[anonymoushn]

Could you point out the problem with such a distribution? It isn't immediately obvious that I cannot satisfy both axioms.

Edit: The helpful explanation linked in a comment on the question you linked is defective because it applies to all continuous probability distributions.

[jerf]

The key is the restriction that in the uniform distribution the probability density must be the same at all points, and if it covers infinity, it can be neither 0 nor anything greater than 0 if it's going to sum to 1. It's perfectly legal to have a probability distribution across all the reals. In fact most if not all of the well-known ones are; the Gaussian/normal distribution is defined on all reals, for instance. But it varies, and the integration from negative infinity to positive infinity sums to 1.

In fact everything that we refer to as "normal" distributions in the real world technically aren't, as the finite nature of the universe means the probability of the extremes is simply zero (give or take being totally wrong about the nature of the universe in which case all bets are off anyhow) rather than very, very small, and in many cases there's a sharp cutoff at 0, or some other arbitrary boundary, which a true normal distribution doesn't have. But it's often still the best mathematical approximation, with negligible error. (... until it isn't.... caveat emptor.)

[waps]

Why isn't the answer P = { Inf -> 1, otherwise 0 } ?

Axiom 1: P(E) elem N => P(E) >= 0, for all E

Trivially satisfied

Axiom 2: P(Omega) = 1

Satisfied: Omega = N { Inf } elem N

Axiom 3: Sigma additivity. Trivially satisfied since it either includes { Inf } or it doesn't, making the outcome 0 or 1.

Where is the problem ?

I think it's pretty clear that this is the only possible solution, because since N is not closed, there is no way to keep a uniform density other than 0.

This does not seem like it's a very useful solution, but it does seem to satisfy the axioms.

[jerf]

In addition to anonymoushn's correct point, I would also observe that even if we do treat infinity as a specific real number, the integration of a function zero at everywhere except one specific point from -inf to +inf is still 0. And further if we are treating it as a real number, you've now violated the uniformity constraint; P(0) != P(inf), therefore it is no longer a uniform distribution, the thing you are putatively building.

For an analysis of this and some other interesting things, see the Dirac delta (psuedo-)function: http://en.wikipedia.org/wiki/Dirac_delta_function

[anonymoushn]

Infinity isn't a positive real :(

[waps]

In most definitions I've seen of it, it is. In fact, it's a natural number.

[krcz]

Using uniform distribution definition from Wikipedia ("all intervals of the same length on the distribution's support are equally probable") we get P(X \in [0,1)) = P(X \in [1, 2)) = P(X \in [2, 3)) = ...

By countable additivity P(\Omega) = P(X \in [0, \infty)) = P(X \in [0, 1)) + P(\X \in [1, 2)) + ... = P(X \in [0, 1)) + P(X \in [0, 1)) + ...

And this evaluates to 0 if P(X \in [0, 1)) = 0 and to \infty if P(\X \in [0, 1)) > 0.

[consz]

>Edit: The helpful explanation linked in a comment on the question you linked is defective because it applies to all continuous probability distributions.

No, why do you think so?

[anonymoushn]

It contains the sentence "For any distribution, the sum of probabilities always equals 1" without the caveat that the sum is of a countable number of probabilities. It applies to every continuous distribution using the same argument, but with "the sum across all naturals" replaced with "the sum across the appropriate universe."

[Fargren]

"the sum across the appropriate universe" for Reals is an integral. An integral can go from -Infinite to Infinite and still equal 1. This is the reason, for instance, why Zeno's Paradox of movement doesn't really forbid all movement.

[anonymoushn]

This comment is more or less irrelevant. The third axiom of probability is about sums of countably many terms: http://en.wikipedia.org/wiki/Axioms_of_probability#Third_axi...

[lwat]

The probability of any finite interval P(a, b) = 0