Point A is picked randomly in a circle with a radius of 1, and center O. What is the variance of length OA?

Question

Point A is picked randomly in a circle with a radius of 1, and center O. What is the variance of length OA?

I believe the CDF has to found first, then we need differentiate it, find the expected value and use a variance property. It doesn't seem like I can use any of the probability distributions to find the CDF though. A hint in the right direction should do the job

Answer 1

The probability to pick a point inside the smaller circle $\|z\|=r$ is clearly $\frac{r^2}{R^2}$, hence the probability density function of $X=\overline{OA}$ is supported on $[0,1]$ and simply given by $f(x)=2x$. It follows that: $$ \mu=\mathbb{E}[X]=\int_{0}^{1}2x^2\,dx = \frac{2}{3}\tag{1} $$ and:

$$ \text{Var}[X]=\mathbb{E}[(X-\mu)^2] = \int_{0}^{1}2x\left(x-\frac{2}{3}\right)^2\,dx = \color{red}{\frac{1}{18}}.\tag{2}$$

Answer 2

Assuming all points on the circle are equally likely, the probability of a point to have a certain radius $r_0$ is

$P(r=r_0)=\frac{(r_0+dr)^2\pi -r_0^2\pi}{\pi}=2r_0dr+(dr)^2$

powers of the differential of $r$ are ignored. Therefore

$P(r)=2rdr$

According to this Pdf, you can calculate both mean and variance. If I have not made any mistakes, mean is $2/3$ and variance is $1/18$.