7 ms·

U[0,∞]

by lmgftp 13y ago

U[0,∞]

ezyang 13y ago
Expected wait time: infinity!
- krcz 13y ago
  Not really, it's silly to talk about expected value here. There's just no such distribution.
- anologwintermut 13y ago
  If you believe that then I have a lottery ticket to sell you. http://en.wikipedia.org/wiki/St._Petersburg_paradox http://en.wikipedia.org/wiki/St._Petersburg_paradox
  - krcz 13y ago
    Can I play as many times as I want?
jerf 13y ago
No such thing: http://math.stackexchange.com/questions/14777/why-isnt-there-a-uniform-probability-distribution-over-the-positive-real-number http://math.stackexchange.com/questions/14777/why-isnt-there... Get far enough into Reflection on Relativity and the author makes some interesting observations based on this tidbit: http://www.mathpages.com/rr/rrtoc.htm http://www.mathpages.com/rr/rrtoc.htm (But it is quite a ways in there.)
- lmgftp 13y ago
  Acknowledged. Mostly a facetious comment, as any known distribution could be found (over infinite time) and you'd only become pseudononymous (which is exactly what the original comment wouldn't like!) On the other hand... It would be secure :) never withdraw. Anonymity through one way function/flow.
- anonymoushn 13y ago
  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.
  - krcz 13y ago
    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.
  - lwat 13y ago
    The probability of any finite interval P(a, b) = 0
  - jerf 13y ago
    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 13y ago

U[0,∞]

by lmgftp 13y ago

U[0,∞]

ezyang 13y ago

Expected wait time: infinity!

krcz 13y ago
Not really, it's silly to talk about expected value here. There's just no such distribution.
anologwintermut 13y ago
If you believe that then I have a lottery ticket to sell you. http://en.wikipedia.org/wiki/St._Petersburg_paradox http://en.wikipedia.org/wiki/St._Petersburg_paradox
- krcz 13y ago
  Can I play as many times as I want?

jerf 13y ago

No such thing: http://math.stackexchange.com/questions/14777/why-isnt-there-a-uniform-probability-distribution-over-the-positive-real-number http://math.stackexchange.com/questions/14777/why-isnt-there... Get far enough into Reflection on Relativity and the author makes some interesting observations based on this tidbit: http://www.mathpages.com/rr/rrtoc.htm http://www.mathpages.com/rr/rrtoc.htm (But it is quite a ways in there.)

lmgftp 13y ago
Acknowledged. Mostly a facetious comment, as any known distribution could be found (over infinite time) and you'd only become pseudononymous (which is exactly what the original comment wouldn't like!) On the other hand... It would be secure :) never withdraw. Anonymity through one way function/flow.
anonymoushn 13y ago
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.
- krcz 13y ago
  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.
- lwat 13y ago
  The probability of any finite interval P(a, b) = 0
- jerf 13y ago
  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 13y ago