1. Joined
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    28 Jan '16 21:147 edits
    Originally posted by twhitehead
    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.
    As I understand it, the mode of a probability distribution is, by the very definition of mode, the value of the random variable of the distribution that has the highest probability. But to be more accurate in the case of a continuous distribution, the mode is the value of the random variable that has the greatest probability density (which itself is not a true probability because you can only get a true probability over some integral of the probability density function ).

    See this apparently confirmed at:

    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, ..."

    But if that said mode has a value that the random variable has exactly zero probability density then I don't see how it would comply with that definition of mode. That's why I think there may be a need to, not to say it isn't a mode because I changed my mind about that since my OP, but rather say it is a different 'kind' of mode that doesn't exactly fit with the usual formal meaning of a mode.

    I am afraid I am always inclined to be rather pedantic like that.
  2. Cape Town
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    29 Jan '16 11:18
    Originally posted by humy
    As I understand it, the mode of a probability distribution is, by the very definition of mode, the value of the random variable of the distribution that has the highest probability.
    But what if there are two 'highest' points? What if the line is flat?

    But if that said mode has a value that the random variable has exactly zero probability density then I don't see how it would comply with that definition of mode.
    I was assuming that the point that we were talking about was outside the range over which your function is defined. Therefore the probability is not zero but undefined at the mode. If the probability is zero, then your function is not continuous and therefore doesn't fit the definition anyway.

    I am afraid I am always inclined to be rather pedantic like that.
    I agree that stretching a definition can lead to confusion just as stretching the definition of equality in the case of infinite series often causes confusion where people don't realise that the sum of an infinite series is not really the same thing as the sum of a finite series, even though they are typically named the same and use the same notation.
    But given that that is the common practice when dealing with limits (ie just tag on the special cases to the definition) I don't see why you can't do so for node for uniformities sake.
  3. Joined
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    29 Jan '16 16:1912 edits
    Originally posted by twhitehead
    But what if there are two 'highest' points? .
    Then there are two modes. Personally I don't have a problem with that.
    What if the line is flat?

    then you can either say there are an infinite number of modes within that range of flat line or say the mode isn't any one point but is a range of values, or, if you strictly define a mode as being a highest 'point' and insist that cannot be part of a plateau, you may say there is no mode; I haven't yet personally decided which I will go for; but it definitely be one of the three above. But, whichever one of these three I choose, don't see how it would effect my definition of my "unreachable mode" kind of mode as it isn't part of a 'flat line'.
    But if that said mode has a value that the random variable has exactly zero probability density then I don't see how it would comply with that definition of mode.
    I was assuming that the point that we were talking about was outside the range over which your function is defined.

    You are talking here about the 'point' being the value of x that is the mode thus that 'point' being in the domain of the function, NOT its codomain, right?
    -Not sure if I got your meaning here.

    Therefore the probability is not zero but undefined at the mode.

    don't understand your inference here with that above "Therefore":
    we are not talking here about pure mathematics but a probability distribution of some actual random variable X.
    Bearing in mind a mode is an x value, why cannot we mathematically define a value of x that has zero probability density i.e. has no chance of ever being observed? Why would that make that value of x "undefined" mathematically?
    For example, if you look at a uniform probability distribution as define in this link:

    https://en.wikipedia.org/wiki/Uniform_distribution_%28continuous%29

    it clearly implies that the x values ARE defined for f(x) that are in the ranges: x<a and x>b even though f(x) outputs 0 probability density for those x values.

    If the probability is zero, then your function is not continuous ...

    Unless I completely misunderstand what you say, this is not true. here is an example of a continuous distribution that has an x value (an infinite number of them in fact) in its domain with zero probability density in the codomain is a uniform probability distribution

    https://en.wikipedia.org/wiki/Uniform_distribution_%28continuous%29

    so, if you can have an x value with 0 density and the distribution is still continuous, why cannot you have a distribution that is such that its mode, which is an x value, has 0 density (but the closer you get to it without reaching it, the higher the density ) and have the distribution still continuous?

    -but I think I might have got your meaning totally wrong.
  4. Cape Town
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    29 Jan '16 21:212 edits
    Originally posted by humy
    Then there are two modes. Personally I don't have a problem with that.
    But you didn't make up a new name for it.

    then you can either say there are an infinite number of modes within that range of flat line or say the mode isn't any one point but is a range of values, or, if you strictly define a mode as being a highest 'point' and insist that cannot be part of a plateau, you may say there is no mode; I haven't yet personally decided which I will go for; but it definitely be one of the three above.
    But you wouldn't make up a new name.

    But, whichever one of these three I choose, don't see how it would effect my definition of my "unreachable mode" kind of mode as it isn't part of a 'flat line'.
    You can call it an 'unreachable mode', but only in the same sense as you might call the earlier ones 'multiple mode' and 'modal range'.

    don't understand your inference here with that above "Therefore":
    You have a function that is defined over a given range. That range is an open set with a limit at zero. The function is not defined at zero.

    Unless I completely misunderstand what you say, this is not true.
    https://en.wikipedia.org/wiki/Continuous_function

    Definition in terms of limits of functions
    The function f is continuous at some point c of its domain if the limit of f(x) as x approaches c through the domain of f exists and is equal to f(c).


    In your case the limit of f(x) as x approaches 0 is its maximum and not zero. But you are claiming that f(0) =0. Therefore the function is not continuous.

    but I think I might have got your meaning totally wrong.
    I am a programmer by trade not a mathematician. Its been over 20 years since I did my degree. So I could be entirely wrong about what you are talking about. But I don't think so.

    https://en.wikipedia.org/wiki/Probability_distribution#Continuous_probability_distribution
    A continuous probability distribution is a probability distribution that has a cumulative distribution function that is continuous.

    I am fairly certain that your cumulative distribution function is not continuous and therefore neither is your probability distribution.
  5. Joined
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    29 Jan '16 22:179 edits
    Originally posted by twhitehead
    But you didn't make up a new name for it.

    [b]then you can either say there are an infinite number of modes within that range of flat line or say the mode isn't any one point but is a range of values, or, if you strictly define a mode as being a highest 'point' and insist that cannot be part of a plateau, you may say there is no mode; I haven't yet per ...[text shortened]... distribution function is not continuous and therefore neither is your probability distribution.
    I think you have misunderstood what I have been saying;
    I am not saying x=0 is "not logically possible" in the mathematical sense as in not allowed in the domain of the function f! I am saying x=0 is"not logically possible" in the sense that it is impossible to observe x=0 in the real physical world (else this would result in a very subtle epistemological contradiction not explained here ); But I am saying x=0 is allowed in domain of the function f And f(0) = 0 BUT, as x approaches 0 without reaching 0, f(x) approaches some finite non-zero upper limit (the f(mode) upper limit). Let me put it this way: you can validly write: f(0) = 0 for my function f.


    You can think of the equation for my distribution very similar to that of an exponential distribution except one key difference is that when you mathematically input x=0 which you are mathematically allowed to do, because x=0 is not logically possible in the actual distribution in the real physical world, f(0) outputs 0 probability density.
  6. Standard memberDeepThought
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    29 Jan '16 23:57
    Originally posted by twhitehead
    But you didn't make up a new name for it.

    [b]then you can either say there are an infinite number of modes within that range of flat line or say the mode isn't any one point but is a range of values, or, if you strictly define a mode as being a highest 'point' and insist that cannot be part of a plateau, you may say there is no mode; I haven't yet per ...[text shortened]... distribution function is not continuous and therefore neither is your probability distribution.
    I'm not convinced by that definition of continuous for probability distributions. However if the requirement is that the cumulative probability function is continuous then Humy's distribution fits the requirement. The function was constructed by taking an absolutely continuous function and setting a point to zero. An isolated point has zero measure and will not contribute to the integral (*) so humy's probability density will be as continuous as the one it was constructed from (See [1]).

    (*) One has to be a little careful here since we're talking about distributions, but the isolated point has value zero and isn't a delta spike which is what would cause problems. I assume the original function was finite (or at least not too infinite) at the origin.

    [1] https://en.wikipedia.org/wiki/Lebesgue_integration
  7. Cape Town
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    30 Jan '16 06:35
    Originally posted by humy
    But I am saying x=0 is allowed in domain of the function f And f(0) = 0 BUT, as x approaches 0 without reaching 0, f(x) approaches some finite non-zero upper limit (the f(mode) upper limit). Let me put it this way: you can validly write: f(0) = 0 for my function f.
    I am not disputing whether or not f is a valid function. I am merely saying that f is not continuous. I didn't do statistics but I do know what a continuous function is.
    So the only question is whether or not f being continuous is required for the definition of 'mode'.
    DeepThought seems to suggest that only the integral of f need be continuous - but this leaves me wondering about discrete probabilities which, unless I am mistaken, have an integral of zero.
  8. Joined
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    30 Jan '16 08:339 edits
    Originally posted by twhitehead
    I am not disputing whether or not f is a valid function. I am merely saying that f is not continuous. I didn't do statistics but I do know what a continuous function is.
    So the only question is whether or not f being continuous is required for the definition of 'mode'.
    DeepThought seems to suggest that only the integral of f need be continuous - but thi ...[text shortened]... me wondering about discrete probabilities which, unless I am mistaken, have an integral of zero.
    this isn't exactly my distribution but it is similar to this exponential distribution:

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

    you will see it has a probability density function equations of:

    if x ≥ 0
    then f(λ|x) = λe^(-λx)

    if x < 0
    then f(λ|x) = 0

    and thus its support is:

    x ∈ [0, ∞ ) (so x=0 included in the support. Note domain is still (-∞ , ∞ ) )

    and mode is

    mode=0

    i.e. x=0 is both part of its support and mode = 0



    BUT, suppose we now modify that continuous distribution to become a new distribution D:

    density function equations for D:

    if x > 0 (note the change from ≥ to > )
    then f(λ|x) = λe^(-λx)

    if x ≤ 0 (note the change from < to ≤ )
    then f(λ|x) = 0


    and thus support for D is:

    x ∈ (0, ∞ ) (so x=0 excluded in the support. Note domain is still (-∞ , ∞ ) )

    and, providing you allow a certain subtle flexibility of the meaning of the word 'mode' so to allow for 'mode' not necessarily being where density is at its highest point, mode of D is still:

    mode=0

    i.e. x=0 is both not part of its support but we still have mode=0.

    OK, is D a continuous distribution just like the exponential one?
    If not, why not?

    Is x=0 still allowed in the domain for D? If not, why not?

    Do we still have mode=0? (at first I though the answer to that was "no" but I have changed my mind to "yes". I found that a bit conceptually tricky )
  9. Cape Town
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    30 Jan '16 08:531 edit
    From:
    https://en.wikipedia.org/wiki/Exponential_distribution
    Alternatively, this can be defined using the right-continuous Heaviside step function, H(x) where H(0)=1:

    From:
    https://en.wikipedia.org/wiki/Heaviside_step_function
    The Heaviside step function, or the unit step function, usually denoted by H (but sometimes u or theta ), is a discontinuous function whose value is zero for negative argument and one for positive argument.


    The exponential distribution as defined is not a continuous function - neither is yours.
  10. Joined
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    30 Jan '16 09:0713 edits
    Originally posted by twhitehead


    The exponential distribution as defined is not a continuous function ....
    I am not aware I said anything about a "continuous function".
    I said my function is a "continuous probably distribution".
    I am not sure exactly what you mean by "continuous function" here but I suspect it doesn't mean the same thing as "continuous probably distribution".

    If you are saying an exponential distribution is not a "continuous probably distribution" by saying it is not a "continuous function" then can you give me a link that specifically says exponential distribution is not a "continuous probably distribution" as opposed to being said to be not a "continuous function"?
    I tied to find a link myself to say one way or another but failed to find one specifically for the exponential distribution but managed to find:
    https://en.wikipedia.org/wiki/List_of_probability_distributions
    where it clearly lists the "wrapped exponential distribution", that I think you would agree is similar to the more usual exponential distribution, under "Continuous distributions". Is a "wrapped exponential distribution" also not a "continuous function"?


    If you are not saying an exponential distribution is not a "continuous probably distribution" then not sure what you are disagreeing with.
  11. Cape Town
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    30 Jan '16 12:51
    I was going by this:
    https://en.wikipedia.org/wiki/Probability_distribution#Continuous_probability_distribution
    A continuous probability distribution is a probability distribution that has a cumulative distribution function that is continuous.

    Your exponential function is not a continuous function. If your exponential function is the 'cumulative distribution function' then it doesn't fit the definition. If your exponential function is the 'probability distribution' then I could be wrong as I am not certain what the relationship is between 'probability distribution' and 'cumulative distribution function' .

    I have a sneaky feeling that open sets cannot be integrated.
  12. Joined
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    30 Jan '16 14:143 edits
    Originally posted by twhitehead
    I was going by this:
    https://en.wikipedia.org/wiki/Probability_distribution#Continuous_probability_distribution
    A continuous probability distribution is a probability distribution that has a cumulative distribution function that is continuous.

    Your exponential function is not a continuous function. If your exponential function is the 'cu ...[text shortened]... relationship is between 'probability distribution' and 'cumulative distribution function' .
    .
    I have read your posts again and again and I still don't get what you are saying regarding why neither my or the conventional exponential distribution is a continuous probability distribution.

    OK, why is a continuous uniform distribution a "continuous probability distribution" but an exponential distribution in NOT a "continuous probability distribution"? I mean, exactly what feature does one have that the other doesn't that makes this difference?

    They BOTH are of a continuous random variable, right?
    And they BOTH allow a continuous random variable X over continuous ranges of x values (such as from x=1 to x=2) in their domain, right?

    And is there any difference between a "continuous probability function" and a "continuous probability distribution"?

    I have a sneaky feeling that open sets cannot be integrated

    I don't see why.
    The probability of X∈[a, b] in, lets say a continuous uniform distribution, is mathematically the same as the probability of X∈(a, b) ( or X∈]a, b[ if you prefer like I do ) i.e. P(X∈[a, b] ) = P(X∈(a, b) ), right?
    So both equal the integral ∫[x=a, b] ...(formula here)... dx even though, as far as I am aware and I may be wrong about this bit, you are not allowed to write ∫(x=a, b) ...(formula here)... dx or ∫]x=a, b[ ...(formula here)... dx (can somebody please correct me on that if I got that bit wrong ).
  13. Joined
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    30 Jan '16 14:264 edits
    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 distribution"?

    If nobody answers here, I will take this to a maths forum and ask the real maths experts.
  14. Cape Town
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    30 Jan '16 15:07
    Originally posted by humy
    I have read your posts again and again and I still don't get what you are saying regarding why neither my or the conventional exponential distribution is a continuous probability distribution.
    Well maybe you can answer this for me first:
    Is the exponential function the 'cumulative distribution function ' or is it the 'probability distribution' and what is the relationship between the two?

    I have no doubt that the exponential function is not continuous but if it is not the 'cumulative distribution function' then that might not be a problem.
  15. Standard memberDeepThought
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    30 Jan '16 15:20
    Originally posted by twhitehead
    I am not disputing whether or not f is a valid function. I am merely saying that f is not continuous. I didn't do statistics but I do know what a continuous function is.
    So the only question is whether or not f being continuous is required for the definition of 'mode'.
    DeepThought seems to suggest that only the integral of f need be continuous - but thi ...[text shortened]... me wondering about discrete probabilities which, unless I am mistaken, have an integral of zero.
    You quoted this from Wikipedia: "A continuous probability distribution is a probability distribution that has a cumulative distribution function that is continuous." - since the cumulative distribution is the integral of the probability distribution then what the snippet from Wikipedia seems to be saying is that the definition of continuity for probability distributions depends on the continuity of the cumulative distribution. Although I'm a little skeptical of that definition.
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