# Mathematical Expectation

• Feb 18th 2010, 08:39 AM
Mathematical Expectation
A point is chosen at random on a circle of radius a. Compute the mathematical expectation of its distance from a fixed point on the circle.
• Feb 18th 2010, 07:27 PM
TKHunny
I must be sleeping. Someone else will have to provide the easy way. This is what I did. It is BRUTE FORCE, to be sure.

For notational convenience, it's a few layers deep.

For simplicity, I thought we should do two things.

1) Set up coordinate axes so that fixed point on the circle is at the origin and the center is at (a,0) on the positive x-axis

2) Forget the bottom half of the circle. By symmetry, we have the same solution.

After that, define:

$x(t) = a - a\cdot \cos(t)$

$y(t) = a\cdot \sin(t)$

For $0 \le t \le \pi$

This defines all possible locations of the point chosen at random.

Define $d(t) = \sqrt{(x(t))^{2}+(y(t))^{2}}$

This is the distance from the origin for each point.

Further, calculating the probability directly, define:

$P(m) = \sum_{n=0}^{m} \left[\frac{1}{m}\cdot d \left(\frac{n\cdot\pi}{m}\right)\right]$

Finally, we have $\lim_{m \to \infty} P(m)$

(Yes, this is tantalizingly close to a Reimann integral.)

I get a*1.273240 to about $10^{-6}$

My first few attempts continued to produce results that were independent of 'a'. Since that obviously is wrong, I had to look elsewhere. Obviously, I wandered around quite a bit.

For what it's worth. It is an interesting exercise.
• Feb 19th 2010, 12:07 PM
Moo
Hello,

Let P be the fixed point, and let M be the random point.
Let $\theta$ such that $(a\cos\theta,a\sin\theta)$ are the coordinates of P.
And let $\sigma$ such that $(a\cos\sigma,a\sin\sigma)$ are the coordinates of M.

Draw a circle and M and P (WLOG, we can draw P in the first quadrant). Draw the line $y=\sin\theta$ (the parallel to the x-axis going through P) and draw the line $x=\cos\theta$ (the parallel to the y-axis going through M)
Let H be the intersection of these two lines. Thus $PHM$ is a right angle triangle, with hypotenuse $PM$ (which is the distance we're interested in !)
We have $PM^2=HM^2+HP^2$

But it can be obvious that $HM=a \cdot |\sin\theta-\sin\sigma|$ and that $HP=a\cdot |\cos\theta-\cos\sigma|$
I haven't come up with a formal proof of it yet, but it shouldn't be that difficult !

Hence $PM^2=a^2\cdot((\cos\theta-\cos\sigma)^2+(\sin\theta-\sin\sigma)^2)=\dots=2a^2(1-\cos(\sigma-\theta))$
- after using the following identities :
* $\cos^2x+\sin^2x=1$
* $\cos x\cos y+\sin x\sin y=\cos(x-y)$

So $D=PM=a\sqrt{2}\cdot\sqrt{1-\cos(\sigma-\theta)}$

Since M is randomly chosen on the circle, $\sigma$ follows a uniform distribution over $[0,2\pi[$ (to a given angle, there's a unique point corresponding to this angle on the circle. Would be different if we were considering a random point in the circle)

So $\mathbb{E}[D]=a\sqrt{2}\cdot \mathbb{E}[\sqrt{1-\cos(\sigma-\theta)}]$

Since the pdf of a rv uniform on $[0,2\pi[$ is $f(x)=\frac{1}{2\pi}$, for $x\in[0,2\pi[$, it follows that :

$\mathbb{E}[D]=\frac{a\sqrt{2}}{2\pi}\cdot\int_0^{2\pi}\sqrt{1-\cos(x-\theta)} dx$

Since the integrand of the integral is $2\pi$-periodic, the value of $\theta$ doesn't change the value of the integral.

In particular, $\mathbb{E}[D]=\frac{a\sqrt{2}}{2\pi}\cdot\int_{0}^{2\pi}\sqrt{1-\cos(x)} ~dx$

Now substitute $x=2u$ (dx=2du)

$\mathbb{E}[D]=\frac{a\sqrt{2}}{\pi}\cdot\int_{0}^{\pi}\sqrt{1-\cos(2u)}~du$

But $\cos(2u)=1-2\sin^2(u)$

So $\mathbb{E}[D]=\frac{a\sqrt{2}}{\pi}\cdot\int_{0}^{\pi} \sqrt{2}\cdot\sin(u) ~du$ (since $\sin(u)\geqslant 0, \forall u \in[0,\pi]$)

Finally, $\mathbb{E}[D]=\frac{2a}{\pi}\cdot\int_0^\pi \sin(u) ~du=\frac{2a}{\pi}\cdot 2=\boxed{\frac{4a}{\pi}}$

(which gives the approximate result that TKHunny got)

That's a cool result ! :D

It looks long, but there's nothing complicated at all ! Only basic geometry and Calc I (or II, I don't really know) (Wink)
• Feb 19th 2010, 02:05 PM
Laurent
Quote:

A point is chosen at random on a circle of radius a. Compute the mathematical expectation of its distance from a fixed point on the circle.

The following is essentially equivalent to what Moo did (which you should read first), but a bit shorter:

Let us set the problem in the complex plane, the circle being centered at 0. Due to the radial symmetry, we may assume that the fixed point is $a(\in\mathbb{R})$. Then the distance between $M=a e^{i\theta}$ and $a$ is given by $|a e^{i\theta}-a|=a |e^{i\theta}-1|= 2a\left|\sin\frac\theta2\right|$ (the last equality is because $e^{i\theta}-1=e^{i\theta/2}(e^{i\theta/2}-e^{-i\theta/2})$ $=2ie^{i\theta/2}\sin\frac{\theta}{2}$).

Since a uniform random point on the circle is given by $M=a e^{i\Theta}$ where $\Theta$ is uniformly distributed in $[0,2\pi]$, we conclude that the expected distance is:

$\int_0^{2\pi}2a|\sin\frac{\theta}{2}|\frac{d\theta }{2\pi}=2a\int_0^\pi \sin\frac{\theta}{2}\,d\theta$ $=\frac{4a}{\pi}\int_0^{\pi/2}\sin (u) du=\frac{4a}{\pi}$.
• Feb 19th 2010, 03:03 PM
TKHunny
I could be a bit more delighted, but not much!

There is just nothing quite like a unique result - just sitting there, waiting for us to find it - however we choose to find it.

Note to student, if you don't think of a solution right off, keep thinking. There are likely to be at least a couple of ways to approach it.