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Calculating a moment of a distribution

  1. Oct 3, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    The Lorentz Distribution is given by

    [tex]p(x) = \frac{1}{pi}\frac{ϒ}{(x-a)^2+ϒ^2}[/tex]

    for -∞≤x≤∞.

    a) Sketch the distribution for a=0, ϒ=1, and compare the form to the Gaussian Distribution.
    b) Calculate the first moment of the distribution assuming again that a =0 and ϒ =1.
    c) Does the second moment exist?

    2. Relevant equations

    Definition of moment: https://en.wikipedia.org/wiki/Moment_(mathematics)
    Cauchy (Lorentz) distribution: https://en.wikipedia.org/wiki/Cauchy_distribution#Mean

    First moment is the mean
    Second moment is the variance
    3. The attempt at a solution

    The first moment, being mean, is thus equated by
    [tex]
    \lim_{\substack{a\rightarrow ∞}}
    \int_{-2a}^{a} xp(x) dx = \lim_{\substack{a\rightarrow ∞}}\int_{-2a}^{a} \frac{x}{pi}\frac{1}{x^2+1} dx = \frac{-log(4)} {2*pi}[/tex]


    By looking at part a (I have this easily graphed), the mean should be something like 0.3 where this value gave me -0.22. Am I off by a negative? Am I even doing this correctly?

    Thank you.
     
  2. jcsd
  3. Oct 3, 2016 #2

    Charles Link

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    In computing the first moment for a=0, I think your limits should simply be minus infinity and plus infinity. Meanwhile, a hint: do you recognize the integrand as an "odd" function, i.e. odd in powers of x? i.e. is ## f(x)=f(-x) ## ? What does that tell you?
     
  4. Oct 3, 2016 #3

    Charles Link

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    One additional item: Try evaluating the integral with a straightforward integration. That result is in agreement with the odd function result if you do it properly.
     
  5. Oct 3, 2016 #4

    RJLiberator

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    The function is odd as you replace x with -x you get the same result.
    Therefore, that is how I remedy the negative value of the solution.

    Hm, I see. I was using the definition from https://en.wikipedia.org/wiki/Cauchy_distribution#Mean

    When I use -a to a, I get the value of 0. That doesn't seem right.

    With straightforward integration, do you mean the integral from -infinity to infinity? If so, this value does not converge from my understanding.
     
  6. Oct 3, 2016 #5

    Charles Link

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    The function is odd, so you get zero. The integral of ## x/(x^2+1) ## may not converge when going from zero to infinity, but if you do it from minus infinity to plus infinity, it gives ## (1/2)ln|+\infty /- \infty|=(1/2)ln|1|=0 ##. This one is subject to some interpretation and might be referred to as a principal value, etc. , but it appears a reasonable answer for this moment is zero. editing... minor correction: should read ## (1/2) ln|+ \infty /+\infty | ## because its ## (1/2) ln|x^2+1| ## in both upper and lower limit.
     
  7. Oct 3, 2016 #6

    RJLiberator

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    Ah. The function being odd will give the value of 0 and that results in the first moment being 0. This is the mean value of the probabiltiy distribution.

    I'm trying to understand how this could be.

    Once I plotted it, it appears as a Gaussian distribution (very comparable). Is the mean, or expected value = 0 as there are values of +x and -x such that they all sum up to be 0?
     
  8. Oct 3, 2016 #7

    Charles Link

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    The integral under the curve, (without the x) is 1. When computing the mean, you put an x in the numerator. You do the same with a Gaussian (put an x in the numerator) and the mean is zero. Also, I made a minor correction to the above post: ( +infinity/+infinity), etc. Please look at previous post.
     
  9. Oct 3, 2016 #8

    Ray Vickson

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    Correct. If you take
    $$\int_{-A}^{B} \frac{1}{\pi} \frac{1}{x^2+1}$$
    you can make ##\lim_{A \to \infty} I(A,kA) = \ln(|k|)/ \pi## come out at any value you want, just by choosing the appropriate value of ##k##. You can even make ##A## and ##B## go to to ##+\infty## in such a way that the limit does not exist.

    In other words, ##f(x)## does not have a well-defined moment in the strict mathematical sense. Of course, there is a great temptation to select the so-called Cauchy Principal value, which chooses the equidistant limits ##A = B## (or ##k = 1##) and gives the rather nice value of 0 to the integral. However, that is really an artifact that has little, if any, justification. The fact is that the Cauchy distribution has some bad properties; it violates the Central Limit Theorem, for one. Simulation tests with Cauchy-distributed random variables show just how different the situation is (with Cauchy) from what we are accustomed to with most other distribution.

    BTW: when you use LaTeX, you get π to look nice (like this ##\pi##) instead of ugly like yours (##pi##), just by putting a "\" in front. That tells the TeX process that you want one of its standard commands. So, type "\pi" instead of "pi".
     
  10. Oct 3, 2016 #9

    RJLiberator

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    Thank you for taking the time to explain some of this to me. This (tonight) is the first time I hear of moment terminology.

    I'm understanding that the first moment is related to mean while the second moment is related to variance, however, this does not mean that the second moment = variance.

    Now, we have that the first moment is equal to 0. The question in part c states "Does the second moment exist?" and my intuition is ringing that if the first moment is equal to 0, then the second moment doesn't exist.

    Is there anywhere I can read or look into this to get a more clear picture?
     
  11. Oct 3, 2016 #10

    Charles Link

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    The variance is ## \sigma^2= E( X^2)-(EX)^2 ## where E stands for expectancy. ## (E(X^2) ## is the second moment and ## EX ## is the first moment or mean.) For the second moment of this distribution, you can write the integral as ## x^2/(x^2+1)=(1-1/(x^2+1)) ## and clearly it diverges in integrating from minus infinity to plus infinity. A good introductory statistics book would probably be useful, but I don't know which are the best ones that are currently available.
     
  12. Oct 3, 2016 #11

    Ray Vickson

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    Please put out of your mind forever the thought that the symmetric Cauchy distribution has first moment equal to zero. NO, it does not; it has no first moment at all!

    That issue is discussed in great detail in dozens of web pages devoted to the Cauchy distribution; one enlightening discussion is given in
    http://math.stackexchange.com/quest...bution-have-no-mean-if-its-symmetric-around-0
    It discusses the issue of why we insist that the Cauchy distribution has no first moment, even though the density function is symmetric about 0 and has equal probability weights on opposite points. There are good reasons for this insistence; it is not just "mathematical quibbling".
     
  13. Oct 3, 2016 #12

    RJLiberator

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    Thank you for the link, this is precisely the topic of discussion for me.
     
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