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 rootmeansquare "noise" in the mean score, is (12N)^{1/2}.

The rootmeansquare "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).
Answers
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} (x0.5)^{2}dx = 1/(12N).
d.
The probability density the median lies at m is proportional to
m^{H}(1m)^{H} if N=2H+1.
The mean location of the median is then
∫_{0<m<1}m^{H}(1m)^{H}mdm
/
∫_{0<m<1}m^{H}(1m)^{H}dm
= 1/2
confirming answer (b).
The mean squared noise in the median is
∫_{0<m<1}m^{H}(1m)^{H}(m1/2)^{2}dm
/
∫_{0<m<1}m^{H}(1m)^{H}dm
= (1/4) / (3+2H).
[All the integrals are Euler betafunction integrals.]
This is
⟨Noise^{2}⟩
= 1/(8+4N).
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