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Let $(X,Y)$ be a uniformly distributed random point on the $2\times 2$ square with four corners at $(0,0),(2,0),(2,2)$ and $(0,2).$ Let $R$ be the distance of $(X,Y)$ to the nearest corner. Find the CDF and the PDF of $R$.

Let $r\geq 0.$ I consider the four quarter disks of radius $r$ inside the square with center at the four corners of the square. Now I observe that the event $\{R\leq r\}$ is the event that the point $(X,Y)$ lies in the union of the four disks. Note that each disk has an area of $\pi r^2/4.$

Since the point $(X,Y)$ is uniformly distributed on the square, which has area equal to $4$, we have by the remark above, that $$F_{R}(r)=\mathsf{P}(R\leq r)= \begin{cases}\displaystyle\frac{\pi r^2}{4}&\text{if }0<r\leq 2\\1&\text{if }\,\,\,r>2 \\0&\text{if }\,\,\,r\leq 0.\end{cases}$$ And differentiating with respect to $r$, we find that $$f_{R}(r)= \begin{cases}\displaystyle\frac{\pi r}{2}&\text{if }0<r\leq 2\\0&\text{otherwise. } \end{cases}$$

Is my approach correct?

Any feedback is much appreciated. Thank you for your time.

Stackman
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    The quarter disks are not disjoint if $r>1$. It seems you overlooked that. Also $R\leq\sqrt2$. Make a picture of e.g. the area where $R<1.2$ to get some overview. – drhab Mar 06 '19 at 08:08
  • @drhab Thank you for your feedback. Would the following observation be correct $$F_{R}(r)= \begin{cases} & \pi r^2/4,,\text{if},,0<r\leq 1\ & 1,,,,,,,,,,,,,,,,\text{if},,r>\sqrt{2} \ & 0,,,,,,,,,,,,,,,,\text{if},,r\leq 0. \ \end{cases}$$ I know I am missing the case where the disks are not disjoint, which would be when $1\leq r\leq \sqrt{2}$. In that case, can I add all 4 areas and then subtract the area of their intersections? – Stackman Mar 06 '19 at 08:38
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    Yes, that is correct. The "hard part" ($1<r<\sqrt2$) is yet to be done ;-). A picture will surely help. – drhab Mar 06 '19 at 08:41
  • @drhab I found the following post: https://math.stackexchange.com/questions/402858/area-of-intersection-between-two-circles, which explains how to find the area of the intersection of two circles of equal area. Using this I found that $F_{R}(r)=\frac{\pi r^2}{4}-2r^2\left(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\right)$ if $1<r<\sqrt{2}.$ – Stackman Mar 06 '19 at 09:29
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    hmm... Apply sanity check: if correct then $F_R(1)=\frac{\pi}4$ and $F_R(\sqrt2)=1$. – drhab Mar 06 '19 at 09:46

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By symmetry you may assume that $(X,Y)\in[0,1]^2$ and that $R:=\sqrt{X^2+Y^2}$. It is then obvious that $$F_R(r)={\pi\over4}r^2\qquad(0\leq r\leq 1)\ .$$ When $1< r\leq\sqrt{2}$ the area in question consists of two triangles and a circular sector. The total area of the two triangles is $\sqrt{r^2-1}$, and the central angle $\alpha$ of the sector is $$\alpha={\pi\over2}-2\arctan\sqrt{r^2-1}\ ,$$ and its area is ${1\over2}\alpha r^2$. Put it all together to obtain $F_R(r)$ for $1< r\leq\sqrt{2}$.

Christian Blatter
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