any examples of 'modeless' continuous probability distributions?

Originally posted by humy
ANYONE here with expert knowledge of this; settle this once and for all:

IS the conventional exponential distribution (i.e. as in https://en.wikipedia.org/wiki/Exponential_distribution ) a "continuous probability distribution"?

If answer "no", can someone explain why not and yet why, say, a continuous uniform distribution is a "continuous probability dist ...[text shortened]... n"?

If nobody answers here, I will take this to a maths forum and ask the real maths experts.

It depends on what you think the domain is. It is clearly continuous away from x = 0, at the origin there is a discontinuity, but if one regards the domain as the non-negative reals then it is continuous everywhere. The support is a little tricky. It is defined as the largest closed subset of the domain in which every open neighbourhood of the set has non-zero measure. So if we look at the points -1, 0, +1 then at x = -1 and in any small neighbourhood of x = -1 the exponential distribution is zero so -1 is not in the support. Every open neighbourhood of x = +1 has non-zero measure - i.e. the integral of the distribution over some open set containing x = +1 is non-zero. For the case x = 0 the open set must contain x = 0 and be open so the integral (to find the measure) has to have limits (a, b) where a < 0 and b > 0, which is the same as the integral (0, b) and that measure is non-zero. So I think x = 0 is in the support both of the standard exponential distribution and of your function. This is because the support is a set of the form [a, b] and contains its limit points.

To summarize: I think that the standard exponential distribution is continuous, at least on its support (but you do need to check exactly how they define this in statistics), but that your function is not as its support includes x = 0, but the function has a discontinuity there.

Originally posted by DeepThought
It depends on what you think the domain is. It is clearly continuous away from x = 0, at the origin there is a discontinuity, but if one regards the domain as the non-negative reals then it is continuous everywhere. The support is a little tricky. It is defined as the largest closed subset of the domain in which every open neighbourhood of the ...[text shortened]... your function is not as its support includes x = 0, but the function has a discontinuity there.

I had always assumed that the word "continuous" in the term "continuous probability distribution" was there to indicate that the "probability distribution" was of a "continuous random variable" and not to indicate anything to do with the upper or lower limits of the support or domain. Perhaps I was wrong.

But, I think if I was wrong, then a continuous uniform distribution may not be a "continuous probability distribution" because it has two "discontinuous" points on its support. And yet I see "continuous uniform distribution" listed under "Continuous distributions" for this link for a list of "probability distributions";

https://en.wikipedia.org/wiki/List_of_probability_distributions

-making me wonder if I was right about that "continuous" in the term "continuous probability distribution" merely being there to indicate "continuous random variable" all the long.

I plan to get to the bottom of this matter eventually to clear up all this confusion. Don't mind being wrong about something as long as I am put right in the end.

Originally posted by humy
And yet I see "continuous uniform distribution" listed under "Continuous distributions" for this link for a list of "probability distributions";

And the heading says: "Supported on a bounded interval" suggesting that they are only defined within the interval.

Originally posted by twhitehead
And the heading says: [b]"Supported on a bounded interval" suggesting that they are only defined within the interval.[/b]

saying the support is only within the bounded interval doesn't imply the distribution is not a continuous probability distribution.

Originally posted by twhitehead
I have a sneaky feeling that open sets cannot be integrated.
Originally posted by humy
I don't see why.

I looked it up and it turns out I am correct. The 'definite integral' requires a closed set:
https://en.wikipedia.org/wiki/Integral

open sets result in 'improper integrals', but such things do exist.

Originally posted by twhitehead
I looked it up and it turns out I am correct. The 'definite integral' requires a closed set:
https://en.wikipedia.org/wiki/Integral

open sets result in 'improper integrals', but such things do exist.

Unless I am missing something, the link implies that you can have an improper integral have a definable finite value (with "If the interval is unbounded, for instance at its upper end, then the improper integral is the limit as that endpoint goes to infinity ... ... a limit may provide a finite result." ) and this appears to me to contradict your earlier assertion of "open sets cannot be integrated".

But you have just taught me something new because I didn't know about improper integrals until now and for that I am grateful. I definitely somehow missed that one in my university maths courses.
So I take it then you CAN validly write;

ʃ (x=a, b] ... dx

after all. I did wonder.

Originally posted by humy
Unless I am missing something, the link implies that you can have an improper integral have a definable finite value

Yes, but with the understanding that that definite value is a limit. This comes back to the situation with a mode that doesn't quite fit the standard definition. If you can still calculate it, and the calculation more or less follows from the definition but involves limits, must you give it a new name?

Originally posted by twhitehead
I looked it up and it turns out I am correct. The 'definite integral' requires a closed set:
https://en.wikipedia.org/wiki/Integral

open sets result in 'improper integrals', but such things do exist.

You need to be very careful with that kind of detail in Wikipedia. I'd not heard of proper and improper integrals before, and it may be that the Wiki writer is relying on a single reference and the term isn't widely accepted. I'm a bit skeptical about it for two reasons. The first is that Lebesgue integration is all defined in terms of open sets - although that doesn't stop the actual integral being defined in terms of a closed set. The other reason is that an unbounded integral (i.e. between plus and minus infinity) is on an open set and it seems odd to define some of the most important integrals as improper. It makes no practical difference since the boundary of any integrable function is a zero measure set, so including and excluding the boundary points won't make a difference to the result.

Originally posted by humy
I had always assumed that the word "continuous" in the term "continuous probability distribution" was there to indicate that the "probability distribution" was of a "continuous random variable" and not to indicate anything to do with the upper or lower limits of the support or domain. Perhaps I was wrong.

But, I think if I was wrong, then a continuous unifor ...[text shortened]... ll this confusion. Don't mind being wrong about something as long as I am put right in the end.

I skim read this bit from Wikipedia [1], it seems to indicate that you are right in the sense that any continuous random variable will have non-zero probability density everywhere (since otherwise the random variable would not be able to take values in the region that the density function is zero). I was relying more on the definition for continuity for normal functions.

Note that the writer of the paragraph makes the point that the probability of a continuous random variable coming out at some specific value 'a' is zero. So your exclusion of the zero point may be unnecessary, since although the probability density is non-zero there the probability of getting a result between zero and zero is zero.

[1] https://en.wikipedia.org/wiki/Probability_distribution#Continuous_probability_distribution

Originally posted by DeepThought
I'd not heard of proper and improper integrals before, and it may be that the Wiki writer is relying on a single reference and the term isn't widely accepted.

It is clearly common terminology and you can find it all over the web:
http://mathworld.wolfram.com/ImproperIntegral.html

This MIT textbook appears to define the 'definite integral' as being on a closed set, but not necessarily a continuous function:
http://ocw.mit.edu/ans7870/resources/Strang/Edited/Calculus/Calculus.pdf

I haven't read the whole book, so I could be wrong.

Originally posted by DeepThought
Note that the writer of the paragraph makes the point that the probability of a continuous random variable coming out at some specific value 'a' is zero. So your exclusion of the zero point may be unnecessary, since although the probability density is non-zero there the probability of getting a result between zero and zero is zero.

An interesting point.

Originally posted by DeepThought
I skim read this bit from Wikipedia [1], it seems to indicate that you are right in the sense that any continuous random variable will have non-zero probability density everywhere (since otherwise the random variable would not be able to take values in the region that the density function is zero). I was relying more on the definition for continu ...[text shortened]...
[1] https://en.wikipedia.org/wiki/Probability_distribution#Continuous_probability_distribution

Note that the writer of the paragraph makes the point that the probability of a continuous random variable coming out at some specific value 'a' is zero.

You wont believe this but I have thoroughly researched this very point and concluded that they (in the link ) are totally wrong because it leads to a nonsense contradiction, i.e. a paradox, which proves that actually the probability of some specific value 'a' is NOT zero but rather is "undefined"! -Which is just another way of simply saying we cannot say or know anything about that probability.
Think about this:
say we observe a sample out of an actual distribution a specific random x value of, lets say, its x=2.3497546754...
OK, if the probability of each and every specific value is zero, then the prior probability of any specific x value, i.e. before sampling a real x value, in the real world being exactly whatever it is, is exactly zero probability. But here is the paradox; it is a contradiction for something to be both logically possible and that same something to have zero probability! If something has zero probability then it cannot happen i.e. its impossible and therefore we would never observe it in the real world. That would mean that, according to that naive interpretation of the maths that assumes a probability exists even if it doesn't, the x value we sample, whatever that exact value is, will be impossible because it has exactly zero probability of being whatever it is; And yet we have sampled an x value from the real world and that x value is exactly whatever that x value is thus proving its probability couldn't have ever been exactly zero!

I have given this kind of nonsense probability of 0 from the naive application of the maths that assumes incorrectly that there must exist a probability, the name of an "improper probability", which I express thus:
P(x) =(≠ ) 0 (but without the spacer between '≠' and ' )' -I had to avoid getting a smiley )
where I call the compound symbol of =(≠ ) the "improper equality symbol"

So your exclusion of the zero point may be unnecessary,

I think you could desire it! Because here we can make the distinction between the nonsense improper probability of zero ( P(x) =(≠ ) 0 ) of what is logically possible, from a proper probability of zero ( P(x) = 0 ) of what is logically impossible; the former being a bit of a misnomer because "improper probability" means a kind of fake probability because it is undefined so you could say that, actually, no real probability exists for that outcome!

Originally posted by humy
i.e. a paradox, which proves that actually the probability of some specific value 'a' is NOT zero but rather is "undefined"! -Which is just another way of simply saying we cannot say or know anything about that probability.

But you can know something about that probability.

i.e. before sampling a real x value, in the real world being exactly whatever it is, is exactly zero probability.
What if we don't say it is 'exactly zero' but the limit of the probability as you make the interval smaller tends to zero? You are trying to deal with infinities but not accepting limits.

But here is the paradox; it is a contradiction for something to be both logically possible and that same something to have zero probability!
We are dealing with infinities here. With an infinity of zeros, we can get 1. 🙂

Originally posted by twhitehead
But you can know something about that probability.

[b]i.e. before sampling a real x value, in the real world being exactly whatever it is, is exactly zero probability.
What if we don't say it is 'exactly zero' but the limit of the probability as you make the interval smaller tends to zero? You are trying to deal with infinities but not accepting l ...[text shortened]... bability! [/b]
We are dealing with infinities here. With an infinity of zeros, we can get 1. 🙂[/b]

But you can know something about that probability.

No, you cannot. Not in this case! At least not without some additional information. Not every theory has a probability under all circumstances! That is one of the many things I have already proved out of my intense research into epistemology over the last year (proof still to be published but I am getting there slowly! ).

I call a theory that cannot currently rationally be assigned a probability an "unassignable theory" and any theory that can an "assignable theory" just two of the ~130 new terms, words and concepts I have so far invented but not yet published.

What if we don't say it is 'exactly zero' but the limit of the probability as you make the interval smaller tends to zero?

So what value does it have in this case if not zero?
The answer is no value! -because there is no probability for that even though there is a probability for a interval.
Saying it "tends" to zero but isn't zero doesn't answer the simple question "what value it is if not zero?".

Originally posted by humy
No, you cannot. Not in this case!

Yes, we can as it is clearly defined.

Not every theory has a probability under all circumstances!
But this one does. And by 'this one' I am referring to any continuous probability distribution. It is practically defined that way.

Saying it "tends" to zero but isn't zero doesn't answer the simple question "what value it is if not zero?".
What is the sum of the series:
1/2 + 1/4 + 1/8 ....
It isn't 1. It only tends to 1. And asking 'what is the value if it isn't one?' doesn't give you licence to say 'it is therefore undefined'.