Foundations of Data Science

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3 4 1 2 q 1 p h 1t 2 h 1t 2 2 3 3 1 2 1 4 The initial distribution is α(q) = 1 and α(p) = 0. At each step a change of state occurs followed by the output of heads or tails with probability determined by the new state. We consider three problems in increasing order of difficulty. First, given a HMM what is the probability of a given output sequence? Second, given a HMM and an output sequence, what is the most likely sequence of states? And third, knowing that the HMM has at most n states and given an output sequence, what is the most likely HMM? Only the third problem concerns a ”hidden” Markov model. In the other two problems, the model is known and the questions can be answered in polynomial time using dynamic programming. There is no known polynomial time algorithm for the third question. How probable is an output sequence Given a HMM, how probable is the output sequence O = O 0 O 1 O 2 · · · O T of length T +1? To determine this, calculate for each state i and each initial segment of the sequence of observations, O 0 O 1 O 2 · · · O t of length t + 1, the probability of observing O 0 O 1 O 2 · · · O t ending in state i. This is done by a dynamic programming algorithm starting with t = 0 and increasing t. For t = 0 there have been no transitions. Thus, the probability of observing O 0 ending in state i is the initial probability of starting in state i times the probability of observing O 0 in state i. The probability of observing O 0 O 1 O 2 · · · O t ending in state i is the sum of the probabilities over all states j of observing O 0 O 1 O 2 · · · O t−1 ending in state j times the probability of going from state j to state i and observing O t . The time to compute the probability of a sequence of length T when there are n states is O(n 2 T ). The factor n 2 comes from the calculation for each time unit of the contribution from each possible previous state to the probability of each possible current state. The space complexity is O(n) since one only needs to remember the probability of reaching each state for the most recent value of t. Algorithm to calculate the probability of the output sequence The probability, Prob(O 0 O 1 · · · O T , i) of the output sequence O 0 O 1 · · · O T state i is given by ending in Prob(O 0 , i) = α(i)p(O 0 , i) 304
for t = 1 to T Prob(O 0 O 1 · · · O t , i) = ∑ j Prob(O 0 O 1 · · · O t−1 , j)a ij p(O t+1 , i) Example: What is the probability of the sequence hhht by the HMM in the above two state example? t = 3 3 t = 2 1 t = 1 1 1 1 + 5 3 1 = 19 32 2 2 72 4 2 384 1 1 + 1 3 1 = 3 8 2 2 6 4 2 32 1 1 = 1 2 2 2 8 t = 0 1 2 0 q 3 1 1 + 5 1 1 = 37 32 2 3 72 4 3 64×27 1 1 2 + 1 1 2 = 5 8 2 3 6 4 3 72 1 1 2 = 1 2 2 3 6 For t = 0, the q entry is 1/2 since the probability of being in state q is one and the probability of outputting heads is 1 . The entry for p is zero since the probability of starting in 2 state p is zero. For t = 1, the q entry is 1 since for t = 0 the q entry is 1 and in state q 8 2 the HMM goes to state q with probability 1 and outputs heads with probability 1. The 2 2 p entry is 1 since for t = 0 the q entry is 1 and in state q the HMM goes to state p with 6 2 probability 1 and outputs heads with probability 2 3 . For t = 2, the q entry is which 2 3 32 consists of two terms. The first term is the probability of ending in state q at t = 1 times the probability of staying in q and outputting h. The second is the probability of ending in state p at t = 1 times the probability of going from state p to state q and outputting h. From the table, the probability of producing the sequence hhht is 19 384 + 37 1728 = 0.0709. p The most likely sequence of states - the Viterbi algorithm Given a HMM and an observation O = O 0 O 1 · · · O T , what is the most likely sequence of states? The solution is given by the Viterbi algorithm, which is a slight modification to the dynamic programming algorithm just given for determining the probability of an output sequence. For t = 0, 1, 2, . . . , T and for each state i, calculate the probability of the most likely sequence of states to produce the output O 0 O 1 O 2 · · · O t ending in state i. For each value of t, calculate the most likely sequence of states by selecting over all states j the most likely sequence producing O 0 O 1 O 2 · · · O t and ending in state i consisting of the most likely sequence producing O 0 O 1 O 2 · · · O t−1 ending in state j followed by the transition from j to i producing O t . Note that in the previous example, we added the probabilities of each possibility together. Now we take the maximum and also record where the maximum came from. The time complexity is O(n 2 T ) and the space complexity is O(nT ). The space complexity bound is argued as follows. In calculating the probability of the most likely sequence of states that produces O 0 O 1 . . . O t ending in state i, we remember the 305
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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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Page 333 and 334: In the discrete case, x = [x 0 , x
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Lemma 11.1 If a dilation equation i
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The Haar Wavelet ⎧ ⎨ 1 0 ≤ x
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11.3 Wavelet Systems So far we have
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11.5 Conditions on the Dilation Equ
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the support of both sides of the eq
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and the wavelet functions are ortho
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11.7 Sufficient Conditions for the
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Example of orthogonality when wavel
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11.9 Designing a Wavelet System In
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3. f(x) = 1 2 f(2x) + 1 2 f(2x −
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12 Appendix 12.1 Asymptotic Notatio
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these equalities by derivatives.
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Gaussian and related integrals To v
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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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