Originally posted by @eladar
I'm not sure how you can have a finite group of numbers and not be able to find a median.

in this case it isn't a finite group of numbers.

Originally posted by @humy
in this case it isn't a finite group of numbers.

How can any infinte set have a median?

Originally posted by @humy
Well you would think so! I thought so!

This particular distribution is for a discrete random variable g that can is a natural number and is thus this is for a probability mass function.
It always normalizes for all allowed input values.

However, and this is a bit that had me confused for nearly over a week, for certain input values (other than g itself ...[text shortened]... 1/2 (or over) for ANY allowed g input value! ?
-I would say this makes nonsense of the median.

You've probably balls up the integral. The median point m of a distribution is the point such that the integral of the distribution to the left of said point (i.e. between -infinity and m) is equal to the integral to the right. For that not to be well defined on a normalized function you'd have to have a function that is integrable over the whole number line, but not over subintervals, which is a contradiction, or regions where the distribution is negative, producing more than one median, which produces problems trying to interpret it as a probability density (although note that solutions to the Klein-Gordon equation have this property). So I suggest you check you've done the integral correctly.

Originally posted by @deepthought
You've probably balls up the integral. The median point m of a distribution is the point such that the integral of the distribution to the left of said point (i.e. between -infinity and m) is equal to the integral to the right. For that not to be well defined on a normalized function you'd have to have a function that is integrable over the whole numb ...[text shortened]... Gordon equation have this property). So I suggest you check you've done the integral correctly.

So he is calculating areas.

Originally posted by @deepthought
The median point m of a distribution is the point such that the integral of the distribution to the left of said point (i.e. between -infinity and m) is equal to the integral to the right.

I knew that.
But what if, for some particular input values, the distribution ceases to be a purely numerical one because it becomes partly a categorical one because there is a non-zero probability greater than 1/2 of the outcome being not represented by ANY number (not even x=0 ) thus that outcome is, metaphorically speaking, "off the cumulative distribution function's (CDF) radar" ? Then there is no median, right?
At least for a continuous distribution (this one is actually a discrete one but lets not complicate this so lets pretend it is a continuous one ), the median is defined as x being such that CDF(x) = 1/2. But if there is no x that is such that CDF(x) = 1/2, what is the median?
The distribution still normalizes (like all distributions MUST to be valid) but only when you ADD the probability of a none numerical outcome with the limit of the CDF as the numerical outcome x tends to +infinity BUT that limit is some number less than 1/2 so never 'reaches' the "CDF(x) = 1/2" output requirement for there to be a median.

Let me put it this way; there is a particular allowed set of input values that is such that it results in a 2/3 probability mass of there being that none numerical outcome and a 1/3 probability mass of there being any one of the possible numerical outcomes thus, as random variable x tends to +infinity, CDF(x) tends to 1/3. So, with THAT input set, how can there be an allowed x that is such that CDF(x) = 1/2 ? The codomain of CDF in that particular case is, extremely unusually, the [0, 1/3) interval rather than the much much more usual (0, 1) or (0, 1] or [0, 1) or [0, 1].

Originally posted by @humy
I knew that.
But what if, for some particular input values, the distribution ceases to be a purely numerical one because it becomes partly a categorical one because there is a non-zero probability greater than 1/2 of the outcome being not represented by ANY number (not even x=0 ) thus that outcome is, metaphorically speaking, "off the cumulative distribution ...[text shortened]... the [0, 1/3) interval rather than the much much more usual (0, 1) or (0, 1] or [0, 1) or [0, 1].

This is from my son in law:


'About your question:

There is something not right about infinite medians.
Infinite MEAN values are commonplace. But infinite medians are a problem...
The reason is the following:
If the MEAN is infinite, all this signifies is that there are enough large events that more than compensate the the many more smaller events. For example, the Cauchy distribution has infinite mean and variance.
But an infinite MEDIAN means that only HALF the random numbers are ltess than infinite, while the other half are equal to or greater than infinite.
Obviously this is impossible for a NORMALIZABLE probability distribution.
By normalizable, I mean that the sum or integral of all probabilities equals 1 exactly.
Does my answer help?"

best wishes
Gandhi

Originally posted by @sonhouse
This is from my son in law:


'About your question:

There is something not right about infinite medians.
Infinite MEAN values are commonplace. But infinite medians are a problem...
The reason is the following:
If the MEAN is infinite, all this signifies is that there are enough large events that more than compensate the the many more smaller event ...[text shortened]... r integral of all probabilities equals 1 exactly.
Does my answer help?"

best wishes
Gandhi

Large events meaning infintely large or just large?

Originally posted by @humy
I knew that.
But what if, for some particular input values, the distribution ceases to be a purely numerical one because it becomes partly a categorical one because there is a non-zero probability greater than 1/2 of the outcome being not represented by ANY number (not even x=0 ) thus that outcome is, metaphorically speaking, "off the cumulative distribution ...[text shortened]... the [0, 1/3) interval rather than the much much more usual (0, 1) or (0, 1] or [0, 1) or [0, 1].

If some results are not describable with numbers it's not obvious to me how mathematics can say anything about it. If you are saying that the median ends up infinite then your distribution is not normilisable, so it's not a distribution. My intuition is that this is an if and only if relationship - there is a finite median if and only if the distribution is normilisable.

Originally posted by @deepthought
If some results are not describable with numbers it's not obvious to me how mathematics can say anything about it. If you are saying that the median ends up infinite then your distribution is not normilisable, so it's not a distribution. My intuition is that this is an if and only if relationship - there is a finite median if and only if the distribution is normilisable.

That's pretty much what Gandhi said.

Originally posted by @sonhouse
This is from my son in law:


'About your question:

There is something not right about infinite medians.
Infinite MEAN values are commonplace. But infinite medians are a problem...
The reason is the following:
If the MEAN is infinite, all this signifies is that there are enough large events that more than compensate the the many more smaller event ...[text shortened]... r integral of all probabilities equals 1 exactly.
Does my answer help?"

best wishes
Gandhi

well one would naturally think that. That is why I had such a hard time getting my head around it.
I believe I have found an exception to the implicit " this is impossible for a NORMALIZABLE probability distribution. " 'rule'.

Originally posted by @deepthought
- there is a finite median if and only if the distribution is normilisable.

I think you misunderstand. There are some (not all) possible combination of input values that result in the distribution being without a median and even without a median it DOES normalize because you need to ADD the probability mass of the possible NONE numerical outcome (which cannot be even called "x=0" ) with the probability mass of all the possible NUMERICAL outcomes to see how it normalizes (those two things sum to 1 as required) BUT because the probability mass of the possible NONE numerical outcome is greater than 1/2 there is no meaningful median.

ANYONE;

Just a new thought; perhaps I shouldn't say that "there is no median" in that situation but rather say "the median is the none numerical outcome that consists of NO outcome from the random sampling process because the probability mass of the distribution has been modified by the previous random sampling to now become unsamplable is greater than 1/2".
But would that make any sense saying that?
I mean, would saying that be just gibberish because of the way the median is defined?

Originally posted by @humy
I think you misunderstand. There are some (not all) possible combination of input values that result in the distribution being without a median and even without a median it DOES normalize because you need to ADD the probability mass of the possible NONE numerical outcome (which cannot be even called "x=0" ) with the probability mass of all the possible NUMERIC ...[text shortened]... ng that?
I mean, would saying that be just gibberish because of the way the median is defined?

I just want to do a reality check. You have some random variable x, which can be any positive semi-definite real number, f(x) is a function which gives the probability of the variable being in some small range of width ∆x centred on x as f(x)∆x + O(∆x^2).

For the function to be normilisable your function must approach zero as the random variable diverges (f(x) -> 0 in the limit that x -> ∞.). You seem to have f(∞ ) ≠ 0, which is fine provided you have extended the real number line and added a point at infinity at which your function is discontinuous. So you seem to have a multiple of a Dirac delta function at the point at infinity. Is this what you are saying?

Originally posted by @deepthought
I just want to do a reality check. You have some random variable x, which can be any positive semi-definite real number, f(x) is a function which gives the probability of the variable being in some small range of width ∆x centred on x as f(x)∆x + O(∆x^2).

For the function to be normilisable your function must approach zero as the random variable div ...[text shortened]... ave a multiple of a Dirac delta function at the point at infinity. Is this what you are saying?

You have some random variable x, which can be any positive semi-definite real number

random variable x in this case is always a fully defined real number. But the confusing part here is that sometimes an x as an outcome isn't always the only possible outcome. Sometimes a non-x outcome can be so which you cannot even represent it with "x=0" because it is actually NO x.

For the function to be normilisable your function must approach zero as the random variable diverges (f(x) -> 0 in the limit that x -> ∞.).

I'm not sure if that is a misedit. Are you referring to the limit of the cumulative distribution function (CDF) of random variable x as x tends to +infinity? If so, I think you are trying to say "(f(x) -> 1 in the limit that x -> ∞.). " and not "(f(x) -> 0 in the limit that x -> ∞.). " ( and where "f(x)" means "CDF(x) " ).
What I am saying is, for certain inputs (other than x itself) lim{x -> ∞} CDF(x) < 1 but the function still normalizes because the non-numerical part is, metaphorically speaking, 'off the CDF radar". To see how it normalizes, you need to do this;

probability_mass(non_numerical_outcome) + lim{x -> ∞} CDF(x) = 1 (as required to normalize)

when with much more typical numerical distributions you would, of course, only have to do this to see how it normalizes;

lim{x -> ∞} CDF(x) = 1 (as required to normalize)

You seem to have f(∞ ) ≠ 0

correct. x must be finite in this case else total nonsense.

So you seem to have a multiple of a Dirac delta function at the point at infinity. Is this what you are saying?

I admit I always really struggle with the Dirac delta function so don't know.

Originally posted by @humy

You have some random variable x, which can be any positive semi-definite real number

random variable x in this case is always a fully defined real number. But the confusing part here is that sometimes an x as an outcome isn't always the only possible outcome. Sometimes a non-x outcome can be so which you cannot even represent it with "x=0 ...[text shortened]... ying? [/quote]
I admit I always really struggle with the Dirac delta function so don't know.

The CDF is the indefinite integral of the function f(x). So I'm saying that for limit x-> ∞ CDF(x) -> 1 we need limit x->∞ f(x) -> 0. Otherwise the integral won't converge, f(x) is our distribution and this is a strong constraint on it.

Suppose we are conducting some experiment the outcome of which is a real number or "fail". So after 100 trials we have 94 real numbers and 6 instances of "fail". Is this a good example of what you mean by "non-x outcomes"?

Originally posted by @deepthought
The CDF is the indefinite integral of the function f(x). So I'm saying that for limit x-> ∞ CDF(x) -> 1 we need limit x->∞ f(x) -> 0. Otherwise the integral won't converge, f(x) is our distribution and this is a strong constraint on it.

OH RIGHT! Sorry about that but I completely misunderstood what you were saying but I am with you now;
Yes, for my distribution, it is indeed limit x->∞ f(x) -> 0 as required.

Suppose we are conducting some experiment the outcome of which is a real number or "fail". So after 100 trials we have 94 real numbers and 6 instances of "fail". Is this a good example of what you mean by "non-x outcomes"?

Yes. That is one valid way to view it.
Note if we have a certain combination of input values (other than x itself) then there is a greater probability of "fail" as in NO x than there is of "numerical outcome" as in an x.