# Posts tagged *stat*

One popular model for classification is nearest neighbors. It's broadly applicable and unbiased as it makes no assumptions about the generating distribution of the data. Suppose we have $N$ observations. Each observation can be written as $(\mathbf{x}_i,y_i)$, where $0 \leq i \leq N$ and $\mathbf{x}_i = (x_{i1},x_{i2},\ldots,x_{ip})$, so we have $p$ features, and $y_i$ is the class to which $\mathbf{x}_i$ belongs. Let $y_i \in C$, where $C$ is a set of possible classes. If we were given a new observation $\mathbf{x}$. We would find the $k$ closest $(\mathbf{x}_i, y_i)$, say $(\mathbf{x}_{j_1}, y_{j_1}), (\mathbf{x}_{j_2}, y_{j_2}),\ldots, (\mathbf{x}_{j_k}, y_{j_k})$. To classify $\mathbf{x}$, we would simply take a majority vote among these $k$ closest points. While simple and intuitive, as we will see, nearest neighbors runs into problems when $p$ is large.

Consider this problem:

Consider $N$ data points uniformly distributed in a $p$-dimensional unit ball centered at the origin. Find the median distance from the origin of the closest data point among the $N$ points.

Let the median distance be $d(p, N)$. First to keep things simple consider a single data point, so $N = 1$. The volume of a $p$-dimensional ball of radius $r$ is proportional to $r^p$, so $V(r) = Kr^p$. Let $d$ be the distance of the point, so $P(d \leq d(p,1)) = 0.5$. Viewing probability as volume, we imagine a smaller ball inside the larger ball, so \begin{align*} \frac{1}{2} &= P(d \leq d(p, 1)) \\ &= \frac{V(d(p,1))}{V(1)} \\ &= d(p,1)^p \\ &\Rightarrow d(p,1) = \left(\frac{1}{2}\right)^{1/p}, \end{align*} and in general, $P(d \leq t) = t^p$, where $0 \leq t \leq 1$. For example when $p = 1$, we have

For $p=2$,Now, consider the case when we have $N$ data points, $x_1, x_2, \ldots, x_N$. The distance of the closest point is $$d = \min\left(\Vert x_1 \Vert, \Vert x_2 \Vert, \ldots, \Vert x_N \Vert\right).$$ Thus, we'll have \begin{align*} \frac{1}{2} &= P(d \leq d(p,N)) \\ &= P(d > d(p,N)),~\text{since $P(d \leq d(p,N)) + P(d > d(p,N)) = 1$} \\ &= P\left(\Vert x_1\Vert > d(p,N)\right)P\left(\Vert x_2\Vert > d(p,N)\right) \cdots P\left(\Vert x_N\Vert > d(p,N)\right) \\ &= \prod_{i=1}^N \left(1 - P\left(\Vert x_i \Vert \leq d(p,N)\right)\right) \\ &= \left(1 - d(p,N)^p\right)^N,~\text{since $x_i$ are i.i.d and $P(\Vert x_i\Vert \leq t) = t^p$}. \end{align*} And so, \begin{align*} \frac{1}{2} &= \left(1 - d(p,N)^p\right)^N \\ \Rightarrow 1 - d(p,N)^p &= \left(\frac{1}{2}\right)^{1/N} \\ \Rightarrow d(p,N)^p &= 1 - \left(\frac{1}{2}\right)^{1/N}. \end{align*} Finally, we obtain $$\boxed{d(p,N) = \left(1 - \left(\frac{1}{2}\right)^{1/N}\right)^{1/p}}.$$

So, what does this equation tell us? As the dimension $p$ increases, the distance goes to 1, so all points become far away from the origin. But as $N$ increases the distance goes to 0, so if we collect enough data, there will be a point closest to the origin. But note that as $p$ increases, we need an exponential increase in $N$ to maintain the same distance.

Let's relate this to the nearest neighbor method. To make a good prediction on $\mathbf{x}$, perhaps, we need a training set point that is within distance 0.1 from $\mathbf{x}$. We would need 7 data points for there to be a greater than 50% chance of such a point existing if $p = 1$. See how $N$ increases as $p$ increases.

p | N |
---|---|

1 | 7 |

2 | 69 |

3 | 693 |

4 | 6932 |

5 | 69315 |

10 | 6,931,471,232 |

15 | $\scriptsize{6.937016 \times 10^{14}}$ |

The increase in data needed for there to be a high probability that there is a point close to $\mathbf{x}$ is exponential. We have just illustrated the *curse of dimensionality*. So, in high dimensions, other methods must be used.