise~x)

The ans object contains all the information on the fit. For example:

ans$breakpoints
gives the changepoint locations (100 in this example) using the BIC information criterion for the decision on number of changepoints.  If you want to use another criterion you can get the fitted values under `ans$RSS.table`.  You can get the fit using the standard `lm` function using

lm(ynoise~x+x*breakfactor(ans))

I've put x*breakfactor(ans) here as we have a break in both the intercept and trend, just putting breakfactor(ans) here would only put a break in the intercept.

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  • $\begingroup$ Hi I've been looking into this package. Initial results however, have produced regressions that are not connected at the break point. Ideally, I would like to produce a model where the breakpoints are 'kinks' in the line instead of gaps between segments. I haven't found a way to add that constraint to the breakpoint package, but the segmented package looks promising for that. $\endgroup$ – Mir Henglin Aug 23 '16 at 4:08
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    $\begingroup$ The segmented package is another good choice but you need care when working with data that has potentially multiple changepoints. $\endgroup$ – adunaic Aug 24 '16 at 9:44
  • $\begingroup$ If you are still looking for regressions that are connected at the breakpoint then you might want to email Paul Fearnhead as a PhD student of his has just finished his thesis on precisely this problem. He will likely share the code: maths.lancs.ac.uk/~fearnhea $\endgroup$ – adunaic Oct 4 '16 at 12:06

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