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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).

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  • $\begingroup$ There are a lot of distributions with mass on $(0,1)$ and a given mean: beta, triangular, uniform on subsets of $(0,1)$ symmetric around your mean... If you can assign a probability to each of these distributions (e.g., using a particular lognormal distibution for one parameter of the beta - the other one can then be calculated using the predefined mean), then you could in principle simulate your mixture. Otherwise, "averaging over all distributions" makes little sense. $\endgroup$ – Stephan Kolassa Oct 1 '14 at 10:37
  • $\begingroup$ Hm. I mean all conceivable functions on (0,1) that have area=1. $\endgroup$ – Elias Hasle Oct 1 '14 at 11:27
  • $\begingroup$ Since there are an infinite number of these functions, how are you going to assign prior probabilities to them? $\endgroup$ – Glen_b Oct 1 '14 at 11:51
  • $\begingroup$ Hm. Equal probabilities, as far as possible. $\endgroup$ – Elias Hasle Oct 1 '14 at 12:58
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    $\begingroup$ "Equal probabilities" isn't well-defined when there are infinitely many possibilities (don't forget that each family of distributions (e.g. beta, triangular, uniform) each have infinitely many distributions with the given mean, each with different standard deviations). I think the maximum entropy solution suggested below is the closest to something you're looking for. $\endgroup$ – DavidR Oct 1 '14 at 17:18
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What is commonly used in your situation is the maximum entropy solution:

http://en.wikipedia.org/wiki/Maximum_entropy_probability_distribution

In particular see http://www.math.uconn.edu/~kconrad/blurbs/analysis/entropypost.pdf

theorem 5.1 page 8.

basically it is $q_\alpha(x)=C_\alpha(x)\exp(\alpha x)$ for $x$ in $[0,1]$ and zero elsewhere. and $\alpha$ and $C_\alpha$ are set to match your mean and so that the density integrates to 1.

The Wallis derivation of Maximum entropy principle see http://en.wikipedia.org/wiki/Principle_of_maximum_entropy seems similar to your simulations...

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