In Elements of Statistical Learning, a problem is introduced to highlight issues with k-nn in high dimensional spaces. There are $N$ data points that are uniformly distributed in a $p$-dimensional unit ball.

The median distance from the origin to the closest data point is given by the expression:

$$d(p,N) = \left(1-\left(\frac{1}{2}\right)^\frac{1}{N}\right)^\frac{1}{p}$$

When $N=1$, the formula breaks down to half the radius of the ball, and I can see how the closest point approaches the border as $p \rightarrow \infty$, thus making the intuition behind knn break down in high dimensions. But I can't grasp why the formula has a dependence on N. Could someone please clarify?

Also the book addresses this issue further by stating: "...prediction is much more difficult near the edges of the training sample. One must extrapolate from neighboring sample points rather than interpolate between them". This seems like a profound statement, but I can't seem to grasp what it means. Could anyone reword?

  • 1
    $\begingroup$ You need to edit your displayed equation a little. Is that $\frac 1N$ exponent applicable only to that $1$ in the numerator the way it looks now, or did you want it to apply to the whole $\frac 12$? $\endgroup$ Jan 2 '15 at 19:39
  • 1
    $\begingroup$ It would help to distinguish the "hypersphere" (which in $\mathbb{R}^p$ is a manifold of dimension $p-1$) from the "unit ball" (which has dimension $p$). The hypersphere is the boundary of the ball. If, as your title says, all points are sampled from the hypersphere, then--by definition--they all have distance $1$ from the origin, the median distance is $1$, and all are equally close to the origin. $\endgroup$
    – whuber
    Jan 2 '15 at 21:06
  • $\begingroup$ @DilipSarwate It is applied to the whole $\frac{1}{2}$. In the book there is an example where $N=500, p=10$ so $d(p, N) \approx 0.52$ $\endgroup$
    – user64773
    Jan 2 '15 at 22:48

The volume of an $p$-dimensional hyperball of radius $r$ has a volume proportional to $r^p$.

So the proportion of the volume more than a distance $kr$ from the origin is $\frac{r^p-(kr)^p}{r^p}=1-k^p$.

The probability that all $N$ randomly chosen points are more than a distance $kr$ from the origin is $\left(1-k^p\right)^N$. To get the median distance to the nearest random point, set this probability equal to $\frac12$. So $$\left(1-k^p\right)^N=\tfrac12 $$ $$\implies k=\left(1-\tfrac1{2^{1/N}}\right)^{1/p}.$$

Intuitively this makes some sort of sense: the more random points there are, the closer you expect the nearest one to the origin to be, so you should expect $k$ to be a decreasing function of $N$. Here $2^{1/N}$ is a decreasing function of $N$, so $\tfrac1{2^{1/N}}$ is an increasing function of $N$, and thus $1-\tfrac1{2^{1/N}}$ is a decreasing function of $N$ as is its $p$th root.

  • $\begingroup$ Ah, nice way of looking at it. Would you be able to reinterpret the quote in my second question? $\endgroup$
    – user64773
    Jan 2 '15 at 23:09
  • $\begingroup$ I suspect it may be suggesting that in high dimensions, points to predict are effectively a long way from the training data, as if on the edge of a sphere, so you are not really interpolating but rather extrapolating, and so uncertainties are much greater. But I do not really know. $\endgroup$
    – Henry
    Jan 3 '15 at 1:03
  • 1
    $\begingroup$ I don't get it - I understand why this expression is the probability for all points to be farther than kr, but why does setting this probability to 1/2 gives the median distance?? $\endgroup$
    – ihadanny
    Oct 20 '15 at 21:38
  • 1
    $\begingroup$ @ihadanny: the value $k=\left(1-\tfrac1{2^{1/N}}\right)^{1/p}$ gives the fraction of the radius where the probability all $N$ points are further away is $\frac12$, and so where the probability at least one point is closer is $1-\frac12=\frac12$, so $kr$ is the median of the distribution of the distance of the closest point. $\endgroup$
    – Henry
    Oct 20 '15 at 23:22
  • 1
    $\begingroup$ Definition of median, half are bigger and half are smaller. $\endgroup$ Sep 20 '18 at 14:54

And now without hand waving

  1. For any sequence of i.i.d rv's, $$P( \min_{1\le i\le N} Y_i > y ) = (1-F(y))^N,$$ where $F$ is the common CDF

  2. Thus if we have $N$ i.i.d uniformly distributed $X_i$ in the unit ball in $p$ dimensions, then $$P( \min_{1\le i\le N} ||X_i|| > r ) = (1-F(r))^N,$$ where $F$ is the common CDF of the distances, $||X_i||, i=1,2,\ldots,N$. Finally, what is the CDF, $F$, for a uniformly distributed point in the unit ball in $R^p$? The probability that the point lies in the ball of radius r within the ball of unit radius equal the ratio of volumes:

$$F(r) = P ( ||X_i|| \le r ) = C r^p/( C 1^p) = r^p$$

Thus the solution to

$$1/2 = P( \min_{1\le i\le N} ||X_i|| > r ) = (1- r^p)^N$$


$$r = (1 - (1/2)^{1/N})^{1/p}.$$

Also your question about dependence on the sample size, $N$. For $p$ fixed, as the ball fills up with more points, naturally the minimum distance to the origin should become smaller.

Finally, there is something amiss in your ratio of volumes. It looks like $k$ should be the volume of the unit ball in $R^p$.


As concise but in words:

We want to find the median distance of the closest point to the origin in $N$ uniformly distributed points in the ball at the origin of unit radius in $p$ dimensions. The probability that the smallest distance exceeds $r$, (call this quantity expression [1]) is the $N^{th}$ power of the probability that a single uniformly distributed point exceeds $r$, because of statistical independence. The latter is one minus the probability that a single uniformly distributed point is less than $r$. The latter is the ratio of volumes of the ball of radius $r$ to the ball of unit radius, or $r^p$. We can now write expression [1] as

$$ P( \min_{1\le i\le N} ||X_i|| > r ) = (1- r^p)^N.$$

To find the median of the distribution of the minimum of the distances, set the above probability to $1/2$ and solve for $r$, obtaining the answer.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.