Can someone explain like I am 5 year-old about this problem from Hastie's ESL Book? I am working through Hastie's ESL book, and I am having a tough time with Question 2.3.  The question is as follows:

We are considering a nearest neighbor estimate at the origin, and the median distance from the origin to the closest data point is given by this equation.  I have no idea where to begin in terms of trying to derive this.
I know that most data points are closer to the boundary of the sample space, than to any other data point (curse of dimensionality), but I am having trouble translating this into the Linear Algebra / Probability sense.
 A: Let $r$ be distance from the origin, and let $V_0[p]$ be the volume of the unit hypersphere in $p$ dimensions. Then the volume contained in a hypersphere of radius $r$ is
$$V[r]=V_0[p]r^p$$
If we let $P=V[r]/V_0[p]$ denote the fraction of the volume contained within this hypersphere, and define $R=r^p$, then
$$P[R]=R$$
If the data points are uniformly distributed within the unit ball, then for $0\leq R\leq 1$ the above formula is a cumulative distribution function (CDF) for $R$. This is equivalent to a uniform probability density for $R$ over the unit interval, i.e. $p[R]= P'[R] =1$. So, as hinted by Mark Stone in the comments, we can reduce the $p$ dimensional case to an equivalent 1D problem.
Now if we have a single point $R$, then by definition of a CDF we have $\Pr[R\leq \rho]=P[\rho]$ and $\Pr[R\geq \rho]=1-P[\rho]$. If $R_{\min}$ is the smallest value out of $n$ points, and the points are all independent, then the CDF for is given by
$$\Pr[R_{\min}\geq \rho]=\Pr[R\geq \rho]^n=(1-\rho)^n$$
(this is a standard result of univariate extreme value theory).
By definition of the median, we have
$$\frac{1}{2}=\Pr[(R_{\min})_{\mathrm{med}}\geq R]=(1-R)^n$$
which we can rewrite as
$$(1-d^p)^n=\frac{1}{2}$$
which is equivalent to the desired result.
EDIT: Attempt at "ELI5"-style answer, in three parts.


*

*For the 1D case with a single point, the distance is uniformly distributed over $[0,1]$, so the median will be $\frac{1}{2}$.

*In 1D, the distribution for the minimum over $n$ points is the first case to the $n$-th power.

*In $p$ dimensions, the distance $r$ is not uniformly distributed, but $r^p$ is.
