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.
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.