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. Advertisements

  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
    1. Advertisements

  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
  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
  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
  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
  7. Yes it is, sorry about that. Mine isn't even a distribution, if all n are
    This is what it should have been:

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

    Paul Beekhuizen, Nov 11, 2004
    1. Advertisements

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.