Peculiar distribution...

Discussion in 'Recreational Math' started by Risto Lankinen, Nov 9, 2004.

  1. Hi!

    Find a random distribution, whose all outcomes are strictly
    less than the expected [=weighted arithmetic average] value.

    - Risto -
     
    Risto Lankinen, Nov 9, 2004
    #1
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  2. The discrete distribution defined by

    P(X=n) = (1/2)^n, for n>=1.

    It has an infinite expectation and each realisation is finite.

    - Paul
     
    Paul Beekhuizen, Nov 9, 2004
    #2
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  3. [rearranged in bottom-posting order, which seems to work better for
    posts longer than a few lines]
    That distribution has expectation equal to 2.

    Try this:

    P(X=n) = (6/Pi^2) n^(-2), for n=>1.
     
    Christopher J. Henrich, Nov 9, 2004
    #3
  4. Risto Lankinen

    Virgil Guest

    What kind of weights are allowed?

    If the sum (or integral for a continuous distribution) of the weights
    equals 1, as it should, are you sure that this is even possible?
     
    Virgil, Nov 9, 2004
    #4
  5. Risto Lankinen

    Proginoskes Guest

    The expected value of X is 1 / 2^1 + 2 / 2^2 + 3 / 2^3 + 4 / 2^4 + ...,
    which converges to 2. The expected value of 2^X is
    2^1 / 2^1 + 2^2 / 2^2 + 2^3 / 2^3 + ..., which diverges to +infinity.
    I think this is what you meant. (St. Petersburg Paradox?)
    -- Christopher Heckman
     
    Proginoskes, Nov 10, 2004
    #5
  6. Hi!

    This very much resembles the discrete distibution that I had
    in mind (e.g. constant divided by a square yields a finite sum
    [integral] for accumulated probability, but each term multiplied
    by n yields an infinite sum [integral] for the expectation).

    Here's my originally intended answer:

    P(X=n) = 1/n - 1/(n+1) = 1/(n*n+n) for positive integer n .

    It's easy to see that the distribution is proper (i.e. its sum is
    exactly one). The expectation, on the other hand, is the sum
    of 'n/(n*n+n) = 1/(n+1)' which is infinite.

    I truly find this property quite peculiar!

    (Actually, I'm also wondering if there's a natural phenomenon
    that manifests this distribution in reality?)

    - Risto -
     
    Risto Lankinen, Nov 10, 2004
    #6
  7. Yes it is, sorry about that. Mine isn't even a distribution, if all n are
    allowed...
    This is what it should have been:

    P(X=2^n) = (1/2)^n, for all natural n>0.

    Paul
     
    Paul Beekhuizen, Nov 11, 2004
    #7
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