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

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

  1. The mean score is 0.5.
  2. The median score is 0.5.
  3. The root-mean-square "noise" in the mean score, is   (12N)-1/2.
  4. 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-10<x<1 (x-0.5)2dx = 1/(12N).

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

0<m<1mH(1-m)Hmdm /0<m<1mH(1-m)Hdm = 1/2

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

0<m<1mH(1-m)H(m-1/2)2dm /0<m<1mH(1-m)Hdm = (1/4) / (3+2H).

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

⟨Noise2⟩ = 1/(8+4N).

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