Expectation that the Distance Between two Points in the Unit Disc is 1

On pg. 64 of David William's book Wrighing the Odds, there appears the following exercise:

Problem. Two points $A$ and $C$ are chosen independently at random in the unit disc $D$. Let $d$ denote the distance between them. Prove that $E(d^2)=1$, where $E$ denotes expectation.

We have the measure space $D$ with the Lebesgue measure $\mu$, so we can form the product space $D\times D$ with product measure $\lambda=\mu\times \mu$. For any real $r\geq 0$, we can consider the subset $S_r$ of $D\times D$ defined as $S_r=\{(a, b)\in D\times D: d(a, b)\leq r\}$. Then if $f$ denote the distribution function of the random variable $d^2$, we have $\lambda(S_r)=\int_0^r f dx$. This would give us $f$, and then one can calculate $E(d^2)$ by using the fact that $E(d^2)=\int_{\mathbf R}xf(x) dx$.

But the above approach is cumbersome. The author also seems to have something else in mind, for he has given as hint "use the pythagorean theorem to exploit symmetry".

Can somebody see a better approach? Thanks.

Solution 1:

Let the coordinates of $A$ and $C$ be $(x_A, y_A)$ and $(x_C, y_C)$ respectively. Then \begin{align} \mathsf{E}(d^2) &= \mathsf{E}((x_A - x_C)^2 + (y_A - y_C)^2) \\ &= \mathsf{E}(x_A^2 + x_C^2 + y_A^2 + y_C^2) \\ &= \mathsf{E}(x_A^2 + y_A^2) + \mathsf{E}(x_C^2 + y_C^2) \\ &= 2\mathsf{E}(x_A^2 + y_A^2) \end{align} where $\mathsf{E}(x_A^2 + y_A^2)$ is the expected squared distance from $A$ to the origin and its value is $\frac{1}{2}$.