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

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

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

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

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

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

    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 $\displaystyle x_{1}$ and $\displaystyle x_{2}$ uniformely distributed in $\displaystyle [0,1]$. It is well known that the random variable $\displaystyle x=|x_{1}-x_{2}|$ has p.d.f. equal to...

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

    From (1) we derive directly...

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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 $\displaystyle x=|x_{1}-x_{2}|$ is found to be...

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

    ... so that is...

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

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

    In general could be...

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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 $\displaystyle x=|x_{1}-x_{2}|$ 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

    $\displaystyle \chi$ $\displaystyle \sigma$
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