Thread: variance of separations between random parameters

1. 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!

2. Originally Posted by corsica
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

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

3. 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

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

4. 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.

5. 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

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