As soon as you’re dealing with hyperreals you lose completeness, which may or may not be important. It’d be interesting to see this done in detail.
I don’t see where this would lead us! As I pointed in a previous post, in a continuous pdf, the probability of a particular outcome is zero, but the probability of the outcome liyng in a determined interval is generally nonzero. Since the width of the interval can be made arbitrarily small, I don’t see what you need infinitesimals for.
Cabbage: You’re right, I was too hasty.
Let’s step back from this argument a moment. Can I say unarguably that (1) Depending on what form the infinite universe takes it can be ‘almost’ certain or ‘almost’ impossible that there’s another copy of the OP somewhere and (2) one way of illustrating this is with the standard mathematical definition of probability?
Maybe that isn’t actually the best way to undersand it, come to think of it…
As Cabbage mentioned, it would allow you to assign different probabilities to events that must happen and events that have probability 1 now but aren’t necessary. It’s more of an interesting exercise than anything.
If you know what is the pdf of the values of this constant, you can without any problem to assign a probability to the event that its value lies in the interval (a;b), 0<=a<=b<=1.
For instance, if all values in the interval [0;1] are equiprobable, you have a uniform pdf and the probability that the value of your constant belongs to the interval is simply b-a.
With other pdf’s the calculation is less straightforward, but still possible. You may have to use integral calculus, but you don’t need to pe a PhD to do this.
If you have no idea of what the pdf is, there is no means, to my knowledge, to assign any probability to the event.
Sérgio: But, can you assign a probability to both the open and closed intervals? If not, which is ok? If so, mustn’t they be the same and doesn’t that fail the ‘intuitive’ test Ryan was talking about?
Shade
The probability for the closed interval is equal to the probability for the open interval plus the probabilities of the limits of the interval:
p{[a;b]} = p{(a;b)} + p{a} + p{b} = p{(a;b)} + 0 + 0 = p{(a;b)}
So it does not matter if the interval is open or closed, the probability is the same!
Sorry, perhaps I got a bit ahead of myself. I assumed that having two nested events with the same probabilty was one of the things your system was trying to avoid.
Further, we have the problem that there are events U=[a,b] and V=(a,b) and we can find the probability either happening, but you can’t say “the probability of V not U.” That ommission is looks very inelegant to me.
Since V is a subset of U, P(U and not V) = P(U) - P(V). It’s 0.
Yes, but sergio’s system was in response to my observation that ended
Shade
p(V,~U) = p(a) + p(b) = 0 + 0 = 0.
As I said before, the probability of a particular outcome in a continuous pdf is zero.
Note that this refers to a priori probabilities.
The a priori probability of a is zero, so the a priori probability of ~a is 1.
If I perform the experiment and find a, the a posteriori probability of a is 1 and the a posteriori probability of ~a is 0.
So, events with zero a priori probabilities do occur. Once they occur their a posteriori probabilities become one.
I don’t know if it explains your doubt.
As far as I can see your probability system is exactly the same as the standard. And specifically includes possible events with zero probabilities. Your post was a response to my asking if it was possible to avoid this (as apparently Ryan desired.)
Was your post intended as an example of a definition of probability that didn’t involve 0-probability posssible events??
If not, then I apologise, I misunderstood. Though imho understandably, since that’s what you replied to. Is so, I don’t see how it doesn’t!
Shade
I used the standard probability system. My point is: there is no need to postulate different systems, as the one proposed by Cabbage with infinitesimals.
Events with zero probability are totally meaningful. If you need a very small nonzero probability in a continuous pdf all you need is to take a tiny interval.:smack:
OK, I apologise, I totally misunderstood.
In my defense, you didn’t mention infitessimals anywhere, and defined the probability of an interval to be its length, and gave the probability of a point to be zero, so I don’t think it was quite clear that that’s what you meant.
Well, I guess that depends on how you define “need”. I do think that it’s somewhat “intuitively unsatisfying” that a probability function can’t distinguish between a set with probability one that isn’t certain, from a set with probability one that is certain. And regardless, I definitely think it would be interesting to see what results you could get by extending the notion of probability to include infinitesimals.
Now that I think about it, I don’t think that “certain” is defined in a sufficiently rigorous manner to do what you wanted. I suppose you could arbitrarily say that some events are certain (and that certainty is preserved by unions and intersections), and that the complement of a certain event is impossible…
Might be worth looking at.
Now that I think about it, it’s not possible to do what The Ryan wants using infinitesimals. Consider the case of an infinite sequence of coin flips–the limit of 1/2[sup]n[/sup] as n grows unboundedly is 0, and the introduction of infinitesimals won’t change that.
Wouldn’t it be correct to say that the limit of 1/2[sup]n[/sup] is an infinitesimal?
No, the limit is zero!
I seriously doubt it–I can construct a sequence of hyperreals with the nth term corresponding to 1/2[sup]n[/sup] (the nth term is (1/2[sup]n[/sup], 1/2[sup]n[/sup], 1/2[sup]n[/sup], …)), and the limit as n goes to infinity of that is (0, 0, 0, …), a hyperreal that corresponds to 0.
It’s certainly possible that you know more about infinitesimals than I do and can see a flaw in the above reasoning; if so, please explain it.