I have a linear regression problem of the form z ~ x + y + b where z, y and x are continuous variables and b is categorical with two levels A and B. The information if a sample is b=A or b=B got lost for all samples. However I've always got pairs of data from which I know that one of them is A, the other one B (in total 27 pairs). I just don't know which is which.

I want to determine if b is a significant factor to the model and compare the strength of its effect to x and y


So far I've been thinking about a Monte Carlo simulation randomly assigning the classes and then doing the regression analysis and evaluating the p-value and the effect strength. I will end up with a distribution of these variables. Will their median then be the most probable p-value and effect strength? I'm not sure about this. Are there better strategies?

  • $\begingroup$ One note is that this regression would usually be written: $Y = b_0 + b_1x_1 + b_2x_2 + b_3x_3$ rather than as you have it. $\endgroup$
    – Peter Flom
    Feb 20, 2018 at 12:08

1 Answer 1


I'm afraid you are out of luck. In each pair, the observation with the higher z could have b=A and the one with the lower z have b=B, and b would have a significant impact. Or the other way around, and you would again have a significant impact of b - but with the opposite sign! Or, thirdly, z and b could have zero relationship, and within each pair, the values of b could simply be random permutations of A and B. All three possibilities are completely consistent with your setup.

Unless, that is, b is related in some way to x and y. But to assess that, you would again need labeled samples for b. And if you already know that such a relationship exists, then you don't need your missing value dataset.

  • $\begingroup$ I just realise that I didn't point out that each pair has same x and y, they only differ in b. It also doesn't matter in which direction the effect of b points, only if it's there and how strong it is. Does this change things? $\endgroup$
    – ascripter
    Feb 20, 2018 at 11:14
  • 1
    $\begingroup$ No, it doesn't. Note how I didn't refer to x and y in my first paragraph. That's not because I forgot, that's because they are irrelevant (except for a potential relationship between them and b, per my second paragraph). Any arrangement is a priori equally likely, unless you have grounds to assume one is more likely than another. In short: if you don't know anything, then you can't draw any conclusions. $\endgroup$ Feb 20, 2018 at 11:42

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