# Peculiar distribution...

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

1. ### Risto LankinenGuest

Hi!

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

- Risto -

Risto Lankinen, Nov 9, 2004

2. ### Paul BeekhuizenGuest

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

3. ### Christopher J. HenrichGuest

[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. ### VirgilGuest

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. ### ProginoskesGuest

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. ### Risto LankinenGuest

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).

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. ### Paul BeekhuizenGuest

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