Foundations of Data Science

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Independent of irrelevant alternatives – If individuals modify their rankings but keep the order of a and b unchanged, then the global order of a and b should not change. Arrow showed that no ranking algorithm exists satisfying the above axioms. Theorem 10.1 (Arrow) Any algorithm for creating a global ranking from individual rankings of three or more elements in which the global ranking satisfies unanimity and independence of irrelevant alternatives is a dictatorship. Proof: Let a, b, and c be distinct items. Consider a set of rankings in which each individual ranks b either first or last. Some individuals may rank b first and others may rank b last. For this set of rankings, the global ranking must put b first or last. Suppose to the contrary that b is not first or last in the global ranking. Then there exist a and c where the global ranking puts a > b and b > c. By transitivity, the global ranking puts a > c. Note that all individuals can move c above a without affecting the order of b and a or the order of b and c since b was first or last on each list. Thus, by independence of irrelevant alternatives, the global ranking would continue to rank a > b and b > c even if all individuals moved c above a since that would not change the individuals relative order of a and b or the individuals relative order of b and c. But then by unanimity, the global ranking would need to put c > a, a contradiction. We conclude that the global ranking puts b first or last. Consider a set of rankings in which every individual ranks b last. By unanimity, the global ranking must also rank b last. Let the individuals, one by one, move b from bottom to top leaving the other rankings in place. By unanimity, the global ranking must eventually move b from the bottom all the way to the top. When b first moves, it must move all the way to the top by the previous argument. Let v be the first individual whose change causes the global ranking of b to change. We now argue that v is a dictator. First, we argue that v is a dictator for any pair ac not involving b. We will refer to three rankings of v (see Figure 10.1). The first ranking of v is the ranking prior to v moving b from the bottom to the top and the second is the ranking just after v has moved b to the top. Choose any pair ac where a is above c in v’s ranking. The third ranking of v is obtained by moving a above b in the second ranking so that a > b > c in v’s ranking. By independence of irrelevant alternatives, the global ranking after v has switched to the third ranking puts a > b since all individual ab votes are the same as just before v moved b to the top of his ranking. At that time the global ranking placed a > b. Similarly b > c in the global ranking since all individual bc votes are the same as just after v moved b to the top causing b to move to the top in the global ranking. By transitivity the global ranking must put a > c and thus the global ranking of a and c agrees with v. Now all individuals except v can modify their rankings arbitrarily while leaving b in its extreme position and by independence of irrelevant alternatives, this does not affect the 330
b . a . a b b b b . . b b a a b b . c . . b b b b v global . . a . c . b b v . . c . global . c . . b b v . c . . global first ranking second ranking third ranking Figure 10.1: The three rankings that are used in the proof of Theorem 10.1. global ranking of a > b or of b > c. Thus, by transitivity this does not affect the global ranking of a and c. Next, all individuals except v can move b to any position without affecting the global ranking of a and c. At this point we have argued that independent of other individuals’ rankings, the global ranking of a and c will agree with v’s ranking. Now v can change its ranking arbitrarily, provided it maintains the order of a and c, and by independence of irrelevant alternatives the global ranking of a and c will not change and hence will agree with v. Thus, we conclude that for all a and c, the global ranking agrees with v independent of the other rankings except for the placement of b. But other rankings can move b without changing the global order of other elements. Thus, v is a dictator for the ranking of any pair of elements not involving b. Note that v changed the relative order of a and b in the global ranking when it moved b from the bottom to the top in the previous argument. We will use this in a moment. The individual v is also a dictator over every pair ab. Repeat the construction showing that v is a dictator for every pair ac not involving b only this time place c at the bottom. There must be an individual v c who is a dictator for any pair such as ab not involving c. Since both v and v c can affect the global ranking of a and b independent of each other, it must be that v c is actually v. Thus, the global ranking agrees with v no matter how the other voters modify their rankings. 10.2 Hare System for Voting One voting system would be to have everyone vote for their favorite candidate. If some candidate receives a majority of votes, he or she is declared the winner. If no candidate receives a majority of votes, the candidate with the fewest votes is dropped from the slate and the process is repeated. 331
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Foundations of Data Science ∗ Avr
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4.4 Branching Processes . . . . . .
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8.2.2 Structural properties of the
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12.4.7 Median . . . . . . . . . . .
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densities, discrete optimization, e
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2 High-Dimensional Space 2.1 Introd
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Also, if x and y are independent, t
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1 1 − ɛ Annulus of width 1 d Fig
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integral gives ∫ ∞ 0 e −r2 r
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The volume of the hemisphere below
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points onto the circle. In higher d
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Using the substitution 2z = y 2 , |
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For the nearest neighbor problem, i
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√ √ 2d+O(1) ≤ 2d + ∆2 −O(
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2.10 Bibliographic Notes The word v
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Exercise 2.9 A 3-dimensional cube h
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Exercise 2.26 Explain how the volum
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Exercise 2.40 Consider a non orthog
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1 Exercise 2.50 Use the probability
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This decomposition of A can be view
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3.3 Singular Vectors We now define
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orthonormal basis w 1 , w 2 , . . .
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A n × d = U n × r D r × r V T r
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3.6 Left Singular Vectors Theorem 3
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Lemma 3.10 (Analog of eigenvalues a
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Proof: Let A = r∑ σ i u i vi T i
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a i , let dist(a i , l) denote its
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such models. Mixture models are a v
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1. The best fit 1-dimension subspac
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This leads to the following theorem
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Thus, the maximum cut problem can b
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and this meets the claimed error bo
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(0,3) (1,1) (3,0) ⎛ M = ⎝ 1 1 0
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Exercise 3.18 Modify the power meth
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2. What percent of the Forbenius no
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City Bei- Tian- Shang- Chong- Hoh-
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5 10 15 20 25 30 35 40 4 9 14 19 24
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Prob(vertex has degree k) = ( ) n
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and thus k k ≤ n. Since k! ≤ k
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For x to be equal to zero, it must
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No items E(x) ≥ 0.1 At least one
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Our first task is to figure out wha
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√ Thus, the probability for all s
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Figure 4.7: A degree three vertex w
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ftp://ftp.cs.rochester.edu/pub/u/jo
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Size of frontier 0 ln n nθ n Numbe
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For a small number i of steps, the
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same or are disjoint, ⎛( n∑ )
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are O(ln n) in size and the expecte
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f(x) q p 0 f(f(x)) f(x) x Figure 4.
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may be infinite. That is, ∞∑ i=
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Property cycles 1/n giant component
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Keep in mind that the leading terms
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4.6 Phase Transitions for Increasin
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p(n) A symmetric argument shows tha
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simple. We supply some intuition be
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Proof: Say that t is a “busy time
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Consider a graph in which half of t
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problem contributes a minuscule err
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one. On the other hand, for δ = 1,
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for all δ greater than δ critical
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expected degree d i = 1, 2, 3 √ t
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Destination Figure 4.17: For d < 2,
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For at least half of the pairs of v
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S i+1 is (1 − 1 ) |Si| ≤ 1 −
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Exercise 4.7 Let f (n) be a functio
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Exercise 4.19 (Birthday problem) Wh
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Exercise 4.38 Consider a model of a
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Exercise 4.53 Consider graph 3-colo
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5 Random Walks and Markov Chains A
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A B C (a) A B C (b) Figure 5.1: (a)
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in time polynomial in n. A quantity
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5.2 Markov Chain Monte Carlo The Ma
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p(a) = 1 2 p(b) = 1 4 p(c) = 1 8 p(
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5 8 7 12 1 3 1 6 1 8 1 3 3,1 3,2 3,
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Figure 5.4: A network with a constr
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There is a simple interpretation of
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Let γ i = π 1 + π 2 + · · · +
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5.4.1 Using Normalized Conductance
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The Gaussian distribution on the in
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6 6 1 8 1 5 8 4 5 5 Graph with boun
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and then reaching a from y before r
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every vertex? Hitting time The hitt
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clique of size n/2 x y } {{ } n/2 F
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i ↑ ↓ ↑ ↓ ↑ j ↑ =⇒ i
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Theorem 5.13 Let G be an undirected
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0 1 2 3 4 12 20 Number of resistors
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1 2 4 Figure 5.11: Paths obtained f
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whereas, with k self-loops, the equ
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or p[I − (1 − α)A] = α (1, 1,
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Exercise 5.10 Let p be a probabilit
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i 1 i 2 R 1 R 2 R 3 Figure 5.13: An
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(a) 1 2 3 4 (b) 1 2 3 4 (c) 1 2 3 4
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Exercise 5.40 Suppose that the cliq
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Exercise 5.55 Using a web browser b
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will allow us to make mathematical
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Not spam { }} { { Spam }} { x 1 x 2
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1 ∨ x 8 = 1}, or more succinctly,
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described using O(k log d) bits: lo
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In fact, we can show this bound is
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y definition of w ∗ . Similarly,
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y x φ Figure 6.4: Data that is not
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Theorem 6.11 (Online to Batch via R
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function of concept class H, will g
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Corollary 6.16 (VC-dimension sample
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A sphere in d-dimensions is a set o
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then with probability ≥ 1 − δ,
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work on a variety of measures. One
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Since weight(0) = n, the total weig
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Stochastic Gradient Descent: Given:
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sleeping experts problem. Combining
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⎫ ⎪⎬ ⎪⎭ Each gate is conn
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W 1 W 2 W 1 W 2 W 3 (a) (b) Figure
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convolution pooling Image Convoluti
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discard separators in advance that
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6.14.2 Active learning Active learn
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6.16 Exercises Exercise 6.1 (Sectio
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√ L log(L(T )) |S|) will be at mo
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7 Algorithms for Massive Data Probl
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important to know if some popular s
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We now give an example of a 2-unive
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estimate of m. To obtain a coin tha
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expected value of the sum will be z
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δ have values in ±1. There are ex
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mean. (ii) There is “no free lunc
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⎡ ⎢ ⎣ A m × n ⎤ ⎡ ⎥
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One uses the primitive in Section ?
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would require s ≥ n, which clearl
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its that capture sufficient informa
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7.6 Exercises Algorithms for Massiv
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Counting Frequent Elements The Majo
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Exercise 7.30 Blast: Given a long s
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Preliminaries: We will follow the s
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B A Figure 8.1: Example where the n
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Lloyd’s algorithm: Start with k c
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8.2.5 k-means clustering on the lin
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a clustering that is ɛ-close to C
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Theorem 8.5 Assume A satisfies (c,
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probability is q (where, q < p). 32
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1 + x ≤ e x for all real x (1 −
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Thus, n! ≤ n n e −n√ ne. For
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from which the current inequality f
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Example: Let f (x) = x k for k an e
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Random variables x 1 , x 2 , . . .
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12.4.7 Median One often calculates
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The density of y is the unit varian
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Generation of random numbers accord
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Example: Consider flipping a coin 1
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Setting λ = ln(1 + δ) Prob ( s >
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12.5 Bounds on Tail Probability Aft
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Collect terms of the summation with
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The first integral is just the stan
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of a symmetric matrix, counting mul
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capture this situation. Now AA T =
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The above theorem tells us that the
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Important special cases are |x| 0 t
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Lemma 12.23 Let A be a symmetric ma
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etween vertices in different blocks
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∞∑ ∑ ix i = ∞ i=0 i=0 x d d
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Generating functions are useful for
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Thus, u n = (n − 1) u n−2 . The
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f(x) a c b Figure 12.3: Illustratio
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need to argue that no other sequenc
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1. How large must δ be if we wish
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Exercise 12.40 We are given the pro
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Index 2-universal, 240 4-way indepe
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Lagrange, 423 Law of large numbers,
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References [AK] Sanjeev Arora and R
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[MR95b] [MU05] [MV10] Rajeev Motwan
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