RangeVoting.org - Answer to puzzle ?? - Median and Average and Noise in Both

Answer to puzzle ?? - Median and Average and Noise in Both?

Puzzle
Given N scorers who each randomly independently choose a score from the uniform distribution on the real interval [0,1], show:

The mean score is 0.5.
The median score is 0.5.
The root-mean-square "noise" in the mean score, is (12N)^-1/2.
The root-mean-square "noise" in the median score, is (8+4N)^-1/2 if N is odd. In other words, median is noisier than mean for each N>1 (both are the same if N=1).

a. N∫_0<x<1 x = N/2, then divide by N.

b. Reflection symmetry shows median has to be at 0.5.

c. N^-1 ∫_0<x<1 (x-0.5)²dx = 1/(12N).

d. The probability density the median lies at m is proportional to m^H(1-m)^H if N=2H+1. The mean location of the median is then

∫_0<m<1m^H(1-m)^Hmdm / ∫_0<m<1m^H(1-m)^Hdm = 1/2

confirming answer (b). The mean squared noise in the median is

∫_0<m<1m^H(1-m)^H(m-1/2)²dm / ∫_0<m<1m^H(1-m)^Hdm = (1/4) / (3+2H).

[All the integrals are Euler beta-function integrals.] This is

⟨Noise²⟩ = 1/(8+4N).