[EMAIL PROTECTED] (Kload) wrote in message news:<[EMAIL PROTECTED]>... > Why do leverage points have a stronger effect than outliers??
A high-leverage point doesn't necessarily have a strong effect on the line. If the high-leverage point lies on the line you'd get with that point absent, the high-leverage point has no effect on the estimate (though it has a big effect on the standard error of the estimate of the line). > My current suspicion is that in residuals r=y-mx-c the leverage points > (x) are multipled by m hence if m>1, leverage points will have a > stronger effect than outliers. Certainly all the example data I've > seen (yours, Rouseeuw's) has m>1. Am I on the right track?? No. You can get the effect with m<<1. Leverage is a measure of the rate of change of a fitted value with changes in the observation (how sensitive is the fit to this point to a small change in the value?). This is in fact utterly unrelated to m, and is a function of the arrangement of the x's only. If you're familiar with matrix algebra, it isn't hard to show that the vector of these leverages is simply the diagonal of the hat-matrix from regression. If you're not familiar with matrix algebra, you won't know what the hat-matrix is. Glen . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
