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I have some analysis that compares relative velocity shifts between regions. I have a representative sample below (in reality there are more regions -- this is for clarity).

Assume that I have a dataset with 4 regions -- and the possibility of a relative velocity shift between all four. The only method that I have for determining this relative shift involves a process that forces one of the regions to have a shift of "zero" -- and the other values to shift relative to that.

This means that the y-intercept is arbitrary, but the relative y-values between data points is still important. The plot below shows an example of what I'm doing:

two datasets

This plot shows the result of finding the relative shifts by alternately setting the first region to zero (red); and the last region to zero (blue). As you can tell, the slope is the same between the two measurements. I also have an error estimate for each point not set to be zero.

I want to fit for the slope across all of the regions.

My question is: is there a statistically robust way to combine this into one fit? Or to fit two linear regressions with independent y-intercepts but tied slopes? Is there another way to think about this entirely?

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  • $\begingroup$ For other ways to think about the problem (especially versions with more points in the scatterplot) please see the varied answers to a generalization of this situation at stats.stackexchange.com/questions/33078. (In the generalization the information about the colors is lost.) $\endgroup$
    – whuber ♦
    May 16, 2014 at 14:52

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This seems like a pretty straightforward application of dummy variables. All you need to do is include an indicator variable that takes the value 1 for red and value 0 for not-red (or vice versa. The coding scheme is arbitrary). As a result, the regression will fit two intercepts, one for red and one for not-red, but the regression will have a common slope for both groups.

So your model will look like this $$\hat{y}=\beta_0 +\beta_{red}d+\beta_{slope}x$$ where $d$ is our indicatory d ummy variable taking 1 for red and 0 for not-red, and $x$ is your independent variable. Because we've coded reds as binary indicators, the intercept for the red group is $\beta_0+\beta_{red}*1=\beta_0+\beta_{red}$, while it's simply $\beta_0+\beta_{red}*0=\beta_0$ for the not-red group.

To extend this to the general case of $k$ different groups, you will need to create $k-1$ dummy variables which are all 0 except for the group membership, which is 1. (We create $k-1$ groups because $k$ total groups would create linearly dependent columns with the column of 1's for the intercept.)

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  • $\begingroup$ As an aside, if you are using R then you can force the software not to fit an intercept at all. It will still fit the correct number of variables, but each $\beta$ will reflect just the intercept for that dummy coded group (also done automatically by default for factors). For example this fit: summary(lm(decrease~0+treatment+rowpos,data=OrchardSprays)) produces a separate intercept for each treatment, but a common slope for rowpos. $\endgroup$
    – russellpierce
    May 16, 2014 at 13:14
  • $\begingroup$ And, for completeness, summary(lm(decrease~-1+treatment+rowpos,data=OrchardSprays)) does the same thing with a slightly different command. $\endgroup$
    – Sycorax ♦
    May 16, 2014 at 13:15

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