A circle with radius $a$, and two points $A, B$ on the disk are chosen independently and follow uniform distribution. Setting the distance of $\overline{AB} = R$
Now I would like to find the expectation of $R^2$. First, I think the distribution will be like $A \sim B \sim U[0, a^2π]$. Yet, I have no much thought about connecting the distance $R$ with expectation.
To me, it probablity looks like this $E[R^2] = \int_0^{a^2π}x^2f(x)dx$, in this case, what should I replace with $f(x)$ to stand for $R^2$?
If $A$ and $B$ are uniformly and independently distributed on the disk centered at origin and of radius $a$, then \begin{align*} \mathbb E[R^2] &= \mathbb E[(A_1-B_1)^2+(A_2-B_2)^2]\\ &= \mathbb E[A_1^2 - 2 A_1 B_1 + B_1^2 + A_2^2 - 2 A_2 B_2 + B_2^2]\\ &=4\mathbb E[A_1^2]-4\mathbb E[A_1]\cdot \mathbb E [B_1]\\ &=4 \mathbb E[A_1^2]\\ &=2\mathbb E[A_1^2+A_2^2] \end{align*} because $A_1$, $A_2$, $B_1$ and $B_2$ all have the same distribution with $0$ average, $A_1$ and $B_1$ are independent as well as $A_2$ and $B_2$.
Now $A_1^2+A_2^2$ is just the square of the distance of $A$ to $0$, it can be computing by integrating \begin{align*} \mathbb E[A_1^2+A_2^2] &= \int_{\mathbf x\in C_a} \frac{1}{\pi a^2}(x_1^2+x_2^2) ~d\mathbf x\\ &= \int_0^{2\pi}\int_0^a \frac{1}{\pi a^2} r^2\cdot r ~dr d\theta\\ &=\left[ \frac{r^4}{2a^2} \right]_0^a\\ &=\frac{a^2}{2} \end{align*} Therefore $\mathbb E[R^2]=a^2$.