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  1. 27 Jan '16 11:20 / 12 edits
    can anyone give me an example of a continuous probability distribution that has no definable mode or modes?
    I am note talking here about, say, a continuous uniform distribution, which arguably has an infinite number of modes along a 'plateau' on its probability density function thus merely has no unique mode/modes. I am talking here about a continuous distribution that literally has no valid/definable mode/modes whatsoever!

    I believe I may have found (or more like ~invented) such a type of 'modeless' continuous distribution (which I have named an "expon paracav" ) because where the random variable X value has its highest point on the graph for its probability density equation, its 'apparent' mode value (X=0 in this case) if mathematically applied/interpreted naively, is not allowed else this leads to an epistemological contradiction. But I wondered how unusual or unique that property is. So I search the net for another example of a 'modeless' continuous distribution but got absolutely nowhere.
    Anyone?
  2. Standard member sonhouse
    Fast and Curious
    27 Jan '16 13:31 / 3 edits
    Originally posted by humy
    can anyone give me an example of a continuous probability distribution that has no definable mode or modes?
    I am note talking here about, say, a continuous uniform distribution, which arguably has an infinite number of modes along a 'plateau' on its probability density function thus merely has no unique mode/modes. I am talking here about a continuous d ...[text shortened]... or another example of a 'modeless' continuous distribution but got absolutely nowhere.
    Anyone?
    You might find more help going to a math journal than scrounging around on the net. Maybe contacting known mathematicians. I would contact Terrence Tao if I could.

    If a guy with an IQ of 230 can't help, I don't know who could.

    I bet he would be interested in your question too.
  3. 27 Jan '16 17:50
    Originally posted by sonhouse
    ...I would contact Terrence Tao if I could.

    If a guy with an IQ of 230 can't help, I don't know who could.

    I bet he would be interested in your question too.
    He has an IQ of 230!? wow.
    I don't exactly 'need' nor desperate for an answer, just slightly curious, that's all. I think I prefer to wait until if or when I desperately need an answer to an extremely difficult problem I am totally stuck on before asking an astonishing mathematical genius like him.
  4. Standard member DeepThought
    Losing the Thread
    27 Jan '16 21:17
    Originally posted by humy
    can anyone give me an example of a continuous probability distribution that has no definable mode or modes?
    I am note talking here about, say, a continuous uniform distribution, which arguably has an infinite number of modes along a 'plateau' on its probability density function thus merely has no unique mode/modes. I am talking here about a continuous d ...[text shortened]... or another example of a 'modeless' continuous distribution but got absolutely nowhere.
    Anyone?
    When you say mode how exactly are you defining it? If one defines it as the point for which the function is maximal in a range then all finite distributions have at least one mode. If the mode is not unique then one can define a unique mode by taking the average over the set of candidates. If the distribution diverges then provided its integral over the whole range is finite any of the infinite points would do as a mode. If there is more than one of them then one could see if the integral over some small neighbourhood centred on the point is bigger for one of the candidates. So the only way I can see you could get this is if you were to construct some weird distribution, which clearly has an infinite number of maxima all the same, but no sane way of taking an average over the set because it's not finite and doesn't have the same measure as your base space, such as if all the maxima were on the rationals.

    Since you seem to have multiple modes for them not to be identifiable you'd need to have the function diverging and no way of comparing the "height" of the distribution there. I don't see how that is possible. I'm worried about your statement "its 'apparent' mode value ... is not allowed else this leads to epistemological contradiction.". If you've found a contradiction then there is something wrong with what you are doing and it's probably not that the mode is not really a mode.
  5. 27 Jan '16 22:20 / 8 edits
    Originally posted by DeepThought
    When you say mode how exactly are you defining it? If one defines it as the point for which the function is maximal in a range then all finite distributions have at least one mode. If the mode is not unique then one can define a unique mode by taking the average over the set of candidates. If the distribution diverges then provided its integral over t ...[text shortened]... omething wrong with what you are doing and it's probably not that the mode is not really a mode.
    https://en.wikipedia.org/wiki/Mode_%28statistics%29
    "...The mode of a continuous probability distribution is the value x at which its probability density function has its maximum value, so the mode is at the peak...."

    now look at the graph for the probability density function for the exponential distribution at:

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

    which looks very similar to but isn't the same as that of my distribution that I found. If you look along the x axis, you will see that where x=0, P(x) is some definable finite number that depends on what parameter λ of the distribution is. Clearly that is the mode of this distribution because that is the highest point on that curve and the only highest point on that curve. And if you randomly pick an x out of the random variable X out of the actual distribution, it must be logically possible for that x to equal 0.

    But my distribution differs from that because although if you plot the equation for the density function, it appears to clearly make highest point of the curve be at x=0 and clearly with no other possible highest point, unlike with the exponential distribution, x=0 is NOT logically possible but any value greater than x, no matter how close to 0, is logically possible. In other words, the support on my function is not
    x ∈ [0, infinity) like it is the the exponential function, but rather x ∈ (0, infinity).

    So just look at the graph for the probability density function for the exponential distribution and imagine the hypothetical for a moment that x=0 is NOT a logical possibility i.e. randomly picking x with a value of exactly 0 is not logically possible because the actual distribution cannot have x=0: what would the mode of that function be if not 0? I cannot see any other candidate for the mode, can you? This would be exactly the same reason why there doesn't appear to be any definable mode on my distribution.

    (I haven't done anything here to explain WHY x=0 is not logically possible for my distribution; but that is far too complicated to explain here properly and is an entirely different matter. In this case, there existing a x=0 will lead to a contradiction in reality; which of course makes an instance of x=0 in reality nonsense. But any instance of x>0 leads to no contradiction )
  6. Standard member DeepThought
    Losing the Thread
    27 Jan '16 23:39 / 2 edits
    Originally posted by humy
    https://en.wikipedia.org/wiki/Mode_%28statistics%29
    "...The mode of a continuous probability distribution is the value x at which its probability density function has its maximum value, so the mode is at the peak...."

    now look at the graph for the probability density function for the exponential distribution at:

    https://en.wikipedia.org/wiki/Exponentia ...[text shortened]... kes an instance of x=0 in reality nonsense. But any instance of x>0 leads to no contradiction )
    With you now. You've defined your distributions to have domain x > 0 as a consequence of which your function has no largest value. I don't think that your function is unique, in the sense that an infinite number of different functions will have the same property, since any continuous function whose largest value is at zero has no mode in the range x > 0 because any local maximum will still be smaller than the function is in the region x < epsilon for some epsilon sufficiently small eg. 1/(x+1).

    Note that you can still define a mode, in your example x = 0 is not in the domain of the function, but it is a limit point of the domain so I don't see why you shouldn't define the mode to be there even though the function isn't defined there. It is information carrying as it tells one that values close to zero are more likely than values further from zero.
  7. 28 Jan '16 00:00 / 2 edits
    Originally posted by sonhouse to humy
    You might find more help going to a math journal than scrounging around on the net.
    Maybe contacting known mathematicians. I would contact Terrence Tao if I could.

    If a guy with an IQ of 230 can't help, I don't know who could.
    I bet he would be interested in your question too.
    Terence Tao writes an excellent blog on mathematics. That said, higher mathematics is
    extremely specialized, and I don't know if he's particularly interested in probability distributions.
    By the way, some famous mathematicians receive much nuisance mail from cranks,
    so they may lack the time and inclination to reply to every letter.
  8. 28 Jan '16 08:25 / 3 edits
    Originally posted by DeepThought
    You've defined your distributions to have domain x > 0 as a consequence of which your function has no largest value. I don't think that your function is unique, in the sense that an infinite number of different functions will have the same property ...

    Arr, yes, but these are just pure mathematical functions that any idiot can arbitrary contrive. But how many of them are for known probability distribution functions that you know of if any that have some real application (like mine although I haven't explained how so) that have that property?

    Note that you can still define a mode, in your example x = 0 is not in the domain of the function, but it is a limit point of the domain so I don't see why you shouldn't define the mode to be there even though the function isn't defined there. It is information carrying as it tells one that values close to zero are more likely than values further from zero.


    Thanks for that: I appreciate your perspective which was rather different from mine.
    I will now seriously consider defining the mode that way as opposed to saying simply it 'doesn't exist'.
    If I do define the mode that way, I guess I will give it a special name to emphasize that x=mode is not logically possible in reality here. At the moment, I am thinking of calling it in my book something like an "improper mode" so the much more common default mode is a "proper mode"; or something like that.
  9. Standard member sonhouse
    Fast and Curious
    28 Jan '16 12:00
    Originally posted by Duchess64
    Terence Tao writes an excellent blog on mathematics. That said, higher mathematics is
    extremely specialized, and I don't know if he's particularly interested in probability distributions.
    By the way, some famous mathematicians receive much nuisance mail from cranks,
    so they may lack the time and inclination to reply to every letter.
    Sorry, one 'r'. Got it.
  10. Standard member DeepThought
    Losing the Thread
    28 Jan '16 17:15
    Originally posted by humy
    You've defined your distributions to have domain x > 0 as a consequence of which your function has no largest value. I don't think that your function is unique, in the sense that an infinite number of different functions will have the same property ...

    Arr, yes, but these are just pure mathematical functions that any idiot can arbitrary co ...[text shortened]... "improper mode" so the much more common default mode is a "proper mode"; or something like that.
    I studied theoretical physics at university so I tend to think of any suitably normalized function as a viable distribution. From memory I think this applies to the zero angular momentum wavefunctions of the hydrogen atom (certainly for the 1s state, what I can't remember is if this applies to non-ground state wave functions), which would provide an infinite set of distributions with a mode at r = 0.

    You could try some formulation like limiting mode to indicate that it refers to a limit point that's not an interior point.
  11. 28 Jan '16 18:06 / 23 edits
    Originally posted by DeepThought
    ...
    You could try some formulation like limiting mode to indicate that it refers to a limit point that's not an interior point.
    perhaps something along the lines of;

    mode → 0, mode ≠ 0

    ?

    or perhaps I should indicate an "improper mode" with this new unconventional notation instead:

    mode →≠ 0

    which would simply be shorthand for the more conventional notation of;
    mode → 0, mode ≠ 0

    and where " →≠ " means something along the lines of; " tending towards the limit of but never equal to the limit of "
    ?
  12. 28 Jan '16 19:11
    Originally posted by humy
    perhaps something along the lines of;

    mode → 0, mode ≠ 0

    ?

    or perhaps I should indicate an "improper mode" with this new unconventional notation instead:

    mode →≠ 0

    which would simply be shorthand for the more conventional notation of;
    mode → 0, mode ≠ 0

    and where " →≠ " means something along the lines of; " tending towards the limit of but never equal to the limit of "
    ?
    Actually, just thought right now;
    -if I have that then I can simply call that " →≠ " to indicate an "unreachable limit" as apposed to " →= " to indicate a "reachable limit" and then I don't call that mode an "improper mode" but rather more appropriately and very simply call it an "unreachable mode". Simple!
  13. 28 Jan '16 20:03
    I really don't see why you want to name it different or treat it different. Mode is somewhat loosely defined anyway, I see no reason why you shouldn't just extend it to include limits that aren't members of the set.
  14. 28 Jan '16 20:04
    Originally posted by humy
    can anyone give me an example of a continuous probability distribution that has no definable mode or modes?
    From Wikipedia:
    https://en.wikipedia.org/wiki/Mode_(statistics)#Uniqueness_and_definedness
    Certain pathological distributions (for example, the Cantor distribution) have no defined mode at all.
  15. 28 Jan '16 20:13
    Originally posted by twhitehead
    From Wikipedia:
    https://en.wikipedia.org/wiki/Mode_(statistics)#Uniqueness_and_definedness
    Certain pathological distributions (for example, the Cantor distribution) have no defined mode at all.
    oh I think that is weird!