From: ben.usenet@bsb.me.uk   
      
   Ben Bacarisse  writes:   
      
   > Tim Rentsch  writes:   
      
   A couple of further thoughts only one of which is directed at you   
   Tim (at the end)...   
      
   Given the general conception of a mean (rather than any other kind of   
   summary statistic) as minimising the sum of squares of some "distance"   
   metric:   
      
     Sum_{i=1,n} difference(A, t(i))^2   
      
   we can characterise the two contenders by what distance is being   
   minimised.  For the less well discussed "conventional average" (to use   
   your terminology) we are minimising the sum of squares of the shorter   
   arc lengths between A and the t(i).   
      
   For the "vector average", we convert the t(i) to unit vectors u(i) and   
   we calculate the mean if the u(i) to get a vector m.  The "average", A,   
   is just the direction of this vector -- another point on the unit   
   circle.  In this case we are minimising the sum of squares of the   
   /chord/ lengths between A and the t(i).   
      
   The mean vector itself (which may not lie on the unit circle) minimises   
   the sum of the squares of the length of the vector differences m-u(i):   
      
     Sum_{i=1,n} |m - u(i)|^2   
      
   and any other vector along the same line as m (i.e. c*m for real c)   
   minimises   
      
     Sum_{i=1,n} |c*m - u(i)|^2   
      
   This includes, of course, m projected out to the unit circle.   
      
   This distinction between arc lengths and chord lengths helps to   
   visualise where these averages differ, and why the conventional average   
   may seem more intuitive.   
      
   Incidentally, I found another book on statistics on spheres, and that   
   gets the average over in a few paragraphs.  It states, without   
   considering any alternatives, that the average is the vector average.  I   
   can't find anyone using or citing the arc-length minimising average,   
   despite it being natural on a sphere to find the mid-point of, say, a   
   set of points on the Earth's surface.   
      
   Tim, my best shot at calculating this other average sorts the points to   
   find the widest gap.  I suspect your algorithm is similar since you say   
   it is O(n log n).   
      
   --   
   Ben.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)