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Math Help - variance of separations between random parameters

  1. #1
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    variance of separations between random parameters

    There are N points which are randomly distributed (according to a uniform probability distribution) over the circumference of a circle.

    Now consider the angular separation \theta between two neighbouring points. The sum of all these angular separations wil always correspond to one full rotation, i.e.:
    \sum_{i=1}^{N}\theta_{i}=360^{\circ}

    The expected value of \theta can then be found as:
    E(\theta)= \frac{360^{\circ}}{N}

    But the thing what I'm really dying to know is: how do I calculate the variance of \theta?

    I've tried for several hours, but I just can't figure out where to start. Does anyone here have any idea? Thanks in advance!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by corsica View Post
    There are N points which are randomly distributed (according to a uniform probability distribution) over the circumference of a circle.

    Now consider the angular separation \theta between two neighbouring points. The sum of all these angular separations wil always correspond to one full rotation, i.e.:
    \sum_{i=1}^{N}\theta_{i}=360^{\circ}

    The expected value of \theta can then be found as:
    E(\theta)= \frac{360^{\circ}}{N}

    But the thing what I'm really dying to know is: how do I calculate the variance of \theta?

    I've tried for several hours, but I just can't figure out where to start. Does anyone here have any idea? Thanks in advance!
    In order to simplify the problem we can subistute the circle with a segment of lengh 1 and analize first the case n=3. Let suppose to have two independent variables x_{1} and x_{2} uniformely distributed in [0,1]. It is well known that the random variable x=|x_{1}-x_{2}| has p.d.f. equal to...

    f(x) = 2\ (1-x)\ \text{if}\ 0 \le x \le 1\ ,\ = 0\ \text{elsewhere} (1)

    From (1) we derive directly...

    \mu = \int_{0}^{1} 2\ x\ (1-x)\ dx = \frac{1}{3} (2)

    \sigma^{2} = \int_{0}^{1} 2\ (x-\frac{1}{3})^{2}\ (1-x)\ dx = \frac{1}{18} (3)

    That is for n=3... now we have to analyze greater values of n...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor chisigma's Avatar
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    In the case n=4 with a little effort the p.d.f. of the random variable x=|x_{1}-x_{2}| is found to be...

     f(x)= 3\ (1-x)^{2}\ \text{if}\ 0\le x \le 1\ ,\ 0\ \text{otherwise} (1)

    ... so that is...

    \mu = 3\ \int_{0}^{1} x\ (1-x)^{2}\ dx = \frac{1}{4} (2)

    \sigma^{2}= 3\ \int_{0}^{1} (x-\frac{1}{4})^{2}\ (1-x)^{2}\ dx = \frac{3}{80} (3)

    In general could be...

     f(x)= (n-1)\ (1-x)^{n-2}\ \text{if}\ 0\le x \le 1\ ,\ 0\ \text{otherwise} (4)

    \mu = (n-1)\ \int_{0}^{1} x\ (1-x)^{n-2}\ dx = \frac{1}{n} (5)

    \sigma^{2}= (n-1)\ \int_{0}^{1} (x-\frac{1}{n})^{2}\ (1-x)^{n-2}\ dx = \frac{n-1}{(n+1)\ n^{2}} (6)

    Kind regards

    \chi \sigma
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  4. #4
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    Wow, that's absolutely great. Thanks so much!

    Do you by any chance have a reference where the p.d.f. of the random variable x=|x_{1}-x_{2}| is explained in more detail? I'd would like to understand the underlying concepts better.
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  5. #5
    MHF Contributor chisigma's Avatar
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    The lecterature treating the problem You have proposed is enormous... among others ...

    SpringerLink - Journal of Mathematical Sciences, Volume 25, Number 3

    ... my modest opinion is that the 'signature' reported below is ever relevant ...

    Kind regards

    \chi \sigma
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