Average Distance from Center of a Unit Cube to Its Surface

Let’s raise the stakes by a dimension. Now, you start at the center of a unit cube. Again, you pick a random direction to move in, with all directions being equally likely. You move along this direction until you reach a point on the surface of the unit cube. On average, how far can you expect to have traveled?

So unlike shifting the shape that we are trying to hit the perimeter of, in the Extra Credit problem, we instead increase the dimensionality. Here $$B_\infty = \left\{ (x,y,z) \in \mathbb{R}^3 \mid \max \{ |x|, |y|, |z| \} = \frac{1}{2} \right\}.$$ Also, instead of parameterizing the direction of travel by a single bearing with respect to the positive $x$-axis, here we need to have two angles, one $\theta \in [0, \pi]$ with respect to the positive $z$-axis and a second $\varphi \in [0,2\pi)$ with respect to the positive $x$-axis. Since the differential of the solid angle of a region of the surface of the sphere is $d\Omega = \sin \theta \,d\varphi \,d\theta$ and the entire surface area of a unit sphere is $4\pi,$ we see that the join distribution function for $\theta$ and $\varphi$ is $$f(\theta, \varphi) = \frac{\sin \theta}{4\pi}.$$ Now, again assuming a unit speed, and given we get the position vector $$r(t) = ( t \cos \varphi \sin \theta, t \sin \varphi \sin \theta, t \cos \theta ).$$

Since you will hit $B_\infty$ whenever $t \max \{ |\cos \varphi \sin \theta|, |\sin \varphi \sin \theta|, |\cos \theta| \} = \frac{1}{2},$ so therefore the distance traveled is \begin{align*} d(\theta, \varphi) &= \frac{1}{2 \max \{ |\cos \varphi \sin \theta|, |\sin \varphi \sin \theta|, |\cos \theta| \}} \\ & = \frac{1}{2} \min \{ |\sec \varphi| \csc \theta, |\csc \varphi| \csc \theta, |\sec \theta | \}, \end{align*} where the final equation removes $\csc \theta$ from the absolute values since $\sin \theta, \csc \theta \geq 0$ for all $\theta \in [0,\pi]$

Therefore, the average distance is $$\hat{d} = \frac{1}{4\pi} \int_0^\pi \int_0^{2\pi} d(\theta, \varphi) \sin \theta \,d\varphi \, d\theta.$$ Now we can decompose this integral into three regions, two of which seem to have analytical solutions, so let's try even though we could just stop here and do the larger integral right here and right now.

Let's only worry about the upper hemisphere, that is $0 \leq \theta \leq \frac{\pi}{2}$. Here we can then break the range into the following three sections:

• Region I - when $0 \leq \theta \leq \frac{\pi}{4},$ where you will eventually hit the top of the unit cube no matter what the value of $\varphi$;
• Region II - when $\cos^{-1} \frac{1}{\sqrt{3}} \leq \theta \leq \frac{\pi}{2},$ where you will eventually hit one of the vertical sides of the unit cube no matter what the value of $\varphi$; and
• Region III - when $\frac{\pi}{4} \leq \theta \leq \cos^{-1} \frac{1}{\sqrt{3}},$ where depending on the value of $\varphi$ you will either hit the top or the sides.

In other words, in Region I, $$d_1(\theta, \varphi) = \frac{1}{2} |\sec \theta|, \forall \varphi \in [0,2\pi),$$ so we have \begin{align*}I_1 &= \frac{1}{2\pi} \int_0^{\pi/4} \int_0^{2\pi} d_1(\theta, \varphi) \,d\varphi \, \sin \theta \,d\theta \\ &= \frac{1}{2\pi} \int_0^{\pi/4} \pi \tan \theta \, d\theta = \left.-\frac{1}{2} \ln | \cos \theta | \right|^{\pi/4}_0 \\ &= \frac{1}{4} \ln 2 \approx 0.17328679514\dots.\end{align*} In Region II, $$d_2(\theta, \varphi) = \frac{\csc \theta}{2} \min \{ |\sec \varphi|, |\csc \varphi| \},$$ so we have \begin{align*} I_2 &= \frac{1}{2\pi} \int_{\cos^{-1} (1/\sqrt{3})}^{\pi/2} \int_0^{2\pi} \frac{1}{2} \min \{ |\sec \varphi|, |\csc \varphi| \} \,d\varphi \csc \theta \sin \theta \,d\theta \\ &= \int_{\cos^{-1} (1/\sqrt{3})}^{\pi/2} \left( \frac{1}{2\pi} \int_0^{2\pi} \frac{1}{2} \min \{ |\sec \varphi|, |\csc \varphi| \} \,d\varphi \right) \,d\theta\\ & = \frac{2}{\pi} \ln ( 1 + \sqrt{2} ) \left( \frac{\pi}{2} - \cos^{-1} \left(\frac{1}{\sqrt{3}}\right) \right)\\ & = \left( 1 - \frac{ 2 \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) }{\pi} \right) \ln (1 + \sqrt{2}) \approx 0.345345573653\dots,\end{align*} since the integral with respect to $\varphi$ is none other than the expected distance traveled inside a square, which we found the solution for in the Classic Problem. The trickier third region has integral $$I_3 = \int_{\pi/4}^{\cos^{-1} (1/\sqrt{3})} \int_0^{2\pi} d(\theta,\varphi) \,d\varphi \sin \theta \, d\theta \approx 0.0920550322989\dots,$$ which does not seem to have a readily available analytical solution. Putting these regions together we get that the overall average distance traveled to the surface of the unit cube when uniformly randomly choosing a direction is $$\hat{d} = I_1 + I_2 + I_3 = \frac{1}{4} \ln 2 + \left( 1 - \frac{ 2 \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) }{\pi} \right) \ln (1 + \sqrt{2}) + I_3 \approx 0.610687401568\dots.$$