Just to clarify this, for the nth "permeantile", I think you'd weight each
point
on the n side (100-n)^2 and on the (100-n) side it would be n^2 .
________________________________
From: Toby Pereira <[email protected]>
To: electorama list <[email protected]>
Sent: Sat, 9 July, 2011 0:19:30
Subject: [EM] Median-based Proportional Representation
*I was thinking about how you would calculate permeantiles. In a uniform
distribution between 0 and 1, the 25th permeantile would be 0.25. If you weight
the averages of each side 3 to 1 in favour of the smaller side of the
permeantile (0 to 0.25), and average these, then you get 0.25. (3*0.125 +
1*0.625) / 4 = 0.25. So for the 10th permeantile, you have (9*0.05 + 1*0.55) /
10 = 0.1 and so on. I imagine this would work for non-uniform distributions
too.
(Sorry for going off topic.)
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