1. Joined
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    15 Dec '16 19:2810 edits
    I have worked out the integral for the mean average, assuming it is defined, of a probability distribution I have discovered, must be;

    ∫[–∞, ∞] x * |h – u| / ( (h – u)^2 + (x – u)^2 ) dx
    h, u, x ∈ ℝ
    h, u, x ∈ (–∞, ∞ )
    h – u ≠ 0

    h and u are the parameters of this distribution while x is the random variable of this distribution.

    I have already established that the variable u is both the mode and the median for this distribution (mode = median = u) which, like a normal distribution, has a symmetrical density function curve (of probability_density(x) = |h – u| / ( (h – u)^2 + (x – u)^2 ) ) symmetrical on either side of the mode.

    According to WolframAlpha, this integral diverges (implying mean undefined) unless you use a "chuchy principle value" ( https://en.wikipedia.org/wiki/Cauchy_principal_value ) a maths concept I am unfamiliar with and don't understand and seems to often give a mean average contrary to that which I would expect if mean is defined.

    But when I use numerical methods using my usual java program for obtaining averages which as always seemed trustworthy in the past, the mean average outputted converges simply on u, which is just what I would expect if this mean is defined.

    So is the mean defined in this case? If so, it is simply u, right?
    I really want to know because I want to know whether I should say in my book the mean is defined as u or if I should say mean is undefined for this distribution.
  2. Standard memberDeepThought
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    16 Dec '16 01:45
    Originally posted by humy
    I have worked out the integral for the mean average, assuming it is defined, of a probability distribution I have discovered, must be;

    ∫[–∞, ∞] x * |h – u| / ( (h – u)^2 + (x – u)^2 ) dx
    h, u, x ∈ ℝ
    h, u, x ∈ (–∞, ∞ )
    h – u ≠ 0

    h and u are the parameters of this distribution while x is the random variable of this distribution.

    I have already esta ...[text shortened]... book the mean is defined as u or if I should say mean is undefined for this distribution.
    A Cauchy principle value is a way of assigning a finite result to the problem of assigning a number to an integral which is not absolutely convergent.

    The integral isn't difficult. With a change of parameters, the integral is:

    ∫[–∞, ∞] x / ( b^2 + (x – a)^2 ) dx

    where I've ignored the factor of |h - u| as it is just a constant multiplier and I've renamed u as a = u, and introduced b = h - u. Note that the integral is logarithmically divergent as it is written. To find a result for the integral we simply ignore that and proceed.

    Changing variable so that x -> x' = x + a, this can be rewritten as:

    ∫[–∞, ∞] (x + a) / ( x^2 + b^2 ) dx = ∫[–∞, ∞] x / ( x^2 + b^2 ) dx + ∫[–∞, ∞] a / ( x^2 + b^2 ) dx = ∫[–∞, ∞] a / ( x^2 + b^2 ) dx ~ a/|b| (if |b| > 0)

    Where the second equality follows since the integrand x/(x^2 + b^2) is odd so its definite integral vanishes when the limits of integration are symmetric about the origin as they are in this case. The final term is integrated to an inverse tangent (unless b = 0), see [1]. I haven't done this carefully hence the '~' at the end of my working, I think there should be a factor of pi in there as well, I just got back from the pub, so if this disagrees with your results then it's not a surprise...

    [1] https://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions#Integrands_of_the_form_xm_.2F_.28a_x2_.2B_b_x_.2B_c.29n
  3. Joined
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    16 Dec '16 08:0011 edits
    Originally posted by DeepThought
    A Cauchy principle value is a way of assigning a finite result to the problem of assigning a number to an integral which is not absolutely convergent.

    The integral isn't difficult. With a change of parameters, the integral is:

    ∫[–∞, ∞] x / ( b^2 + (x – a)^2 ) dx

    where I've ignored the factor of |h - u| as it is just a constant multiplier and ...[text shortened]... /List_of_integrals_of_rational_functions#Integrands_of_the_form_xm_.2F_.28a_x2_.2B_b_x_.2B_c.29n
    Arr; I have just spotted an edit error I have been repeatedly making which could help explain at least some of my confusion!
    I have edited the equation in wrong here because I kept leaving out pi !

    The integral for the mean average, just as I correctly written it in my computer program but had incorrectly edited in the OP, should have been written;

    ∫[–∞, ∞] x * |h – u| / ( pi((h – u)^2 + (x – u)^2) ) dx

    and this is for a probability distribution with probability density defined by;

    probability_density(x) = |h – u| / ( pi((h – u)^2 + (x – u)^2 )) )

    OK, NOW when I enter that integral into wolframAlpha, although it still says it doesn't converge, it then says using catchy principle value that it has the required value of u just as I would expect if mean average is defined.

    So, given that the catchy principle value says the integral for the mean (IF defined) is u, does that mean, despite the integral for that mean average being divergent, the mean average for this distribution IS defined or not?
    My intuition says it is defined but need to be absolutely sure before writing that as a fact my book. I don't want any falsehoods asserted in my book!
  4. Standard memberDeepThought
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    16 Dec '16 08:491 edit
    Originally posted by humy
    Arr; I have just spotted an edit error I have been repeatedly making which could help explain at least some of my confusion!
    I have edited the equation in wrong here because I kept leaving out pi !

    The integral for the mean average, just as I correctly written it in my computer program, should have been written;

    ∫[–∞, ∞] x * |h – u| / ( [b]pi
    ( ...[text shortened]... ral for the mean is u, does that means the mean average for this distribution is defined or not?[/b]
    That additional factor of pi means the result of the integral is u:

    ∫[–∞, ∞] |h - u| x / (pi ( (x - u)^2 + (h - u)^2 )) dx = (|h - u| / pi) * ∫[–∞, ∞] (x + u) / ( x^2 + (h - u)^2 ) dx = (|h - u| / pi) * ∫[–∞, ∞] u / ( x^2 + (h - u)^2 ) dx = (|h - u|/pi)*u*(pi / |h - u|) = u

    Provided your distribution produces sensible results I wouldn't worry about the integral being formally divergent, fundamental physics is full of divergent integrals.
  5. Joined
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    16 Dec '16 08:582 edits
    Originally posted by DeepThought
    That additional factor of pi means the result of the integral is u:

    ∫[–∞, ∞] |h - u| x / (pi ( (x - u)^2 + (h - u)^2 )) dx = (|h - u| / pi) * ∫[–∞, ∞] (x + u) / ( x^2 + (h - u)^2 ) dx = (|h - u| / pi) * ∫[–∞, ∞] u / ( x^2 + (h - u)^2 ) dx = (|h - u|/pi)*u*(pi / |h - u|) = u
    Fantastic!
    OK, unless you tell me this is not necessarily true, I am going to stick my neck out and say in my book that that means the mean average is defined for this distribution (and is u, of course ).
  6. Standard memberDeepThought
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    16 Dec '16 09:07
    Originally posted by humy
    Fantastic!
    OK, unless you tell me this is not necessarily true, I am going to stick my neck out and say in my book that that means the mean average is defined for this distribution (and is u, of course ).
    See my edit above. I don't know enough about distributions to make a judgement, but they are more general than functions in the sense that integrals over things like the Dirac delta function are defined, which don't make sense in the normal theory. So I don't think there's a problem.
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