Expected distance in R^n

Note: This is not a complete solution, but shows that $n$ has a lower limit of $n \geq 11$ . Thanks to @Brian Moehring for his comments and pointing me to Jensen's inequality.

In $n$ -dimensional space, the distance from the origin to a point $(p_1, p_2, \dots, p_n)$ is $d = \sqrt{p_1^2 + p_2^2 + \dots + p_n^2}$ .

A point that is on an $n$ -dimensional unit hypercube centered at the origin will have one coordinate that is exactly $\pm \frac{1}{2}$ and all the other coordinates ranging between $[-\frac{1}{2}, \frac{1}{2}]$ .

By symmetry, we only need to consider $p_1 = \frac{1}{2}$ , and the squares of the expected values of the other coordinates $p^2$ can be found by calculating the value of $\frac{\int_{-\frac{1}{2}}^{\frac{1}{2}} p^2 dp}{\frac{1}{2} - -\frac{1}{2}} = \frac{1}{12}$ .

Therefore, by Jensen's inequality , the expected distance from a point on the surface of this $n$ -dimensional cube to the center is $\mathbb{E}d < \sqrt{(\frac{1}{2})^2 + (n - 1)\frac{1}{12}}$ which has a greater minimum value of $1$ for $n \geq \boxed{11}$ .

I struggled with integration by hand, and then decided to do simple Monte Carlo integration. Included in the code are estimates for the error (as standard deviation). $\pm 5\sigma$ should be more than enough of an error bar. To be rigorous, one should form and test a hypothesis, but here the error bars are way to small for tests to have any practical significance.

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%% Monte Carlo integration. Problem: Mean distance to a face of a hypercube
N_samples=1000000;
Max_dim=20;
r=rand(N_samples,Max_dim)-0.5;
r(:,1)=0.5;
d=sqrt(cumsum(r.^2,2));
m=mean(d);
s=std(d)/sqrt(N_samples);
fprintf('Number of dim.  Mean distance  Standard deviation\n')
fprintf('%8d   %14.4f   %14.5f\n', [(1:Max_dim)', m', s']')

Considering that the faces of n-hypercube are in dimension of (n-1) hypercube and knowing that the average distance from center of the n- hypercube to any other point inside the cube is √(n/12). the average distance from the center of the n-hypercube to any point on the surface of the n-hypercube is given by

√(1/4 +(n-1)/12) > 1, gives n >10.

Answer=11.

Align the cube with coordinate system. Fix one of the coordinates of out point to the value 1/2, so that it is at the surface. The distance squared to the origin is the summed square over the coordinates. Very conveniently, there is no interaction between the coordinates in this problem. So the fixed coordinate contributes $(\frac{1}{2})^2=\frac{1}{4}$ to the squared distance. Any other coordinate may range between -1/2 and 1/2 and thus contributes on average $\frac{\int_{-1/2}^{1/2}x^2dx}{\frac{1}{2}- \frac{-1}{2}} = \frac{1}{12}$ to the squared distance. We need 10 other coordinates to exceed a expected (squared) distance of 1.