For any point P on the unit circle, what is the average of the distance between P and all points on the unit circle?

My main issue with this question stems from the different answers I get depending on how I partition my intervals when attempting to write this question as an integral. The way I understand calculating averages like this is to use the fact that the average of N values is the sum of the values divided by the total number of values (n), meaning as N goes to infinity, this sequence of averages converges to our desired average. One way to do this is to try and write it as a Riemann sum.

One way to do this is if we have a the distance as function of our angle, which if we simply treat our point as being (1,0) is: $$\ f(θ) =\sqrt{(1-\cos(θ))^{2}+(\sin(θ))^{2}} =\sqrt{2}\sqrt{1-\cos(θ)}$$

Since the circle is symmetrical, all the distances are equal so, we can just partition the angles into N intervals between 0 and pi and can write the average as a Riemann Sum, which becomes an integral as n goes to infinity. $$\dfrac{\sum_{i=0}^n\ f(θ_{i})}{n} =\dfrac{\sum_{i=0}^n\ f(θ_{i})}{\dfrac{\pi}{\Deltaθ}}= \dfrac{\sum_{i=0}^n\ f(θ_{i})\Deltaθ}{\pi}= \frac{1}{\pi}\int_0^{\pi}f(θ)dθ=\dfrac{4}{\pi}$$

However it seems another way we can do this is partition along the x axis from -1 t0 1 and integrate this function, which is the distance of point on the unit cirlce to (1,0) as function of x: $$f(x)=\sqrt{(1-x)^{2}+(\sqrt{1-x^{2}})^{2}}=\sqrt{2}\sqrt{1-x}$$

and write the Riemann sum as $$\dfrac{\sum_{i=0}^n\ f(x_{i})}{n} =\dfrac{\sum_{i=0}^n\ f(x_{i})}{\dfrac{2}{\Delta{x}}}= \dfrac{\sum_{i=0}^n\ f(x_{i})\Delta{x}}{2}= \frac{1}{2}\int_{-1}^{1}f(x)dx=\dfrac{4}{3}$$

It seems like both ways are valid when it comes to getting the average, so I don't know why they result in different values.

• The arc covered by $x\in[-1,-0.99]$ is much longer than the arc covered by $x\in[0,0.01]$, hence in your second method you weigh these intervals differently. Sep 13 '20 at 8:19
• you take a point $P$ and draw a diameter from the point through center. Now your starting point is opposite end of the diameter and you traverse on the circle to point $P$. The distance varies from $2$ to $0$ and the angle between $0$ and $\pi / 2$. Average distance will be $4/ \pi$. Sep 13 '20 at 8:47

the average value of discrete variable $$x_i$$ with probability $$P(x_i)$$ is $$\overline{x} = \sum_{i_1}^n P(x_i)x_i$$ I set a circle with radius $$R$$ such that the center on the $$x$$-axis while its circumference passes through the origin, in polar coordinates $$r(\theta) = 2R\cos(\theta) \hspace{1 cm} \theta \in [-\frac{\pi}{2},-\frac{\pi}{2}]$$ $$\overline{r} = \sum_{i=1}^n P(r_i)r_i = \int \frac{dP(r)}{d\theta}r(\theta)\,d\theta$$ The probability of choosing the second point on the circle is distributed uniformly over the circumference (the first point is at the origin) $$dP(r) = \frac{ds}{\int ds} = \frac{ds}{2 \pi R}$$ where $$ds$$ is the arclenght $$ds = \sqrt{(rd\theta)^2+(r'd\theta)^2} = \sqrt{(r)^2+(r')^2}d\theta = 2Rd\theta$$ $$\overline{r} =\int \frac{dP(r)}{d\theta}r(\theta)\,d\theta = \int \frac{2R}{2 \pi R}2R\cos(\theta)\, d\theta = \frac{2R}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(\theta)\, d\theta = \frac{4R}{\pi} < 2R$$