Expected Distance Of Two Points In A Circle

Here is my solution, which is similar to the one given by Geoff Pilling.

Let $r,\theta$ be the position of a random point (in polar coordinates) inside the circle, with $r\in [0,1),\ \theta\in[0,2\pi)$ . Then the probability that it lies in a slice of area of radial width $dr$ and angular width $d\theta$ , around a position $r,\theta$ , is just the area of the slice divided by the total area of the unit circle, i.e., $p(r,\theta)dr d\theta=\frac{rdr d\theta}{\pi}$ , where $p(r,\theta)$ is the joint probability density function of the random variables $r,\theta$ . Thus, $p(r,\theta)=\frac{r}{\pi}$ . Thus, the expected distance from center is $\int_{0}^{2\pi}\int_{0}^1 r\frac{r}{\pi}dr d\theta=\frac{2}{3}$ producing the answer $2+3=\boxed{5}$ .

I did it a little differently.

$<r>= \int_{0}^{1}\frac{2\pi r \times r}{\text{Circle Area}}dr = \int_{0}^{1}\frac{2\pi r^2}{\pi}dr = 2\int_{0}^{1}r^2dr = \frac{2}{3}$

$2+3=\boxed5$

Divide the unit circle into $n$ parts with the first part being a circle with radius $\frac{1}{n}$ , the second part being a doughnut with thickness $\frac{1}{n}$ and a hole of radius $\frac{1}{n}$ and so on. Denote $p_k$ to be the probability that the point lies on the $k$ th portion, and denote $d_k$ to be the distance of the point on the doughnut part farthest from the center, to the center. We easily get that $d_k = \frac{k}{n}$ . Also, $p_k$ is just the area of the $k$ th portion divided by pi, i.e. $\large \frac{ \pi(\frac{k^2}{n^2} - \frac{(k-1)^2}{n^2})}{ \pi} = \frac{2k-1}{n^2}$ .

We can easily see that the answer we are looking for is $\displaystyle \sum_{k=1} ^ {n} p_k d_k$ for arbitrarily large $n$ . We can expand this to give the answer to be

$\huge \frac{ \sum_{k=1} ^ {n} 2k^2 - k}{n^3} =\frac{ \frac{(n)(n+1)(2n+1)}{3} - \frac{(n)(n-1)}{2}}{n^3} = \frac{4n^3 + 3n^2 - n}{6n^3} = \frac{2}{3}$ .

For arbitrarily large $n$ . Our expected value is then $\frac{2}{3}$ and our answer is then $2+3 = \boxed{5}$ .

This problem is the easiest to approach from the point of view of the cumulative distribution function (CDF). The CDF measures the probability that the value of the random variable is less than or equal to the argument of the function. In this case, we want the probability of $x$ being less than or equal to some radius $r$ to be proportional to the share of the area of the unit circle area occupied by the circle of radius $r$ . That is, $P(X \le r) = \frac{\pi r^2}{\pi} = r^2$ . The probability density function is the derivative of the CDF so we have $f_{X}(x) = 2r$ . To obtain the expected value we integrate PDF times the variable from 0 to 1: $\int_{0}^{1} 2 r^2 dr = \frac{2}{3}$

Assuming a radius of 1 as implied in the question.

Consider the length of concentric circles at points along a radius of the circle.

The length increases uniformly from zero to 2 pi, which can be visualised as a right angle triangle.

On average a randomly selected point will be at the centre (centroid) of the triangle.

This is one third of the distance from the circumference to the centre, or 2/3 of the distance from the centre.

Answer: 2 + 3 = 5

nice solution.