I demand that...
- X must be in the interval (0,1).
- The estimate given is the expectation of the unknown distribution. EDIT: Kind of heavy as an assumption, I would suspect.
Would it be possible to somehow average over all distributions on (0,1) that have this expectation, in order to obtain a reasonable distribution for X?
I have made a Monte-Carlo simulation to demonstrate what I mean:
Input N #Means the number of intervals on x axis and also maximum value of y. Input E #Target expectation. Input M #Number of distributions to be drawn. epsilon=0.5/N #tolerance for average x=[(i-1/2)/N for i in 1..N] ans=N zeros sols=0 repeat M times: y=N randints in 0..N, normalized by float-dividing each int by the sum. avg=sum(x*y pointwise) if abs(avg-E)<epsilon: sols=sols+1 ans=ans+y pointwise ans=ans/sols pointwise Output ans #average of fit distributions
I have tested with N=10, M=100000, E=0.75. It takes time, but gives this: for x in [0.05, 0.15, 0.25, 0.35, 0.45, 0.55, 0.65, 0.75, 0.85, 0.95]
(808 distributions qualified) y: 0.0229, 0.0344, 0.0544, 0.0775, 0.1010, 0.1262, 0.1526, 0.1838, 0.2054
I have also tested it with E=0.5 and M=1000000, which seems to give something close to a uniform distribution: (899643 distributions qualified) y: 0.0995, 0.0999, 0.1002, 0.1002, 0.1003, 0.1003, 0.1001, 0.1002, 0.0998, 0.0996
What I am conceiving, is the limit of this procedure for E when N-->infinity and M-->infinity. (M must be very large when N is large or E is far from 0.5, because of the law of large numbers).