In the particular case I gave, of course frequentists could produce an argument that the mean must be in the given range. But this could not be a statistical argument, it would have to be a deductive logical argument.
The frequentists do have an out here: conditional inference. Obviously, (v2+v1)/2 is sufficient for m, so they don’t need any other information for their inference. But it might occur to them to condition on the ancillary statistic v2-v1. In repeated trials where v2-v1 = 0.9, the interval (v1,v2) always contains m.
Edit: As pragmatist mentions below, this is wrong wrong wrong. The minimal sufficient statistic is (v1,v2), although it is true that v2-v1 is ancillary and moreover it is the ancillary complement to the sample mean. That I was working with order statistics (and the uniform distribution!) is a sign that I shouldn’t just grope for the sample mean and say good enough.
True, but is there any motivation for the frequentist to condition on the ancillary statistic, besides relying on Bayesian intuitions? My understanding is that the usual mathematical motivation for conditioning on the ancillary statistic is that there is no sufficient statistic of the same dimension as the parameter. That isn’t true in this case.
ETA: Wait, that isn’t right… I made the same assumption you did, that the sample mean is obviously sufficient for m in this example. But that isn’t true! I’m pretty sure in this case the minimal sufficient statistic is actually two-dimensional, so according to what I wrote above, there is a mathematical motivation to condition on the observed value of the ancillary statistic. So I guess the frequentist does have an out in this case.
The frequentists do have an out here: conditional inference. Obviously, (v2+v1)/2 is sufficient for m, so they don’t need any other information for their inference. But it might occur to them to condition on the ancillary statistic v2-v1. In repeated trials where v2-v1 = 0.9, the interval (v1,v2) always contains m.
Edit: As pragmatist mentions below, this is wrong wrong wrong. The minimal sufficient statistic is (v1,v2), although it is true that v2-v1 is ancillary and moreover it is the ancillary complement to the sample mean. That I was working with order statistics (and the uniform distribution!) is a sign that I shouldn’t just grope for the sample mean and say good enough.
True, but is there any motivation for the frequentist to condition on the ancillary statistic, besides relying on Bayesian intuitions? My understanding is that the usual mathematical motivation for conditioning on the ancillary statistic is that there is no sufficient statistic of the same dimension as the parameter. That isn’t true in this case.
ETA: Wait, that isn’t right… I made the same assumption you did, that the sample mean is obviously sufficient for m in this example. But that isn’t true! I’m pretty sure in this case the minimal sufficient statistic is actually two-dimensional, so according to what I wrote above, there is a mathematical motivation to condition on the observed value of the ancillary statistic. So I guess the frequentist does have an out in this case.