数据科学基石：高维空间与SVD在机器学习中的应用

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"《数据科学基础》是一本深入探讨现代数据分析和技术的教材，涵盖了从高维空间的理解到复杂机器学习算法的详细讲解。本书由Avrim Blum、John Hopcroft和Ravindran Kannan共同编撰，旨在为读者提供数据科学的核心概念。第1章“介绍”引导读者进入数据科学的世界，强调了该领域在当今信息时代的重要性。随后的章节深入探索了以下几个关键主题： 2.1 高维空间：首先介绍了高维空间的概念，强调在实际应用中，随着维度增加，我们遇到的几何特性会与直观经验大相径庭。这包括概率论中的大数定律在高维中的表现以及球体体积随维度的增长。 2.2 法则与几何：探讨了高维空间的几何特性，如单位球的体积变化和靠近极点区域的体积特点。这部分还涉及如何在高维空间中均匀随机生成点的方法。 2.3 高斯分布：随着维度上升，高斯分布（正态分布）的行为变得尤为重要。随机投影和Johnson-Lindenstrauss引理被用来处理高维数据的降维问题，这对于大规模数据处理至关重要。 2.4 分类和聚类：章节中讨论了如何通过SVD等技术来区分高维空间中的不同数据集，如分离两个独立的高斯分布，并且介绍了如何将数据拟合到一个球形高斯模型。 3.1 最佳近似子空间与Singular Value Decomposition (SVD)：这部分是核心内容之一，讲述了SVD的基础理论，包括预处理步骤、奇异向量和奇异值的定义。通过SVD，可以找到数据的最佳低秩近似，这对于机器学习中的特征选择和降维至关重要。 3.2 SVD的应用：包括如何利用左奇异向量进行特征分析，以及快速求解SVD的Power Method方法和其优化版本。此外，SVD与主成分分析（PCA）和矩阵分解有着紧密联系，对于数据压缩和异常检测有广泛的应用。 4. 机器学习部分：书中介绍了大规模数据问题的解决策略，如流式处理、抽样和近似算法，这些都是大数据时代不可或缺的技术。此外，还有诸如聚类、随机图、主题模型、非负矩阵分解等重要概念。 5. 模型与概率论：包括隐马尔可夫模型（HMM）和图形模型，这些是序列数据建模和复杂结构数据理解的关键工具。 6. 信号处理：波形分析，如wavelets，展示了在处理时间或频率信号时的数学工具。《数据科学基础》提供了坚实的数据分析基础，不仅适用于研究人员，也适合对机器学习和数据分析感兴趣的学生和从业者。通过阅读这本书，读者可以深入理解数据科学中的各种核心原理，并掌握在实际问题中运用这些技术的技巧。"

Condition Tail bound

Markov x ≥ 0 Prob(x ≥ a) ≤

E(x)

Chebychev Any x Prob



|x − E(x)| ≥ a



≤

Var(x)

Chernoﬀ x = x

+ x

+ ··· + x

Prob(|x − E(x)| ≥ εE(x))

∈ [0, 1] i.i.d. Bernoulli; ≤ 3e

−cε

E(x)

Higher Moments r positive even integer Prob(|x| ≥ a) ≤ E(x

)/a

Gaussian x =

+ x

+ ··· + x

Prob(|x −

√

n| ≥ β) ≤ 3e

−cβ

Annulus x

∼ N(0, 1); β ≤

√

n indep.

Power Law x = x

+ x

+ . . . + x

Prob



|x − E(x)| ≥ εE(x)



for x

; order k ≥ 4 x

i.i.d ; ε ≤ 1/k

≤ (4/ε

kn)

(k−3)/2

Figure 2.1: Table of Tail Bounds. The Higher Moments bound is obtained by apply-

ing Markov to x

. The Chernoﬀ, Gaussian Annulus, and Power Law bounds follow from

Theorem 2.5 which is proved in the appendix.

To see that this is true, partition A into inﬁnitesimal cubes. Then, (1 − ε)A is the union

of a set of cubes obtained by shrinking the cubes in A by a factor of 1 − ε. When we

shrink each of the 2d sides of a d-dimensional cube by a factor f, its volume shrinks by a

factor of f

. Using the fact that 1 − x ≤ e

−x

, for any object A in R

we have:

volume



(1 − )A



volume(A)

= (1 − )

≤ e

−d

Fixing  and letting d → ∞, the above quantity rapidly approaches zero. This means

that nearly all of the volume of A must be in the portion of A that does not belong to

the region (1 − )A.

Let S denote the unit ball in d dimensions, that is, the set of points within distance

one of the origin. An immediate implication of the above observation is that at least a

1 − e

−d

fraction of the volume of the unit ball is concentrated in S \ (1 − )S, namely

in a small annulus of width  at the boundary. In particular, most of the volume of the

d-dimensional unit ball is contained in an annulus of width O(1/d) near the boundary. If

the ball is of radius r, then the annulus width is O





Lemma 2.6 The surface area A(d) and the volume V (d) of a unit-radius ball in d di-

mensions are given by

A (d) =

2π

Γ(

)

and V (d) =

2π

d Γ(

)

To check the formula for the volume of a unit ball, note that V (2) = π and V (3) =

(

)

π, which are the correct volumes for the unit balls in two and three dimen-

sions. To check the formula for the surface area of a unit ball, note that A(2) = 2π and

A(3) =

2π

√

= 4π, which are the correct surface areas for the unit ball in two and three

dimensions. Note that π

is an exponential in

and Γ





grows as the factorial of

This implies that lim

d→∞

V (d) = 0, as claimed.

2.4.2 Volume Near the Equator

An interesting fact about the unit ball in high dimensions is that most of its volume

is concentrated near its “equator”. In particular, for any unit-length vector v deﬁning

“north”, most of the volume of the unit ball lies in the thin slab of points whose dot-

product with v has magnitude O(1/

√

d). To show this fact, it suﬃces by symmetry to ﬁx

v to be the ﬁrst coordinate vector. That is, we will show that most of the volume of the

unit ball has |x

| = O(1/

√

d). Using this fact, we will show that two random points in the

unit ball are with high probability nearly orthogonal, and also give an alternative proof

from the one in Section 2.4.1 that the volume of the unit ball goes to zero as d → ∞.

Theorem 2.7 For c ≥ 1 and d ≥ 3, at least a 1 −

−c

fraction of the volume of the

d-dimensional unit ball has |x

| ≤

√

d−1

Proof: By symmetry we just need to prove that at most a

−c

fraction of the half of

the ball with x

≥ 0 has x

≥

√

d−1

. Let A denote the portion of the ball with x

≥

√

d−1

and let H denote the upper hemisphere. We will then show that the ratio of the volume

of A to the volume of H goes to zero by calculating an upper bound on volume(A) and

a lower bound on volume(H) and proving that

volume(A)

volume(H)

≤

upper bound volume(A)

lower bound volume(H)

−

To calculate the volume of A, integrate an incremental volume that is a disk of width

and whose face is a ball of dimension d − 1 and radius

1 − x

. The surface area of

the disk is (1 − x

)

d−1

V (d −1) and the volume above the slice is

volume(A) =

√

d−1

(1 − x

)

d−1

V (d −1)dx

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数据科学基石：高维空间与SVD在机器学习中的应用

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