3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 '14 at 14:52
4
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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 '14 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 '14 at 13:15

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.