I'm trying to find a probability distribution $m(r,\theta)$ defined over a unit disk which represents the probability of a point being the midpoint of two randomly chosen points in said disk. I think that I'm pretty close to the answer, but I'm off by a factor of 16 and I don't understand why. Given that the distribution of points is uniform, i.e. $f(r,\theta) = \frac{1}{\pi}$, my reasoning is as follows:
- The first and second points chosen are $P_{1}$ and $P_{2}$ respectively. If we fix the first point chosen inside the disk, the set of points which are possible midpoints for $P_{1}P_{2}$ form a disk of radius $\frac{1}{2}$. (Edit: see the figure below for an illustration)
- Since for each possible $P_{1}$ there is a unique circle of size $\frac{\pi}{4}$ generated representing all the possible midpoints where $P_{1}$ is one of the points, then $m(r,\theta)$ should be proportional to the total "height" of all the unique circles that overlap onto the point $(r,\theta)$, if we imagine each of the radius-$\frac{1}{2}$ circles as having some infinitesimal "height".
- I found the "height" of the overlapping circles to be $2(\frac{\arccos{r}}{4}-\frac{r}{2}\sqrt{\frac{1}{4}-\frac{r^{2}}{4}}) = \frac{1}{2}(\arccos{r}-r\sqrt{1-r^{2}})$. This is the area of overlap between two circles of size $\frac{\pi}{4}$, reason being that since the center of the radius-$\frac{1}{2}$ disks we formed in step 2 will fall within a radius-$\frac{1}{2}$ disk that is concentric to the unit disk, then the set of radius-$\frac{1}{2}$ disks which do not overlap onto $(r,\theta)$ must be more than a distance of $\frac{1}{2}$ from this point. Hence, the set of radius-$\frac{1}{2}$ disks which do form the "height" at point $(r,\theta)$ will have centers within a distance of $\frac{1}{2}$ from the point, and this is the leaf-shaped area between two circles – essentially the expression above. I'm not sure whether this is the correct way to be thinking about the "height" at the point – or if "height" is even the right way to think about it – so I would appreciate if anyone can point out a more rigorous direction.
- Assuming step 3 is correct, then because $f(r,\theta) = \frac{1}{\pi}$, then since we are choosing 2 points, we scale step 3 by $\frac{1}{\pi^{2}}$. However, if we take this as $m(r,\theta)$, then $\int_{0}^{1}\int_{0}^{2\pi}m(r,\theta)r d\theta dr = \frac{1}{16}$.
I'm not very well versed in geomtery and probability, so there is definitely something wrong with my logic. Can anyone point me to where the missing factor of 16 comes from, or whether my entire chain of logic is incorrect?
Figure for step 1:
We just need to show that the boundary of the set of possible midpoints forms a circle.
Consider the diameter that goes through $P_{1}$ and the center of the circle. If $P_{2}$ is on either $D_{1}$ or $D_{2}$, then the midpoint will be at $M_{1}$ or $M_{2}$ respectively. Consider if $P_{2}$ is anywhere else on the larger circle, let's say $A$ (any point on the circle will form the boundary case), then its midpoint we call $X$. $M_{1}P_{1} = \frac{1}{2}D_{1}P_{1}, XP_{1} = \frac{1}{2}AP_{1}$ and $\angle{M_{1}P_{1}X} = \angle{D_{1}P_{1}A}$, so $M_{1}P_{1}X$ and $D_{1}P_{1}A$ are similar triangles by a factor of $\frac{1}{2}$. A similar argument can be applied for $D_{2}$ and $M_{2}$. Hence we have that $D_{1}D_{2}A$ is similar to $M_{1}M_{2}X$ by a factor of $\frac{1}{2}$. Since $D_{1}D_{2}A$ is a right triangle, so is $M_{1}M_{2}X$, and as A varies, X will trace out a circle of radius $\frac{1}{2}$.
P.S. I am quite sure that other than the factor of 16, everything else should be correct, because this is part of a larger problem that I am solving, which I have shown to be correct assuming that the factor of 16 is there.
Best Answer
The area of the wedge subtended by the green arc is $R^2\cos^{-1}\left(\frac rR\right)$ and the area of the red triangle is $r\sqrt{R^2-r^2}$. Therefore, the area of the gray lozenge is $2\!\left(R^2\cos^{-1}\left(\frac rR\right)-r\sqrt{R^2-r^2}\right)$. The gray lozenge is the locus of endpoints of lines, contained in the left disk, with $P$ as a midpoint. The area of the lozenge is proportional to the probability density of the midpoints of lines whose endpoints are uniformly distributed in the left disk.
We compute the normalization factor for the density with $$ \begin{align} &\int_0^R2\left(R^2\cos^{-1}\left(\frac rR\right)-r\sqrt{R^2-r^2}\right) 2\pi r\,\mathrm{d}r\\ &=4\pi R^4\int_0^1\left(\cos^{-1}(r)-r\sqrt{1-r^2}\right)r\,\mathrm{d}r\tag{1a}\\ &=4\pi R^4\left(\int_0^{\pi/2}\frac12\cos^2(r)\,\mathrm{d}r-\frac12\int_0^1r^{1/2}(1-r)^{1/2}\,\mathrm{d}r\right)\tag{1b}\\ &=4\pi R^4\left(\frac\pi8-\frac12\frac{\Gamma\!\left(\frac32\right)\Gamma\!\left(\frac32\right)}{\Gamma(3)}\right)\tag{1c}\\ &=\frac{\pi^2R^4}4\tag{1d} \end{align} $$ Explanation:
$\text{(1a):}$ substitute $r\mapsto Rr$
$\text{(1b):}$ split the integral
$\phantom{\text{(1b):}}$ substitute $r\mapsto\cos(r)$ and integrate by parts in the left integral
$\phantom{\text{(1b):}}$ substitute $r\mapsto r^{1/2}$ in the right integral
$\text{(1c):}$ integrate using the Beta Function Integral
$\text{(1d):}$ simplify
Here is another way to get $\text{(1d)}$. For any point $p$ in the disk, the midpoint, $m$, to the other endpoint varies over a disk of radius $\frac R2$. Thus, for each $p$, we integrate over the area covered by $m$, and then we integrate over all $p$ in the disk; that is, $\pi\frac{R^2}{4}\cdot\pi R^2=\frac{\pi^2R^4}4$.
So the density would be $$ \frac8{\pi^2R^4}\left(R^2\cos^{-1}\left(\frac rR\right)-r\sqrt{R^2-r^2}\right)\tag2 $$ which, for $R=1$, is $$ \frac8{\pi^2}\left(\cos^{-1}(r)-r\sqrt{1-r^2}\right)\tag3 $$