[math-fun] Paradox
Select two random integers, a and b. Clearly, the probability of b being larger than a is 100%, as the expected value of b is infinite, yet the value of a is finite. Similarly, the probability of a being larger than b is 100%, by symmetry. What have I done wrong? Sincerely, Adam P. Goucher
You need to specify the probability distribution by which the integers are selected. If the distribution is normalizable, e.g. p(n) = 1/2^n, thenthere's no paradox. But you seem to want a uniform distribution, p(n) = constant, and this does not exist. If p(n) = 0, then sum(p(n)) = 0, while if p(n) > 0, sum(p(n)) = infinity. The nonsensical nature of a uniform distribution over the integers can be realized by trying to devise a procedure for generating such integers. -- Gene
________________________________ From: Adam P. Goucher <apgoucher@gmx.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Sunday, June 5, 2011 7:20 AM Subject: [math-fun] Paradox
Select two random integers, a and b. Clearly, the probability of b being larger than a is 100%, as the expected value of b is infinite, yet the value of a is finite. Similarly, the probability of a being larger than b is 100%, by symmetry.
What have I done wrong?
Sincerely,
Adam P. Goucher
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
To clarify, I was referring to a uniform distribution over the positive integers. Obviously, distributions with nonzero probability density functions, such as exponential distributions, are not subject to this paradox. Sincerely, Adam P. Goucher
On Sun, Jun 5, 2011 at 7:57 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
To clarify, I was referring to a uniform distribution over the positive integers. Obviously, distributions with nonzero probability density functions, such as exponential distributions, are not subject to this paradox.
What have I done wrong?
There is no uniform distribution over the positive integers. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
From: Mike Stay <metaweta@gmail.com>
To: math-fun <math-fun@mailman.xmission.com> Sent: Sunday, June 5, 2011 8:36 AM Subject: Re: [math-fun] Paradox
On Sun, Jun 5, 2011 at 7:57 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
To clarify, I was referring to a uniform distribution over the positive integers. Obviously, distributions with nonzero probability density functions, such as exponential distributions, are not subject to this paradox.
What have I done wrong?
There is no uniform distribution over the positive integers.
And more generally, there is no uniform distribution over any countable set. -- Gene
On Sun, Jun 5, 2011 at 9:27 AM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
From: Mike Stay <metaweta@gmail.com>
To: math-fun <math-fun@mailman.xmission.com> Sent: Sunday, June 5, 2011 8:36 AM Subject: Re: [math-fun] Paradox
On Sun, Jun 5, 2011 at 7:57 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
To clarify, I was referring to a uniform distribution over the positive integers. Obviously, distributions with nonzero probability density functions, such as exponential distributions, are not subject to this paradox.
What have I done wrong?
There is no uniform distribution over the positive integers.
And more generally, there is no uniform distribution over any countable set.
Well, no *infinite* countable set. :)
-- Gene
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
I'm getting lost here. Exactly why can't a countable set --- eg. rationals x with 0 <= x < 1 --- be assigned uniform probability? [Admittedly, I can't offhand see quite how to do it!] What has denumerability to do with this anyway? It's also impossible to assign uniform probability to the unbounded reals. Fred Lunnon On 6/5/11, Mike Stay <metaweta@gmail.com> wrote:
On Sun, Jun 5, 2011 at 9:27 AM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
From: Mike Stay <metaweta@gmail.com>
To: math-fun <math-fun@mailman.xmission.com> Sent: Sunday, June 5, 2011 8:36 AM Subject: Re: [math-fun] Paradox ... There is no uniform distribution over the positive integers.
And more generally, there is no uniform distribution over any countable set.
Well, no *infinite* countable set. :)
-- Gene
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
On 6/6/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
I'm getting lost here.
Exactly why can't a countable set --- eg. rationals x with 0 <= x < 1 --- be assigned uniform probability? [Admittedly, I can't offhand see quite how to do it!]
What has denumerability to do with this anyway? It's also impossible to assign uniform probability to the unbounded reals.
Fred Lunnon
On 6/5/11, Mike Stay <metaweta@gmail.com> wrote:
On Sun, Jun 5, 2011 at 9:27 AM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
From: Mike Stay <metaweta@gmail.com>
To: math-fun <math-fun@mailman.xmission.com> Sent: Sunday, June 5, 2011 8:36 AM Subject: Re: [math-fun] Paradox ... There is no uniform distribution over the positive integers.
And more generally, there is no uniform distribution over any countable set.
Well, no *infinite* countable set. :)
-- Gene
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
A measure is generally assumed to be countably additive. If the probability of an integer n is 0, then the total measure is 0. If the probability of n is epsilon, then the total measure is infinity, so it is not a probability measure. ====== Non countably-additive measures are legitimate structures, and provide one of the ways to define amenability for a group, but they are totally weird and I don't think you really want to go into that territory. In theory, there are translation-invariant additive but non-countaby-additive measures on Z, but it is known to be impossible to actually define any particular instance of one. Bill Thurston On Jun 5, 2011, at 9:22 PM, Fred lunnon wrote:
I'm getting lost here.
Exactly why can't a countable set --- eg. rationals x with 0 <= x < 1 --- be assigned uniform probability? [Admittedly, I can't offhand see quite how to do it!]
What has denumerability to do with this anyway? It's also impossible to assign uniform probability to the unbounded reals.
On Sun, Jun 5, 2011 at 9:32 PM, Bill Thurston <wpthurston@mac.com> wrote:
====== Non countably-additive measures are legitimate structures, and provide one of the ways to define amenability for a group, but they are totally weird and I don't think you really want to go into that territory. In theory, there are translation-invariant additive but non-countaby-additive measures on Z, but it is known to be impossible to actually define any particular instance of one.
How can this be? Let the measure of {1} be x > 0, and choose an integer p such that x > 1/p. If the measure is translation invariant, then the measure of any singleton is x. But if the measure is finitely additive, then the measure of {1,2,3,...p} is xp > 1, so the measure can't be a probability measure. I must be misinterpreting some part of your terminology, but I'm not sure what. Andy
On Jun 6, 2011, at 1:41 AM, Andy Latto wrote:
On Sun, Jun 5, 2011 at 9:32 PM, Bill Thurston <wpthurston@mac.com> wrote:
====== Non countably-additive measures are legitimate structures, and provide one of the ways to define amenability for a group, but they are totally weird and I don't think you really want to go into that territory. In theory, there are translation-invariant additive but non-countaby-additive measures on Z, but it is known to be impossible to actually define any particular instance of one.
How can this be?
Let the measure of {1} be x > 0, and choose an integer p such that x > 1/p.
If the measure is translation invariant, then the measure of any singleton is x.
But if the measure is finitely additive, then the measure of {1,2,3,...p} is xp > 1, so the measure can't be a probability measure.
You've given a correct deduction that the measure of any singleton, and in fact the measure of every finite set is 0. The measure must be 1 for every set whose complement is finite. The measure must be 1/2 for all even integers, and 1/2 for all odd integers, etc --- there are certain similar things you can deduce. The craziness comes in extending these definitions to *all* subsets of the integers. Bill
I must be misinterpreting some part of your terminology, but I'm not sure what.
Andy
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Speaking of craziness, what about measures with nonstandard (in the sense of hyperreal or surreal) values? It seems natural to assign a uniform probability measure where the value of any single integer is some infinitesimal epsilon. The lack of the Archimedean property means that the measure of {1, 2, 3, ... n} can still be less than one for any n. In particular, I wonder if you can get one of the translation-invariant but non-countably-additive measures that Bill is talking about by taking the standard part of such a nonstandard uniform measure on Z. -Thomas C On Mon, Jun 6, 2011 at 6:50 AM, Bill Thurston <wpthurston@mac.com> wrote:
On Jun 6, 2011, at 1:41 AM, Andy Latto wrote:
On Sun, Jun 5, 2011 at 9:32 PM, Bill Thurston <wpthurston@mac.com> wrote:
====== Non countably-additive measures are legitimate structures, and provide one of the ways to define amenability for a group, but they are totally weird and I don't think you really want to go into that territory. In theory, there are translation-invariant additive but non-countaby-additive measures on Z, but it is known to be impossible to actually define any particular instance of one.
How can this be?
Let the measure of {1} be x > 0, and choose an integer p such that x > 1/p.
If the measure is translation invariant, then the measure of any singleton is x.
But if the measure is finitely additive, then the measure of {1,2,3,...p} is xp > 1, so the measure can't be a probability measure.
You've given a correct deduction that the measure of any singleton, and in fact the measure of every finite set is 0. The measure must be 1 for every set whose complement is finite. The measure must be 1/2 for all even integers, and 1/2 for all odd integers, etc --- there are certain similar things you can deduce. The craziness comes in extending these definitions to *all* subsets of the integers.
Bill
I must be misinterpreting some part of your terminology, but I'm not sure what.
Andy
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Select two random integers, a and b. Clearly, the probability of b being larger than a is 100%, as the expected value of b is infinite, yet the value of a is finite. Similarly, the probability of a being larger than b is 100%, by symmetry.
What have I done wrong?
sounds like the two envelopes problem. http://en.wikipedia.org/wiki/Two_envelopes_problem (if we assume that the probability for selecting m and -m is the same, then the expectation value of b is zero. But I think that was not your intention) if we restate the problem to selecting two random non-negative integers, we can resolve it by arguing that "selecting a random integer" does not define a proper probability distribution. A possible probability distribution for selecting a random (non-negative) integer could be something like p_n= (1-1/exp(0)) exp(-n). Then the expected values for a, b are finite and identical. Is that what you meant? Christoph
participants (8)
-
Adam P. Goucher -
Andy Latto -
Bill Thurston -
Eugene Salamin -
Fred lunnon -
Mike Stay -
Pacher Christoph -
Thomas Colthurst