Lecture 4: Two-point Sampling, Coupon Collector s problem

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1 Randomized Algorithms Lecture 4: Two-point Sampling, Coupon Collector s problem Sotiris Nikoletseas Associate Professor CEID - ETY Course Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 1 / 36

2 Overview A. Pairwise independence of random variables B. The pairwise independent sampling theorem C. Probability amplification via reduced randomness D. The Coupon Collector s problem Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 2 / 36

3 A. On the Additivity of Variance In general the variance of a sum of random variables is not equal to the sum of their variances However, variances do add for independent variables (i.e. mutually independent variables) In fact, mutual independence is not necessary and pairwise independence suffices This is very useful, since in many situations the random variables involved are pairwise independent but not mutually independent. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 3 / 36

4 Conditional distributions Let X, Y be discrete random variables. Their joint probability density function is f(x, y) = Pr{(X = x) (Y = y)} Clearly f 1 (x) = Pr{X = x} = y f(x, y) and f 2 (y) = Pr{Y = y} = x f(x, y) Also, the conditional probability density function is: Pr{(X = x) (Y = y)} f(x y) = Pr{X = x Y = y} = Pr{Y = y} f(x, y) f(x, y) = = f 2 (y) f(x, y) x = Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 4 / 36

5 Pairwise independence Let random variables X 1, X 2,..., X n. These are called pairwise independent iff for all i j it is Pr{(X i = x) (X j = y)} = Pr{X i = x}, x, y Equivalently, Pr{(X i = x) (X j = y)} = = Pr{X i = x} Pr{X j = y}, x, y Generalizing, the collection is k-wise independent iff, for every subset I {1, 2,..., n} with I < k for every set of values { {a i }, b and j / I, it is Pr X j = b i I X i = a i } = Pr{X j = b} Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 5 / 36

6 Mutual (or full ) independence The random variables X 1, X 2,..., X n are mutually independent iff for any subset X i1, X i2,..., X ik, (2 k n) of them, it is Pr{(X i1 = x 1 ) (X i2 = x 2 ) (X ik = x k )} = = Pr{X i1 = x 1 } Pr{X i2 = x 2 } Pr{X ik = x k } Example (for n = 3). Let A 1, A 2, A 3 3 events. They are mutually independent iff all four equalities hold: Pr{A 1 A 2 } = Pr{A 1 } Pr{A 2 } (1) Pr{A 2 A 3 } = Pr{A 2 } Pr{A 3 } (2) Pr{A 1 A 3 } = Pr{A 1 } Pr{A 3 } (3) Pr{A 1 A 2 A 3 } = Pr{A 1 } Pr{A 2 } Pr{A 3 } (4) They are called pairwise independent if (1), (2), (3) hold. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 6 / 36

7 Mutual vs pairwise independence Important notice: Pairwise independence does not imply mutual independence in general. Example. Let a probability space including all permutations of a, b, c as well as aaa, bbb, ccc (all 9 points considered equiprobable). Let A k = at place k there is an a (for k = 1, 2, 3). It is Pr{A 1 } = Pr{A 2 } = Pr{A 3 } = = 1 3 Also Pr{A 1 A 2 } = Pr{A 2 A 3 } = Pr{A 1 A 3 } = 1 9 = thus A 1, A 2, A 3 are pairwise independent. But Pr{A 1 A 2 A 3 } = 1 9 Pr{A 1} Pr{A 2 } Pr{A 3 } = 1 27 thus the events are not mutually independent Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 7 / 36

8 Variance: key features Definition: V ar(x) = E[(X µ) 2 ] = x (x µ) 2 Pr{X = x} where µ = E[X] = x x Pr{X = x} We call standard deviation of X the σ = V ar(x) Basic Properties: (i) V ar(x) = E[X 2 ] E 2 [X] (ii) V ar(cx) = c 2 V ar(x), where c constant. (iii) V ar(x + c) = V ar(x), where c constant. proof of (i): V ar(x) = E[(X µ) 2 ] = E[X 2 2µX + µ 2 ] = E[X 2 ] + E[ 2µX] + E[µ 2 ] = E[X 2 ] 2µE[X] + µ 2 = E[X 2 ] µ 2 Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 8 / 36

9 The additivity of variance Theorem: if X 1, X 2,..., X n ( are pairwise independent random n ) n variables, then: V ar X i = V ar(x i ) i=1 Proof: V ar(x X n ) = E[(X X n ) 2 ] E 2 [X X n ] = n n n n = E Xi 2 + X i X j µ 2 i + µ i µ j = = i=1 1 i j n n (E[Xi 2 ] µ 2 i ) + i=1 n 1 i j n i=1 i=1 1 i j n (E[X i X j ] µ i µ j ) = (since X i pairwise independent, so 1 i j n E(X i X j ) = E(X i )E(X j = µ i µ j ) n V ar(x i ) Note: As we see in the proof, the pairwise independence suffices, and mutual (full) independence is not needed. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 9 / 36 i=1

10 B. The pairwise independent sampling theorem Another Example. Birthday matching: Let us try to estimate the number of pairs of people in a room having birthday on the same day. Note 1: Matching birthdays for different pairs of students are pairwise independent, since knowing that (George, Takis) have a match tell us nothing about whether (George, Petros) match. Note 2: However, the events are not mutually independent. Indeed they are not even 3-wise independent since if (George,Takis) match and (Takis, Petros) match then (George, Petros) match! Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 10 / 36

11 Birthday matching Let us calculate the probability of having a certain number of birthday matches. Let B 1, B 2,..., B n the birthdays of n independently chosen people and let E i,j be the indicator variable for the event of a (i, j) match (i.e. B i = B j ). As said, the events E i,j are pairwise independent but not mutually independent. Clearly, Pr{E i,j } = Pr{B i = B j } = i j). Let D the number of matching pairs. Then D = 1 i<j n E i,j By linearity of expectation we have E[D] = E = E[E i,j ] = 1 i<j n E i,j 1 i<j n = (for ( ) n Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 11 / 36

12 Birthday matching Since the variances of pairwise independent variables E i,j add up, it is: V ar[d] = V ar = V ar[e i,j ] = = ( ) n ( ) 1 i<j n E i,j 1 i<j n As an example, for a class of n = 100 students, it is E[D] 14 and V ar[d] < 14( ) < 14. So by Chebyshev s inequality we have Pr{ D 14 x} 14 x 2 Letting x = 6, we conclude that with more than 50% chance the number of matching birthdays will be between 8 and 20. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 12 / 36

13 The Pairwise Independent Sampling Theorem (I) We can actually generalize and not restrict to sums of zero-one (indicator) valued variables neither to variables with the same distribution. We below state the theorem for possibly different distributions with same mean and variance (but this is done for simplicity, and the result holds for distributions with different means and/or variances as well). Theorem. Let X 1,..., X n pairwise independent variables with the same mean µ and variance σ 2. Let S n = n i=1 X i Then Pr { S n n µ } ( x 1 σ ) 2 n x Proof. Note that E[ S n n ] = nµ n = µ, V ar[ S n n ] = ( ) 1 2 n nσ 2 = σ2 n and apply Chebyshev s inequality. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 13 / 36

14 The Pairwise Independent Sampling Theorem (II) Note: This Theorem actually provides a precise general evaluation of how the average of pairwise independent random samples approaches their mean. If the number n of samples becomes large enough we can arbitrarily close approach the mean with confidence arbitrarily close to 100% (n > σ2 ) i.e. a x 2 large number of samples is needed for distributions of large variance and when we want to assure high concentration around the mean). Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 14 / 36

15 C. Reduced randomness at probability amplification Motivation: Randomized Algorithms, for a given input x, actually choose n random numbers ( witnesses ) and run a deterministic algorithm on the input, using each of these random numbers. intuitively, if the deterministic algorithm has a probability of error ϵ (e.g. 1 2 ), t independent runs reduce the error probability to ϵ t 1 (e.g. 2 ) and amplify the correctness t probability from 1 2 to t however, true randomness is quite expensive! What happens if we are constrained to use no more than a constant c random numbers? The simplest case is when c = 2 e.g. we choose just 2 random numbers (thus the name two-point sampling) Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 15 / 36

16 C. Reduced randomness at probability amplification Problem definition: If our randomized algorithm reduced the error probability to ϵ t with t random numbers, what can we expect about the error probability with t = 2 random numbers only? an obvious bound is ϵ 2 (e.g. when ϵ = 1 2 reducing the error probability from 1 2 to 1 4 ) can we do any better? it turns out that we can indeed do much better and reduce the error probability to 1 t (which is much smaller than the constant ϵ 2 ) Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 16 / 36

17 C. Reduced randomness at probability amplification High level idea: generate t ( pseudo-random ) numbers based on the chosen 2 truly random numbers and use them in lieu of t truly independent random numbers. these generated numbers are dependent on the 2 chosen numbers. Hence they are not independent, but are pairwise independent. This loss of full independence reduces the accuracy of the algorithmic process but is still quite usable in reducing the error probability. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 17 / 36

18 C. Reduced randomness at probability amplification The process (high level description): 1 Choose a large prime number p (e.g. Mersenne prime, ). 2 Define Z p as the ring of integers modulo p (e.g 0, 1, 2,..., ) 3 Choose 2 truly random numbers, a and b from Z p (e.g. a = and b = ). 4 Generate t pseudo-random numbers y i = (ai + b) mod p e.g. y 0 = y 1 = ( ) y 2 = ( ) and so on. 5 Use each of the y i s as t witnesses (in lieu of t purely independent random witnesses) Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 18 / 36

19 C. Reduced randomness at probability amplification Performance (high level discussion) As we will see: for any given error bound ϵ, the error probability is reduced from ϵ 2 to 1 t. however, instead of requiring O(log 1 ϵ ) runs on the deterministic algorithm (in the case of t independent random witnesses) we will require O( 1 ϵ ) runs in the case of 2 independent random witnesses. thus, we gain in the probability amplification but loose on some efficiency. we need significantly less true randomness. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 19 / 36

20 The class RP (I) Definition. The class RP (Random Polynomial time) consists of all languages L admitting a randomized algorithm A running in worst case polynomial time such that for any input x: Notes: x L Pr{A(x) accepts} 1 2 x / L Pr{A(x) accepts} = 0 language recognition computational decision problems the 1 2 value is arbitrary. The success probability needs just to be lower-bounded by an inverse polynomial function of the input size (a polynomial number of algorithm repetitions would boost it to constant, in polynomial time). RP actually includes Monte Carlo algorithms than can err only when x L (one-sided error) Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 20 / 36

21 The RP algorithm An RP algorithm, for given input x, actually picks a random number r from Z p and computes A(x, r) with the following properties: - x L A(x, r) = 1, for half of the possible values of r - x / L A(x, r) = 0, for all possible choices of r As said, - if we run algorithm A t times on the same input x, the error probability is 1 2 but this requires t truly random t numbers. - if we restrict ourselves to t = 2 true random numbers a, b from Z p and run A(x, a) and A(x, b), the error probability can be as high as but we can do better with t pseudo-random numbers: Let r i = a i + b (mod p), where a, b are truly random, as above, for i = 1, 2,..., t. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 21 / 36

22 Modulo rings pairwise independence (I) Let p a prime number and Z p = {0, 1, 2,..., p 1} denote the ring of the integers modulo p. Lemma 1. Given y, i Z p and choosing a, b randomly uniformly from Z p, the probability of y a i + b (mod p) is 1 p Proof. Imagine that we first choose a. Then when choosing b, it must be y a i b (mod p). Since we choose b uniformly and modulo p can take p values, the probability is clearly 1 p indeed. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 22 / 36

23 Modulo rings pairwise independence (II) Lemma 2. Given y, z, x, w Z p such that x w, and choosing a, b randomly uniformly from Z p, the probability of y a x + b (mod p) and z a w + b (mod p) is 1 p 2 Proof. It is y z a (x w) (mod p). Since x w 0, the equation holds for a unique value of a. This in turn implies a specific value for b. The probability that a, b get those two specific values is clearly 1 p 1 p = 1. p 2 Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 23 / 36

24 Modulo rings pairwise independence (III) Lemma 3. Let i, j two distinct elements of Z p,and choose a, b randomly uniformly from Z p. Then the two variables Y i = a i + b (mod p) and Y j = a j + b (mod p) are uniformly distributed on Z p and are pairwise independent. Proof. From lemma 1, it is clearly Pr{Y i = α} = 1 p, for any α Z p, so Y i, Y j indeed have uniform distribution. As for pairwise independence, note that: Pr{Y i = α Y j = β} = Pr{Y i=α Y j =β} Pr{Y j =β} = Thus, Y i, Y j are pairwise independent. 1 p 2 1 p = Pr{Y i = α} Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 24 / 36

25 Modulo rings pairwise independence (IV) Note: This is only pairwise independence. Indeed, consider the variables Y 1, Y 2 Y 3, Y 4 as defined above. Every pair of them are pairwise independent. But, if you give the value of Y 1, Y 2 then we know the values of Y 3, Y 4 immediately, since the values of Y 1 and Y 2 uniquely determine a and b, and thus we can compute all Y i variables. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 25 / 36

26 Modulo rings pairwise independence (V) Thus Y = t i=1 A(x, r i) is a sum of random variables which are pairwise independent, since the r i values are pairwise independent. Assume that x L, then E[Y ] = t 2 and V ar[y ] = t i=1 V ar[a(x, r i)] = t = t 4 The probability that all those t executions failed corresponds to the event Y = 0, and Pr{Y = 0} Pr{ Y E[Y ] E[Y ]} = Pr{ Y t 2 t 2 } V ar[y ] = t ( 2) t 2 4 = 1 ( 2) t 2 t from Chebyshev s inequality. Thus the use of pseudo-randomness (t pseudo-random numbers) allows to exploit our 2 random bits to reduce the error probability from 1 4 to 1 t. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 26 / 36

27 D. The Coupon Collector s problem There are n distinct coupons and at each trial a coupon is chosen uniformly at random, independently of previous trials. Let m the number of trials. Goal: establish relationships between the number m of trials and the probability of having chosen each one of the n coupons at least once. Note: the problem is similar to occupancy (number of balls so that no bin is empty). Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 27 / 36

28 The expected number of trials needed (I) Let X the number of trials (a random variable) needed to collect all coupons at least once each. Let C 1, C 2,..., C X the sequence of trials, where C i {1,..., n} denotes the coupon type chosen at trial i. We call the ith trial a success if coupon type chosen at C i was not drawn in any of the first i 1 trials (obviously C 1 and C X are always successes). We divide the sequence of trials into epochs, where epoch i begins with the trial following the ith success and ends with the trial at which the (i + 1)st success takes place. Let r.v. X i (0 i n 1) be the number of trials in the ith epoch. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 28 / 36

29 The expected number of trials needed (II) n 1 Clearly, X = i=0 X i Let p i the probability of success at any trial of the ith epoch. This is the probability of choosing one of the n i remaining coupon types, so: p i = n i n Clearly, X i follows a geometric distribution with parameter p i, so E[X i ] = 1 p i and V ar(x i ) = 1 p i p 2 i By linearity of expectation: E[X] = E [ n 1 ] n 1 X i = E[X i ] = i=0 i=0 n 1 i=0 n n n i = n = nh n But H n ln n + Θ(1) E[X] n ln n + Θ(n) i=1 1 i = Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 29 / 36

30 The variance of the number of needed trials Since the X i s are independent, we have: n 1 n 1 V ar(x) = V ar(x i ) = = n 2 n i=1 Since lim n i=0 1 n i 2 n 1 i i=1 n i=1 1 i 2 = π2 6 i=0 ni n (n i) 2 = i=1 n(n i) i 2 = we get V ar(x) π2 6 n2 Concentration around the expectation The Chebyshev inequality does not provide a strong result: For β > 1, Pr{X > βn ln n} = Pr{X n ln n > (β 1)n ln n} Pr{ X n ln n > (β 1)n ln n} V ar(x) (β 1) 2 n 2 ln 2 n n2 n 2 ln 2 n = 1 ln 2 n Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 30 / 36

31 Stronger concentration around the expectation Let Ei r the event: coupon type i is not collected during the first r trials. Then Pr{Ei r} = (1 1 n )r e r n For r = βn ln n we get By the union bound we { have n Pr{X > r} = Pr i=1 Pr{E r i E r i } βn ln n } e n = n β (i.e. at least one coupon is not selected), so n Pr{X > r} Pr{Ei r } n n β = n (β 1) = n ϵ, i=1 where ϵ = β 1 > 0 Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 31 / 36

32 Sharper concentration around the mean - a heuristic argument Binomial distribution (#successes in n independent trials each one with success probability p) X B(n, p) Pr{X = k} = ( ) n k p k (1 p) n k (k = 0, 1, 2,..., n) E(X) = np, V ar(x) = np(1 p) Poisson distribution) X P (λ) Pr{X = x} = e E(X) = V ar(x) = λ λ λx x! (x = 0, 1,... ) Approximation: It is B(n, p) P (λ), where λ = np. For large n, the approximation of the binomial by the Poisson is good. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 32 / 36

33 Towards the sharp concentration result Let N r i = number of times coupon i chosen during the first r trials. Then E r i is equivalent to the event {N r i = 0}. Clearly Ni r B ( r, n) 1, thus Pr{Ni r = x} = ( ( r 1 ) x ( ) x) n 1 1 r x n Let λ a positive real number. A r.v. Y is P (λ) Pr{Y = y} = e λ λy y! As said, for suitability small λ and as r approaches, P ( ) ( r n is a good approximation of B r, 1 n). Thus Pr{Ei r} = Pr{N i r = 0} e λ λ0 0! = e λ = e r n (fact 1) Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 33 / 36

34 An informal argument on independence We will now claim that the Ei r (1 i n) events are almost independent, (although it is obvious that there is some dependence between them; but we are anyway heading towards a heuristic). Claim 1. For 1 i n, and any set if indices {j 1,..., j k } not containing { i, Pr Ei r } k l=1 Er j l Pr{Ei r} { } ( Proof: Pr Ei r k Pr {E Ej r i r k )} ( ) l=1 Er jl 1 k+1 r l = { n k = ( ) Pr l=1 Er 1 k r jl} n l=1 r(k+1) e n e rk n = e r n Pr{E r i } Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 34 / 36

35 An approximation of the probability Because { of fact 1 and Claim 1, we have: n } { n } Pr = Pr (1 e m n ) n e ne m n i=1 E m i i=1 E m i For m = n(ln n + c) = n ln n + cn, for any constant c R, we then get { n } { n } Pr{X > m = n ln n + cn} = Pr Pr = 1 e e c i=1 E m i i=1 The above probability: - is close to 0, for large positive c - is close to 1, for large negative c Thus the probability of having collected all coupons, rapidly changes from nearly 0 to almost 1 in a small interval cantered around n ln n (!) Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 35 / 36 E m i

36 The rigorous result Theorem: Let X the r.v. counting the number of trials for having collected each one of the n coupons at least once. Then, for any constant c R and m = n(ln n + c) it is lim n Pr{X > m} = 1 e e c Note 1. The proof uses the Boole-Bonferroni inequalities for inclusion-exclusion in the probability of a union of events. Note 2. The power of the Poisson heuristic is that it gives a quick, approximative estimation of probabilities and offers some intuitive insight towards the accurate behaviour of the involved quantities. Sotiris Nikoletseas, Associate Professor Randomized Algorithms - Lecture 4 36 / 36