> What makes a leverage point a leverage point is that it has high leverage. > > Leverage is the ability of the point to "pull the line" toward itself. > > A sufficiently high leverage point may therefore have a *small* residual. > An M-estimator that has nice properties in location estimation will not > solve the problem if the leverage was high enough that the original > residual was small. > > consider the data > > x y > 1 112 > 2 111 > 3 113 > 4 125 > 5 124 > 6 135 > 105 1 > > The point with x at 105 has very high leverage. > > The residual for the high leverage point from a linear regression through > these points is approximately -1, while the smallest residual (ignoring > sign) among the other points is about 6. So the high leverage point will > not get downweighted - it has the residual closest to zero.
Thank you Glen, you have cleared that up for me! I read a very similar explanation in Rousseeuw's book, but somehow it didn't click - I had convinced myself the answer was in the influence function. While we are on the topic, could you please answer one further question for me? Why do leverage points have a stronger effect than outliers?? 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?? Thanks again for taking the time to help! Julian . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
