logical and causal possibilities and proofs

My opinion is this:
1. Probability is all about how frequently you expect an outcome that has a random component to give a particular result.
2. It is a trivial observation that if you take out the random component and make it deterministically always give a single result then the 'frequency' becomes either 'every single time' or 'never' depending on which result you are referring to.
3. One can therefore assign probability values of 1 and 0 to these results.
4. When laymen use the word probability in conjunction with the 1 and 0 values they may mean one of two things: either that it is deterministic or that there is an infinitesimal probability of one result.
5. I see no benefit to applying probability values to deterministic results as the result is trivial. All probability calculations will have a 1 or 0 and thus the result will trivially proceed as you carry out subsequent calculations.
6. The real focus of probability is in the open set (1,0).
7. I believe it is more useful to define probability as 'undefined' outside this range.
8. Probability can be applied to any situation real or imagined, whereas your definition appeared to be only about its application to the problem of knowledge of the real world. In addition your focus seemed to be on situations where the result is as close to 1 or 0 as possible ie you are focused on either logical proofs or experiments that yield exclusively one result to the point that you can reasonably rule out any other results as ever happening. It seems to me that probability theory has no place in either of the latter situations.

If you have a use for applying probability values of 1 and 0 to deterministic situations (or even static situations as is the case for logical statements) then I would be interested to know what that use is.

Originally posted by twhitehead
My opinion is this:
1. Probability is all about how frequently you expect an outcome that has a random component to give a particular result.
2. It is a trivial observation that if you take out the random component and make it deterministically always give a single result then the 'frequency' becomes either 'every single time' or 'never' depending on wh ...[text shortened]... ions as is the case for logical statements) then I would be interested to know what that use is.

why does something have to have a use to be logically valid?
Absence of use is completely irrelevant to logical validity.
If I where to say a particular premise p logically implies, via an inference i, a particular conclusion c, but we have no use for c, how would that logically imply there must be something logically erroneous with either p or i and thus c?

Originally posted by humy
why does something have to have a use to be logically valid?

It doesn't, and at I thought we agreed that definitions don't have logical validity anyway. They just are.

If I where to say a particular premise p logically implies, via an inference i, a particular conclusion c, but we have no use for c, how would that logically imply there must be something logically erroneous with either p or i and thus c?
Except the premise does not 'imply via an inference'. It is entirely a matter of choice. My issue is with choosing to go with something that has no value - especially if it differs from the already widespread standard definition as it not only has no value but is likely to cause confusion.

Originally posted by twhitehead
... My issue is with choosing to go with something that has no value ...

It appears your choice also has no value for it requires me to add a superfluous extra criterion (the superfluous 4th one with my definition; and I now considering whether I should reject the 4th one after all as it is superfluous ) to the definition of probability just to basically say "oh, by the way, probability cannot take on binary values -no particular reason" thus making the definition more complex than it needs to be.

I should add ( for the other subject ) that, with the many years experience I have with computer programming, for some limited applications, I sometimes find it convenient to lump together certain truth values of certain variables into the same basket of probabilities for certain other variables by simply representing truth values as 1 or 0 probability and group them together with all other probabilities that are between 0 and 1. Failure to do this will require making the program more complex than it needs to be by having separate Boolean variables and then inefficiently making them interact with probabilities -no such interaction would be required if they are placed in the same basket (on request, I will try and dig up a concrete example of that ).
So there is an example of one use for treating a truth value as a probability.

Here is an example of where you must sometimes accept binary values for some probabilities:

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

To calculate that probability, you may have to sometimes accept a few P(A|Bx)=1 or a few P(A|Bx)=0 as inputs.
(don't know how to edit the x as subscript after B as I should there )
Are they also truth values because they take on binary values?
If so, there is yet another example of a use for representing truth values as probabilities.

Originally posted by humy
It appears your choice also has no value for it requires me to add a superfluous extra criterion (the superfluous 4th one with my definition; and I now considering whether I should reject the 4th one after all as it is superfluous ) to the definition of probability just to basically say "oh, by the way, probability cannot take on binary values -no particular reason" thus making the definition more complex than it needs to be.

Although I concede that there may be benefits to including binary values in your definition and that your initial definition may be smaller, I foresee problems down the line where most other definitions in the realm of probability do not include them. The result will be that you will also have to rewrite all those definitions and cause even more confusion. I am not saying you will be wrong, merely that you should think twice about it for the sake of clarity.

As an example, in the other thread, I believe there was some discussion about a "continuous probability distribution". Although some definitions I have seen quite readily allow zero values, I see a problem with that as it fails to differentiate between discrete distributions and continuous ones. Although, I guess one could differentiate by looking at the condition that continuous distributions have a probability of zero (infinitesimal or zero) at each point.
But still, I have failed to find any evidence that a continuous probability function may have gaps in it that are truly zero.
So for example, if you said: pick at random any real number between 0 and 2 but excluding the interval (1,root 2) then would you have a continuous probability distribution? I say no. Your definition would suggest yes. What does the standard definition say? I am not certain and have failed to find clear a unequivocal definition on the matter.

Originally posted by twhitehead
Although I concede that there may be benefits to including binary values in your definition and that your initial definition may be smaller, I foresee problems down the line where most other definitions in the realm of probability do not include them. The result will be that you will also have to rewrite all those definitions and cause even more confusion ...[text shortened]... tion say? I am not certain and have failed to find clear a unequivocal definition on the matter.

As an example, in the other thread, I believe there was some discussion about a "continuous probability distribution". Although some definitions I have seen quite readily allow zero values, I see a problem with that as it fails to differentiate between discrete distributions and continuous ones. Although, I guess one could differentiate by looking at the condition that continuous distributions have a probability of zero (infinitesimal or zero) at each point.

NO NO NO that is NOT what defines the difference between a continuous distribution and and a discrete distribution. The fact that all continuous distribution have zero probability at a point has nothing to do with it!
I thought you already got that?


The word "continuous" in "continuous distribution" has NOTHING to do with whether or not it is a "continuous function" (which is where you got both me and you confused earlier in that thread until that real maths expert explained it to me and then I explained it to you ) but rather is there to indicate that the random variable is a "continuous variable" as opposed to a "discrete variable" and THAT is what distinguishes between a "continuous probability distribution" and a "discrete probability distribution".

So for example, if you said: pick at random any real number between 0 and 2 but excluding the interval (1,root 2) then would you have a continuous probability distribution?

yes. It would be a "continuous distribution" NOT to be confused with a "continuous function". Many continuous distributions are defined with discontinuous functions. Just one example; The uniform probability distribution. It has zero probability density for much of its allowed domain values.
('domain' not to be confused with its 'sample space' here and 'probability density' NOT to be confused with 'probability' here because a probability density is not a true probability but something you can derive a probability of directly via an integration of part/all of its curve )

I am not certain and have failed to find clear a unequivocal definition on the matter.

This definition is incomplete but, just for starters;
All continuous distributions allow all real number values in the sample space between at least one interval [a, b] (or (a, b] or (a, b) or [a, b) ) where a<b thus making the distribution one of a continuous variable.
No discrete distribution allows that else it wouldn't be of a discrete variable but a continuous distribution.

One trivial small complication; I have heard of rare cases where distributions that are of both discrete variable over a certain range of values in its sample space and continuous variable over a different range of values in its sample space, in which case they are both. I would contrive an example on request. But you won't see many of them so, for now, I would advise you to ignore them until you have learned of the definition of the more common types.

Originally posted by humy
NO NO NO that is NOT what ...

Actually you appear to largely agree with the bit you quoted.

The fact that all continuous distribution have zero probability at a point has nothing to do with it!
Actually, it is part of the definition and does not follow trivially from the rest of the definition, so it does have a lot to do with it.

The word "continuous" in "continuous distribution" has NOTHING to do with whether or not it is a "continuous function"
And you will note that I didn't mention 'continuous function' in the post above.

but rather is there to indicate that the random variable is a "continuous variable" as opposed to a "discrete variable"
And how exactly do you define 'continuous variable'?

yes. It would be a "continuous distribution"
Except that it is not a continuous variable, and that is where I think your decision to put zeros in may cause confusion.

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

If you claim that x in my example can take on values in the excluded range but has the value zero, then you can call it a 'continuous variable', but you must also accept that discrete probabilities are also 'continuous variables' which seems to destroy the whole point of the distinction between the two.

The uniform probability distribution. It has zero probability density for much of its allowed domain values. ('domain' not to be confused with its 'sample space' here and 'probability density' NOT to be confused with 'probability' here because a probability density is not a true probability but something you can derive a probability of directly via an integration of part/all of its curve )
So you would agree that within the sample space, it is in fact a continuous function.

This definition is incomplete but, just for starters;
All continuous distributions allow all real number values in the sample space between at least one interval [a, b] (or (a, b] or (a, b) or [a, b) ) where a<b thus making the distribution one of a continuous variable.
Then my example is not a continuous distribution as its sample space is two distinct intervals.
Are you starting to see where confusion creeps in?

One trivial small complication; I have heard of rare cases where distributions that are of both discrete variable over a certain range of values in its sample space and continuous variable over a different range of values in its sample space, in which case they are both.
Or neither. Maybe 'hybrid' would be a better way to put it.

Originally posted by twhitehead
Actually you appear to largely agree with the bit you quoted.

[b]The fact that all continuous distribution have zero probability at a point has nothing to do with it!
Actually, it is part of the definition and does not follow trivially from the rest of the definition, so it does have a lot to do with it.

The word "continuous" in "continuou ...[text shortened]... , in which case they are both.
Or neither. Maybe 'hybrid' would be a better way to put it.[/b]

And how exactly do you define 'continuous variable'?

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

"...A continuous variable over a particular range of the real numbers is one whose value in that range must be such that, if the variable can take values a and b in that range, then it can also take any value between a and b..."

so even if we have an open rather than a closed interval of real numbers of (2, 9), it is still a continuous variable because you can randomly pick any two values for a and b between 2 and 9 but excluding 2 and 9 and such that a<b and define and interval thus: [a, b], and that interval will fit the part of the definition of:
"if the variable can take values a and b in that range, then it can also take any value between a and b". So, for example, you could randomly arbitrarily pick a=3.5, b=4.4, so the interval is [3.5, 4.4] and that would fit that part of the definition.

Except that it is not a continuous variable,

why not?

If you claim that x in my example can take on values in the excluded range ...

I don't. But that is irrelevant to the part of the parts on the interval that isn't excluded.

Then my example is not a continuous distribution as its sample space is two distinct intervals.

No. How does having two distinct intervals in the sample space and all values excluded between the two intervals imply it is not a continuous distribution?
That is irrelevant.
All that is required for it to be a continuous distribution is to define at least ONE interval somewhere, ANYWHERE, along the support of the domain where it can take on all values between some value a and some value b; that's all! Nowhere in that link nor in any other will you find that it says you cannot have more than one such interval in a continuous distribution with no support in a gap between the two. That is because the word "continuous" in "continuous distribution" is referring to "continuous variable", which has nothing to do with whether there is a gap between intervals.

Originally posted by humy

Except that it is not a continuous variable,

why not?

OK, it looks like I have misunderstood the definition. So it is a continuous variable over two distinct ranges.

Originally posted by humy
All that is required for it to be a continuous distribution is to define at least ONE interval somewhere, ANYWHERE, along the support of the domain where it can take on all values between some value a and some value b; that's all!

Actually its a little bit more than that. Anywhere where it does take a value, it must be within one such interval. So hybrid versions are not continuous.

Originally posted by twhitehead
Actually its a little bit more than that. Anywhere where it does take a value, it must be within one such interval. So hybrid versions are not continuous.

hybrid versions are very rare. Out of all the conventional distribution I have come across or researched, plus the ones I made myself, not a single one of them was such a hybrid of the kind I spoke of there.

OK, now that we have settled what a 'continuous variable' is, I have two issues:
1. I would like to know if you can find an example of a probability density function that is not a continuous function over the intervals where the variable is continuous.
2. I would like to draw your attention to the wording of the definition where it says 'if the variable can take values'. Clearly there is an interval in my example where the variable cannot take a value. We then define a probability density function that typically maps values of the variable to a number between 0 and 1. Now where the variable does not 'take values' I would normally be inclined to say 'it is undefined'. I would then go further and say a function that maps those values where it is undefined, would also be undefined. Now I fully realise that it is not 'wrong' to instead assign a zero value, but it does feel wrong to me.

Originally posted by humy
hybrid versions are very rare. Out of all the conventional distribution I have come across or researched, plus the ones I made myself, not a single one of them was such a hybrid of the kind I spoke of there.

Yes, I am sure we can safely ignore them.

Originally posted by twhitehead
OK, now that we have settled what a 'continuous variable' is, I have two issues:
1. I would like to know if you can find an example of a probability density function that is not a continuous function over the intervals where the variable is continuous.
2. I would like to draw your attention to the wording of the definition where it says 'if the variable ...[text shortened]... lly realise that it is not 'wrong' to instead assign a zero value, but it does feel wrong to me.

1. I would like to know if you can find an example of a probability density function that is not a continuous function over the intervals where the variable is continuous.

Are you asking if there can be a probability density function that is a discontinuous function over the allowed values in sample space and therefore the support? i.e. can there be one which is a discontinuous function over a range of random variable x input where all values outputted by f(x) of the range are non-zero?

If so, I can contrive one for you:

A probability density function with support (0, 2] where, over the interval x ∈ (0, 1] , PDF(x) = 1/4, and over the interval x ∈ (1, 2] , PDF(x) = 3/4.
(PDF is the standard abbreviation for probability density function, not to be confused with probability! )

So here we have a probability density function that is not only a discontinuous function at points x=0 and x=2, but also at x=1 even though f(1)>0 ( PDF(1) = 1/4 to be more precise ).

2. I would like to draw your attention to the wording of the definition where it says 'if the variable can take values'.

it is 'if the variable can take values a and b in that range, ...' and you don't have to restrict your only options to a=0 and b=2. For example, why not let a=1/4 and b=1/2? You only need to give one such arbitrary chosen example of such an interval to then be able to say it is a continuous distribution.

Originally posted by humy
If so, I can contrive one for you:

I guess I should have thought of that.

Even more fun would be to let all irrational numbers be twice as likely as rationals. I wonder if a probability density function can be defined in that case.

it is 'if the variable can take values a and b in that range, ...' and you don't have to restrict your only options to a=0 and b=2. For example, why not let a=1/4 and b=1/2? You only need to give one such arbitrary chosen example of such an interval to then be able to say it is a continuous distribution.
No, it is not sufficient to find one such example. For it to be a continuous distribution, then every single point in the sample space must be part of an interval of non-zero width. That is what ensures that any single point probability is zero (the condition you dismissed as being irrelevant earlier).