An around the origin centered disk with radius $r>0$ is halved along the horizontal axis of coordinates. At the edge of the upper half, a point p is randomly selected and and the semicircle is devided again along the line that goes through the origin and p. Calculate the density and the expected value of the random variable that describes the ratio between the smaller and the greater segment.
So we have:
$A=tr^2, B=(\pi -t)r^2, p=(rcos(t),rsin(t)), U\sim\mathscr U([0,\pi]), t \in [0,\pi]$
I don't see how to proceed here. Any help is greatly appreciated!
The ratio of areas of the two sectors is equal to the ratio of their arc lengths; both of these ratios are distributed as $U/(1-U)$ when $U$ is uniformly distributed on $[0,1]$. The ratio $R$ between the smaller and greater sector areas is given by $R = \min(U/(1-U), (1-U)/U)$. One can work out the expectation of $R$ as follows, noting that $U/(1-U)<(1-U)/U$ just when $U<1/2$: $$ ER = \int_0^{1/2} \frac u {1-u} du + \int_{1/2}^1 \frac {1-u} u du = 2\int_{1/2}^1 (\frac 1 u - 1) du = 2\ln 2 - 1.$$ A similar calculation gives the cdf and then the density.