We are interested in finding the density function for the difference between $X_1$ and $X_2$ ( $\left|X_1-X_2\right|$). If $X_1$ and $X_2$ are independent and uniformly distributed on (0,1) in the unit square, we then have the following probabilistic statement:

$P\left(\left|X_1-X_2\right|>d\right)={\left(1-d\right)}^2$(5)

Here, ${\left(1-d\right)}^2$ is the total area of the two shaded right triangles, as shown in the unit square in Figure 2 [3] - [5] .

The total area that is non-shaded is $1-{\left(1-d\right)}^2$, which can be related to $P\left(\left|X_1-X_2\right|<d\right)$ via the following:

$P\left(\left|X_1-X_2\right|<d\right)=1-{\left(1-d\right)}^2$(6)

The right-hand side is a cumulative density function, since it accounts for the totality of the non-shaded area. Taking the first derivative of the right-hand side with respect to “ $d$”, we have the following:

$P\left(\left|X_1-X_2\right|=d\right)=D_d\left[1-{\left(1-d\right)}^2\right]=2\left(1-d\right)$(7)

Thus, the density function for $\left|X_1-X_2\right|$ on the unit square appears as follows in Figure 3 .

There are two different distance measures in the unit square that we are interested in: the Euclidean distance and the Rectilinear distance. For each, both points are uniformly distributed and in the (0, 1) interval, and shown in Figure 4(a) and Figure 4(b) .

For Euclidean distance, our distance measure is the following [3] :

$d=\sqrt{{\left(X_1-X_2\right)}^2+{\left(Y_1-Y_2\right)}^2}$(8)

This distance is on the (0, $\sqrt{2}$) interval. This equation can also be shown as follows:

$d^2={\left(X_1-X_2\right)}^2+{\left(Y_1-Y_2\right)}^2$(9)

The reasoning for deriving the density function for the unit line in Equation (9) and Figure 2 can be extrapolated as follows:

$P\left({\left(X_1-X_2\right)}^2+{\left(Y_1-Y_2\right)}^2>d\right)={\left(1-d\right)}^2$(10)

From this, we can see that two components are added, permitting us to use the convolution to find the density function for $d$ [6] :

$\left(f\ast g\right)\left(d\right)={\displaystyle {\int }_{-\infty }^{\infty }f\left(t\right)\ast g\left(d-t\right)\text{d}t}$(11)

This provides us with the density function:

$d=\{\begin{array}{ll}0,\hfill & d\le 0\hfill \\ 2d\left(d^2+\pi -4d\right),\hfill & d\le 1\hfill \\ -2d\left(d^2+2{\sin }^{-1}\left(\frac{d^2-2}{d^2}\right)+2+\frac{4-4d^2}{\sqrt{d^2-1}}\right),\hfill & d\le \sqrt{2}\hfill \\ 0,\hfill & d>\sqrt{2}\hfill \end{array}$(12)

For Rectilinear Distance, our distance measure is on the (0, 2) interval and is as follows:

$d=\left|X_1-X_2\right|+\left|Y_1-Y_2\right|$(13)

Because two components are added, we can once again convolve the two components directly above to obtain the density function for the rectilinear distance in the unit square, which is as follows:

$f\left(d\right)=\{\begin{array}{ll}0,\hfill & d\le 0\hfill \\ \frac{2}{3}\left(d^3-6d^2+6d\right),\hfill & d\le 1\hfill \\ \frac{-2}{3}{\left(d-2\right)}^3,\hfill & d\le 2\hfill \\ 0,\hfill & d>2\hfill \end{array}$(14)

A comparison of these two density functions is shown in Figure 5 below.

In the unit cube, we have two points each with X, Y and Z coordinates, all uniformly distributed in the (0, 1) interval. This is shown in Figure 6(a) and Figure 6(b) .

The Euclidean Distance between two points, each point having three coordinates, is as follows

$d=\sqrt{{\left(X_1-X_2\right)}^2+{\left(Y_1-Y_2\right)}^2+{\left(Z_1-Z_2\right)}^2}$(15)

This can also be written as follows:

$d^2={\left(X_1-X_2\right)}^2+{\left(Y_1-Y_2\right)}^2+{\left(Z_1-Z_2\right)}^2$(16)

One will note that the above now has three components and has been added. As such, the convolution can be applied twice here to obtain the density function for the distance measure above. The first convolution involves the distance density involving X and Y from the unit square. The second convolution involves the convolution obtained by employing (X and Y) with Z. This density function is as follows:

$f\left(d\right)=\{\begin{array}{ll}0,\hfill & d\le 0\hfill \\ d\left(d^3+6d\pi -8d^2-4\pi \right),\hfill & d\le 1\hfill \\ -2d\left(a_1+a_2+a_3+a_4\right),\hfill & d\le \sqrt{2}\hfill \\ -d\left(b_1-b_2+b_3+b_4+b_5-b_6+b_7\right),\hfill & d\le \sqrt{3}\hfill \\ 0,\hfill & d>\sqrt{3}\hfill \end{array}$(17)

Where:

$a_1=4d^2{\sin }^{-1}\left(\frac{d^2-2}{d^2}\right)$ (18)

$a_2=4d^2{\tan }^{-1}\left(\sqrt{d^2-1}\right)$ (19)

$a_3=d^4+\left(2\pi +3\right)d^2-4d\pi +3\pi -\frac{1}{2}$ (20)

$a_4=\frac{-8d^4+4d^2+4}{\sqrt{d^2-1}}$ (21)

$b_1=8d{\tan }^{-1}\left(\frac{d^3+d^2-3d-1}{\sqrt{d^2-2}\cdot \left(d^3-d^2-d-1\right)}\right)$ (22)

$b_2=-4d^2{\cot }^{-1}\left(\frac{d^2-d-1}{\sqrt{d^2-2}}\right)$ (23)

$b_3=4d\left(d+2\right){\tan }^{-1}\left(\frac{d^2+d-1}{\sqrt{d^2-2}}\right)$ (24)

$b_4=\left(8d^2-4\right){\sin }^{-1}\left(\frac{d^2-3}{d^2-1}\right)$ (25)

$b_5=16{\sin }^{-1}\left(\frac{\sqrt{d^2-2}}{\sqrt{d^2-1}}\right)$ (26)

$b_6=16{\sin }^{-1}\left(\frac{1}{\sqrt{d^2-1}}\right)$ (27)

$b_7=\frac{\left(d+5\right)\left(\sqrt{d^2-2}-8d^2+16\right)\left(d-1\right)}{\sqrt{d^2-2}}$ (28)

As is the case for rectilinear distance in the unit square, we can continue the logic for the three-dimensional case. The rectilinear distance between two points in the unit cube is as follows:

$d=\left|X_1-X_2\right|+\left|Y_1-Y_2\right|+\left|Z_1-Z_2\right|$(29)

The density function is as follows:

$f\left(d\right)=\{\begin{array}{ll}0,\hfill & d\le 0\hfill \\ \frac{1}{15}{d}^2\cdot \left(d^3-15d^2+60d-60\right),\hfill & d\le 1\hfill \\ \frac{1}{15}\left(-2d^5+30d^4-150d^3+330d^2-315d+93\right),\hfill & d\le 2\hfill \\ \frac{1}{15}{\left(d-3\right)}^2,\hfill & d\le 3\hfill \\ 0,\hfill & d>3\hfill \end{array}$(30)

Figure 7 shows this density function.

In the unit circle, we have the center at the origin (0, 0) and a radius limited to 1. We are not guaranteed a feasible point by random points on the (0, 1) interval. We must generate random points such that coordinates do not exceed a radius of 1. For any point, the x and y coordinates are limited to the maximum radius of 1:

$x^2+y^2\le 1$(31)

Because of this, we employ polar coordinates, such that the random angle is shown as $\theta$ on the (0, 2π) interval, and $r$, the random radius, is on the (0, 1) interval. Our $X$ and $Y$ coordinates are as follows:

$X_1=\sqrt{r_1}\cos {\theta }_1,Y_1=\sqrt{r_1}\sin {\theta }_1$(32)

$X_2=\sqrt{r_2}\cos {\theta }_2,Y_2=\sqrt{r_2}\sin {\theta }_2$ (33)

Take note of the radius component in the square root form. This must be done to guarantee a random distribution of a radius on the (0, 1) interval. If the square root were omitted, the random radius would be skewed toward the center of the circle. We would like a radius that is proportional to the circle’s area. Since the area ( $A$) of a circle is proportional to $\pi r^2$, we desire a radius that takes on the following form:

$r=\sqrt{\frac{A}{\pi }}$(34)

This should justify taking the square root of the random number in the (0, 1) interval for the radius. These distances are shown in Figure 8(a) and Figure 8(b) .

Using Equations (32) and (33) above, we have the Euclidean Distance between two points in the unit circle:

$d=\sqrt{{\left(\sqrt{r_1}\cdot \text{cos}\left({\theta }_1\right)-\sqrt{r_2}\cdot \text{cos}\left({\theta }_2\right)\right)}^2+{\left(\sqrt{r_1}\cdot \text{sin}\left({\theta }_1\right)-\sqrt{r_2}\cdot \text{sin}\left({\theta }_2\right)\right)}^2}$(35)

Unfortunately, we are unable to obtain a density function via the distance function above. The required integration of Equation (35) above does not permit a symbolic solution. Fortunately, there is a clever proof available for the density function of the Euclidian distance between two points in the unit circle, which is as follows [7] :

$f\left(d\right)=\frac{4d}{\pi }\left({\cos }^{-1}\left(\frac{d}{2}\right)-\frac{d}{2}\sqrt{1-{\left(\frac{d}{2}\right)}^2}\right)$(36)

Using Equations (32) and (33) above, we have the Rectilinear Distance between two points in the unit circle.

$d=\left|\sqrt{r_1}\cdot \text{cos}\left({\theta }_1\right)-\sqrt{r_2}\cdot \text{cos}\left({\theta }_2\right)\right|+\left|\sqrt{r_1}\cdot \text{sin}\left({\theta }_1\right)-\sqrt{r_2}\cdot \text{sin}\left({\theta }_2\right)\right|$(37)

As is the case with the Euclidean distance between two points in the unit circle, the Rectilinear Distance is also difficult to obtain, and there are no exact solutions known to the authors. As such, a numerical approximation is employed. The Pearson Distribution is exploited to obtain an approximation to density function of a non-symmetric nature. This approximation is detailed in a subsequent section of this paper. The interval for possible rectilinear distances in the unit circle is (0, $2\sqrt{2}$). For this problem, we approximate the distance density via the following:

$f\left(d\right)=1.3940{\left(-0.2912\cdot d^2+0.8089\cdot d+0.0316\right)}^{1.7168}{\text{e}}^{-0.8992\cdot {\tanh }^{-1}\left(0.7006\cdot d-0.9730\right)}$(38)

The density functions for the unit circle measurements are as shown in Figure 9 below:

In the unit sphere, we have the center at the origin (0, 0) and a radius limited to 1. We are not guaranteed a feasible point by random points on the (0, 1) interval. We must generate random points such that coordinates do not exceed a radius of 1. For any point, the x, y and z coordinates are limited to the maximum radius of 1:

$x^2+y^2+z^2\le 1$(39)

Because of this, we again employ polar coordinates, such that the random altazimuth angle is shown as $\theta$ on the (0, 2π) interval. The declination angle, $\phi$, is in the (0, π) interval, and $r$, the random radius, is on the (0, 1) interval. Our $X$, $Y$ and $Z$ coordinates are as follows:

$X_1=\sqrt{r_1}\cos {\theta }_1\sin {\phi }_1,Y_1=\sqrt{r_1}\sin {\theta }_1\sin {\phi }_1,Z_1=\sqrt{r_1}\cos {\phi }_1$(40)

$X_2=\sqrt{r_2}\cos {\theta }_2\sin {\phi }_2,Y_2=\sqrt{r_2}\sin {\theta }_2\sin {\phi }_2,Z_2=\sqrt{r_2}\cos {\phi }_2$(41)

As was the case for the unit circle, we must take the square root of the radius so that the distribution of the points is uniform throughout the unit sphere. Figure 10(a) and Figure 10(b) are shown in Figure 10(a) and Figure 10(b) .

The Euclidean Distance between two points in the unit sphere is as follows:

$d=\sqrt{{\left(X_1-X_2\right)}^2+{\left(Y_1-Y_2\right)}^2+{\left(Z_1-Z_2\right)}^2}$(42)

where the $X$, $Y$ and $Z$ coordinates are determined by Equations (40 and 41) above. A proof by Lellouche and Souris shows that the density function of a unit sphere has an interval of (0, 2) and is as follows [8] [9] :

$f\left(d\right)=\frac{3d^2}{2}\left({\left(\frac{d}{2}\right)}^3-3\left(\frac{d}{2}\right)+2\right)$(43)

The Rectilinear Distance between two points in the unit sphere is as follows:

$d=\left|X_1-X_2\right|+\left|Y_1-Y_2\right|+\left|Z_1-Z_2\right|$(44)

where $X$, $Y$ and $Z$ coordinates are determined by equations above. As was the case with the Rectilinear Distance in the unit circle, the author is not aware of an exact mathematical solution. As such, the Pearson distribution was again used to obtain the following density function approximation:

$f\left(d\right)=3.628{\left(-0.1748\cdot d^2+0.5828\cdot d+0.0476\right)}^{2.8599}{\text{e}}^{-0.5473\cdot {\tanh }^{-1}\left(0.5726\cdot d-0.9543\right)}$(45)

This density function spans the (0, $2\sqrt{3}$) interval, and is shown in Figure 11 below.

The Unit triangle has vertices at (0, 0), $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ and (1, 0). The X coordinate must be on the (0, 1) interval, and the Y coordinate must be on the $\left(0,\frac{\sqrt{3}}{2}\right)$ interval. Additionally, to guarantee the random point is in the unit triangle, the following two conditions must be met:

$y\le \sqrt{3}x$ (46)

$y\le \sqrt{3}-\sqrt{3}x$ (47)

Figure 12(a) and Figure 12(b) show an example of Euclidean and Rectilinear distances in the unit triangle.

After feasible points are generated in the unit triangle, the Euclidean distance is as follows:

$d=\sqrt{{\left(X_1-X_2\right)}^2+{\left(Y_1-Y_2\right)}^2}$(48)

The density function is:

$f\left(d\right)=132.8{\left(-0.3047d^2+0.2844d+0.0012\right)}^{1.6407}{\text{e}}^{-1.144{\tanh }^{-1}\left(2.124d-0.9912\right)}$(49)

For the points generated above, the Rectilinear distance is as follows:

$d=\left|X_1-X_2\right|+\left|Y_1-Y_2\right|$(50)

The density function is:

$f\left(d\right)=140.8{\left(-0.2295d^2+0.3033d+0.0048\right)}^{2.1791}{\text{e}}^{-1.851{\tanh }^{-1}\left(1.478d-0.9771\right)}$(51)

Figure 13 shows the density function.