Consider the experiment of choosing at random a chord of the unit circle in $\mathbb{R}^{2}$. Assume that the circle is given parametrically by $(cos2\pi t, sin2\pi t),\ 0\leq t<1$. Let $\Omega=[0,1)^{2}$. Let $\mathcal{A}$ denote de borel $\sigma$-algebra of subsets of $\Omega$ and let $P$ the uniform distribution in $\Omega$. Let $\Psi$ be the space of chords of the given circle and let $\mathcal{B}$ denote the Borel $\sigma$-algebra of subsets of $\Psi$, with $\Psi$ being regarded as a subspace of the metric space of all compact subsets of $R^{2}$(with the Hausdorff metric).
For $\omega= (\alpha, \beta)\in\Omega$, let $X_{2}(\omega)$ be the chord that both passes through the point whose polar coordinates are $(2\alpha-1, \pi\beta)$ and is perpendicular to the line segment from the origin to that point. In case $\alpha=1/2$, this line consist of a single point, but we can still define its direction to be $\pi\beta$. Compute the probability of the event that the random chord intersects the positive vertical and negative horizontal axes.
My Approach. If my analysis is accurate, the points that satisfy that a chord intersects the positive vertical and negative horizontal axes are in some subregion of the region in the unit square at the top right, more precisely, the set $A=\left\{\omega=(\alpha, \beta) : \frac{1}{2} \leq \alpha< 1 \text { and } \frac{3}{4}<\beta < 1 \right\}$ and the reflection of $A$ about the line $\alpha=\beta$.
In the picture $\omega '=(\frac{1}{2},\frac{7\pi}{8})$.
Now, spousing this is correct, the probability should be the area of the subregion but i don't know what is the shape or the equation of the subregion.
How can I find the shape or the equation of this region?
Best Answer
The “good” region for $\omega$ in your solution doesn’t look quite right to me. An equation of the line is $x\cos{\pi\beta}+y\sin{\pi\beta}=2\alpha-1$, with $x$- and $y$-intercepts $(2\alpha-1)/\cos{\pi\beta}$ and $(2\alpha-1)/\sin{\pi\beta}$, respectively. For these the have the correct signs, the point must be in the second quadrant, so we must have $1/2\lt\alpha\le1$ and $1/2\lt\beta\lt1$. The region is symmetric, but not about the line $\alpha=\beta$, as we shall soon see.
Not all values of $\alpha$ in this range produce intercepts in $(0,1]$, which is what we need for the chord to intersect both coordinate axes. For a fixed $\beta$, the maximal value of $\alpha$ is the least value for which the line passes through either the point $(-1,0)$ or $(0,1)$, so $$\alpha\le\min\left(\frac12(1-\cos{\pi\beta}),\frac12(1+\sin{\pi\beta})\right),$$ or equivalently, $$\begin{cases}\alpha\le\frac12(1-\cos{\pi\beta}), &\text{if }\frac12\lt\beta\le\frac34 \\ \alpha\le\frac12(1+\sin{\pi\beta}), &\text{if }\frac34\lt\beta\lt1.\end{cases}$$ The region looks approximately like this:
(the horizontal axis represents $\alpha$).
This region is symmetric about the line $\beta=3/4$, so its area is equal to $$2\int_{3/4}^1\sin{\pi\beta}-\frac12\,d\beta.$$