John K. answered 08/12/15
Tutor
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Math and Engineering Tutor, Professional Engineer
I some cases yes. They lie in [0, 1]. Consider, as an example, the k-nearest neighbour smoother, which is the average of the k nearest measured values to the given point. Then, at each of the n measured points, the weight of the original value on the linear combination that makes up the predicted value is just 1/k. Thus, the trace of the hat matrix is n/k. Thus the smooth effort costs n/k effective degrees of freedom. The subject requires some research that is more than just one answer.