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For a uniform random point in a disk, the event $R<x$ occupies a disk of area $\pi x^2$ when $0<x<1$. ThusThe radial probability density function is $f_R(r)=2r$ on $(0,1)$. Integrating givesChoose the polar angle $\Theta \in [0,2\pi )$. The Cartesian density is $1/\pi$ in the disk, and the polar coordinates Jacobian determinant is $r$. HenceThe product support and product density prove that $R$ and $\Theta$ are independent, with uniform angle.

Reflection symmetry gives $\mathbb{E}X=\mathbb{E}Y=0$ and $\mathbb{E}(XY)=0$, so $\mathrm{Cov}(X,Y)=0$. They are nevertheless not independent. The events $X>3/4$ and $Y>3/4$ each have positive probability, but their intersection is empty inside the disk, because it would require $X^2+Y^2>18/16>1$. Independence would give a positive product probability for that intersection.

Except at the probability-zero center, $X/R=\cos \Theta$ and $Y/R=\sin \Theta$. ThereforeAn arbitrary definition at the center does not change these expected values. This example distinguishes uncorrelated Cartesian coordinates from independent polar coordinates.

Uniform area density in the unit disk is $1/\pi$. The polar coordinates Jacobian gives the joint densityThus the coordinates are independent and the uniform random point in a disk has

For independent points $A,B$, the area of triangle $OAB$ isNowIndependence therefore givesConditioned on $A,B$, the probability that $C$ lies inside $OAB$ is its area divided by the disk area $\pi$. Hence

Four points in general position fail to form a convex quadrilateral exactly when one lies inside the triangle formed by the other three. Each of $A,B,C$ lies inside the triangle formed by $O$ and the other two with probability $4/(9{\pi }^2)$, and these three events are disjoint. The remaining possibility is that $O$ lies inside $ABC$. By origin in a triangle of three radial random points, this has probability $1/4$. Therefore

Conditioned on being inside the circle, the point is a uniform random point in a disk of radius $1/2$. The annulus from $r$ to $r+dr$ has area $2\pi rdr+O(d{r}^2)$. Dividing by the disk's area givesThe radial probability density function is $8r$ on $[0,1/2]$ and integrates to one. Without conditioning, the corresponding annulus probability for a point in the original square is $2\pi rdr+O(d{r}^2)$.