Appendix A
Appendix
A.1 Probability theory
A random variable is a mapping from a probability space to R.Togivean example, the probability space could be that of all 2n possible outcomes of n tosses of a fair coin, and Xi is the random variable that is 1 if the ith toss is a head, and is 0 otherwise. An event is a subset of the probability space. The following simple bound —called the union bound—is often used in the book. For every set of events B1,B2,...,Bk,
k ∪k ≤ Pr[ i=1Bi] Pr[Bi]. (A.1) i=1
A.1.1 The averaging argument We list various versions of the “averaging argument.” Sometimes we give two versions of the same result, one as a fact about numbers and one as a fact about probability spaces. Lemma A.1
If a1,a2,...,an are some numbers whose average is c then some ai ≥ c. Lemma A.2 (“The Probabilistic Method”) If X is a random variable which takes values from a finite set and E[X]=μ then the event “X ≥ μ” has nonzero probability. Corollary A.3 If Y is a real-valued function of two random variables x, y then there is a choice c for y such that E[Y (x, c)] ≥ E[Y (x, y)].
403
DRAFT -- DRAFT -- DRAFT -- DRAFT -- DRAFT -- 404 APPENDIX A. APPENDIX
Lemma A.4
If a1,a2,...,an ≥ 0 are numbers whose average is c then the fraction of ai’s that are greater than (resp., at least) kc is less than (resp, at most) 1/k. Lemma A.5 (“Markov’s inequality”) Any non-negative random variable X satisfies 1 Pr (X ≥ kE[X]) ≤ . k Corollary A.6 ∈ − If a1√,a2,...,an [0, 1] are numbers whose√ average is 1 γ then at least 1 − γ fraction of them are at least 1 − γ.
Can we give any meaningful upperbound on Pr[X If a1,a2,...,an are numbers in the interval [0, 1] whose average is ρ then at least ρ/2 of the ai’s are at least as large as ρ/2. Proof: Let γ be the fraction of i’s such that ai ≥ ρ/2. Then γ +(1− γ)ρ/2 must be at least ρ/2, so γ ≥ ρ/2. 2 More generally, we have Lemma A.8 If X ∈ [0, 1] and E[X]=μ then for any c<1 we have 1 − μ Pr[X ≤ cμ] ≤ . 1 − cμ Example A.1 Suppose you took a lot of exams, each scored from 1 to 100. If your average score was 90 then in at least half the exams you scored at least 80. A.1.2 Deviation upperbounds Under various conditions, one can give upperbounds on the probability of an event. These upperbounds are usually derived by clever use of Markov’s inequality. The first bound is Chebyshev’s inequality, useful when only the variance is known. DRAFT -- DRAFT -- DRAFT -- DRAFT -- DRAFT -- A.1. PROBABILITY THEORY 405 Lemma A.9 (Chebyshev inequality) If X is a random variable and E[X]=μ and E[(X − μ)2]=σ2 then Pr[|X − μ| >kσ] ≤ 1/k2. Proof: Apply Markov’s inequality to the random variable (X −μ)2,noting that by definition of variance, E[(X − μ)2]=σ2. 2 The next inequality has many names, and is widely known in theoretical computer science as Chernoff bounds. It considers scenarios of the following type. Suppose we toss a fair coin n times. The expected number of heads is n/2. How tightly is this number concentrated? Should we be very surprised if after 1000 tosses we have 625 heads? The bound we present is more general, and concerns n different coin tosses of possibly different biases (the bias is the probability of obtaining “heads”; for a fair coin this is 1/2). A sequence of {0, 1} random variables is Poisson, as against Bernoulli, if the expected values can vary between trials. Theorem A.10 (“Chernoff” bounds) Let X1,X2,...,Xn be independent Poisson trials and let pi = E[Xi],where n n 0 where the last equality holds because the Xi r.v.s are independent. Now, t E[exp(tXi)] = (1 − pi)+pie , therefore, t t E[exp(tXi)] = [1 + pi(e − 1)] ≤ exp(pi(e − 1)) i i i t t =exp( pi(e − 1)) = exp(μ(e − 1)), i (A.5) DRAFT -- DRAFT -- DRAFT -- DRAFT -- DRAFT -- 406 APPENDIX A. APPENDIX as 1 + x ≤ ex. Finally, apply Markov’s inequality to the random variable exp(tX), viz. E[exp(tX)] exp((et − 1)μ) Pr[X ≥ (1+δ)μ]=Pr[exp(tX) ≥ exp(t(1+δ)μ)] ≤ = , exp(t(1 + δ)μ) exp(t(1 + δ)μ) using lines (A.4) and (A.5) and the fact that t is positive. Since t is a dummy variable, we can choose any positive value we like for it. Simple calculus shows that the right hand side is minimized for t =ln(1+δ)and this leads to the theorem statement. 2 By the way, if all n coin tosses are fair (Heads has probability√ 1/2) then the the probability of seeing N heads where |N − n/2| >a n is at most e−a2/2. The chance of seeing at least 625 heads in 1000 tosses of an unbiased coin is less than 5.3 × 10−7. A.1.3 Sampling methods Typically, to take n independent samples the amount of random bits required increases linearly with n. At a few places in the book we encounter pairwise independent sampling. This is very frugal with its use of random bits, since it allows n samples to be taken using 2 log n random bits. The downside is that one can only use Chebyshev’s inequality for deviation bounds, not Chernoff bounds. Definition A.1 Random variables x1,...,xm are pairwise independent if ∀i, j ∀a, b Pr[xi = a, xj = b]=Pr[xi = a]Pr[xj = b]. (A.6) Pairwise independent random variables are useful because of the Chebyshev inequality. Suppose that x1,...,xm are pairwise independent with E(xi)= 2 μ, Var(xi)=σ .Thenwehave 2 − 2 2 − 2 Var( xi)=E[( xi) ] E[ xi] = (E[xi ] E[xi] ), i i i where we have used pairwise independence in the last step.√ Thus Var( xi)= 2 2 mσ . By the Chebyshev inequality, Pr[| xi − mμ| >k mσ] ≤ 1/k .So, the sum of the variables is somewhat concentrated about the mean. This is in contrast with the case of complete independence, when Chernoff bounds would give an exponentially stronger concentration result (the 1/k2 would be replaced by exp(Θ(−k2))). DRAFT -- DRAFT -- DRAFT -- DRAFT -- DRAFT -- A.2. POLYNOMIALS 407 Example A.2 In Zp,choosea and b randomly and independently. Then the random variables a + b, a +2b,...,a+(p − 1)b are pairwise independent. Indeed, for any t, s, j = k ∈ Zp, Pr[a + jb = t]=1/p,andPr[a + jb = t, a + kb = s]=1/p2, because this linear system is satisfied by exactly one (a, b)-pair out of p2. Example A.3 Let m =2k − 1. The set [1 ...m]isin1− 1 correspondence with the set 2[1...k] \∅. We will construct m random variables corresponding to all nonempty subsets of [1 ...k]. Pick uniformly at random k binary ⊂ strings t1,...,tk of length n and set YS = i∈S ti (mod 2), where S ∅ [1 ...k], S = . For any S1 = S2, the random variables YS1 and YS2 are independent, because one can always find an i such that i ∈ S2 \ S1,so the difference between YS1 and YS2 is always a sum of several uniformly distributed random vectors, which is a uniformly distributed random vector itself. That is, even if we fix the value of YS1 , we still have to toss a coin for each position in YS2 ,and 1 Pr[Y = s, Y = t]=Pr[Y ⊕ Y = t ⊕ s|Y = s]= . S1 S2 S1 S2 S1 22n A.1.4 The random subsum principle The following fact is used often in the book. Let a1,a2,...,ak be elements of a finite field GF (q). We do not know the elements, but we are told that at least one of them is nonzero. Now we pick a subset S of {1, 2,...,k} uniformly at random. Then the probability that i∈S ai = 0 is at least 1/2. Here is the reason. Suppose that aj is nonzero. Then we can think of S as having been picked as follows. First we pick a random subset S of {1, 2,...,k}\{j} and then we toss a fair coin to decide whether or not to add j to it. If i∈S ai = 0 then with probability 1/2 S = S ,inwhichcase ∪{ } i∈S ai =0.If i∈S ai = 0 then with probability 1/2 S = S j ,in which case i∈S ai =0. A.2 Polynomials We list some basic facts about univariate polynomials. DRAFT -- DRAFT -- DRAFT -- DRAFT -- DRAFT -- 408 APPENDIX A. APPENDIX Theorem A.11 A nonzero polynomial of degree d has at most d distinct roots. Proof: d i Suppose p(x)= i=0 cix has d +1distinctrootsα1,...,αd+1 in some field F .Then d i · αj ci = p(αj)=0, i=0 for j =1,...,d+ 1. This means that the system Ay = 0 with ⎛ ⎞ 2 d 1 α1 α1 ... α1 ⎜1 α α2 ... αd ⎟ A = ⎜ 2 2 2 ⎟ ⎝...... ⎠ 2 d 1 αd+1 αd+1 ... αd+1 has a solution y = c.ThematrixA is a Vandermonde matrix, hence det A = (αi − αj), i>j which is nonzero for distinct αi. Hence rankA = d + 1. The system Ay = 0 has therefore only a trivial solution — a contradiction to c = 0. 2 Theorem A.12 For any set of pairs (a1,b1),...,(ad+1,bd+1) there exists a unique polynomial g(x) of degree at most d such that g(ai)=bi for all i =1, 2,...,d+1. Proof: The requirements are satisfied by Lagrange Interpolating Polyno- mial: d+1 (x − aj) b · j=i . i (a − a ) i=1 j= i i j If two polynomials g1(x),g2(x) satisfy the requirements then their difference p(x)=g1(x) − g2(x)isofdegreeatmostd, and is zero for x = a1,...,ad+1. Thus, from the previous theorem, polynomial p(x) must be zero and poly- nomials g1(x),g2(x)identical.2 The following elementary result is usually attributed to Schwartz and Zippel in the computer science community, though it was certainly known earlier (see e.g. DeMillo and Lipton). DRAFT -- DRAFT -- DRAFT -- DRAFT -- DRAFT -- A.2. POLYNOMIALS 409 Theorem A.13 (Schwartz-Zippel) If a polynomial p(x1,x2,...,xm) over F = GF (q) is nonzero and has total degree at most d,then d Pr[p(a ..a ) =0] ≥ 1 − , 1 m q where the probability is over all choices of a1..am ∈ F . Proof: We use induction on m.Ifm = 1 the statement follows from Theorem A.11. Suppose the statement is true when the number of variables is at most m − 1. Then p can be written as d i p(x1,x2,...,xm)= x1pi(x2,...,xm), i=0 where pi has total degree at most d − i.Sincep is nonzero, at least one of pi is nonzero. Let k be the largest i such that pi is nonzero. Then by the inductive hypothesis, d − k Pr [pi(a2,a3,...,am) =0] ≥ 1 − . a2,a3,...,am q Whenever pi(a2,a3,...,am) =0, p(x1,a2,a3,...,am) is a nonzero uni- variate polynomial of degree k, and hence becomes 0 only for at most k values of x1. Hence k d − k d Pr[p(a ..a ) =0] ≥ (1 − )(1 − ) ≥ 1 − , 1 m q q q and the induction is completed. 2 A.2.1 Representing data by multivariate polynomials Lemma A.14 (Polynomial extension lemma) For some integer h let H = {u0,u1, ··· ,uh} be a subset of h +1elements from GF (q). For each function f : Hm → F there exists a polynomial fˆ(x1,x2, ··· ,xm) whose degree in each xi is h and which satisfies fˆ(v)=f(v) for all v ∈ Hm. This polynomial can be computed in time poly(hm). Proof: The requirements are satisfied by the Lagrange interpolating poly- nomial: − ˆ ··· m Πj= ixi uj f(x1,x2, ,xm)= f(u)Πi=1 (A.7) Π ui − uj u∈Hm j=i 2 DRAFT -- DRAFT -- DRAFT -- DRAFT -- DRAFT -- 410 APPENDIX A. APPENDIX A.3 Eigenvalues and expanders An eigenvalue of an n × n symmetric real matrix M is a real number λ such that there exists a vector x ∈$n such that M · x = λx.Itisknownthat for every such matrix there are n eigenvalues λ1 ≥ λ2 ≥ ··· ≥ λn and n n corresponding eigenvectors u1,u2,...,un ∈$ that form an orthonormal n basis of $ .TheCourant Fisher theorem characterizes λ1 as xT Mx λ1 =min (A.8) x∈n |x|2 and more generally λi as xT Mx λi =min (A.9) ∈ 2 x Li−1 |x| where Li−1 is the space of vectors orthogonal to u1,u2,...,ui−1. The adjacency matrix of an n-node undirected graph G =(V,E)isan n × n matrix AG whose (i, j)entryis1if{i, j} is an edge and 0 otherwise. In a few places in the book we are interested in constant degree ex- panders, and these have more than one definition. (The definitions are related.) We define them using eigenvalues. We recall that for a d-regular graph, the largest eigenvalue of the adjacency matrix is d. Definition A.2 (Expander family) If λ An explicit construction of a (5,λ)-expander family is known for some λ<5 (Margulis, simplified in Gabber-Galil). For every positive integer n, the family contains a graph vertex-set has size close to n.Inthiscontext, “explicit construction” means that there exists a polynomial-time machine that, given any vertex number i, gives the 5 neighbors of i. (Note that the running time is polynomial in the representation of i,inotherwords, poly(log i).) An explicit construction of expanders (for some constant d) also follows from the algorithm in Section 18.6. Let G be any (d, λ)-expander on n nodes, with adjacency matrix A.Let M be the Markov chain defined by G (that is, M is a n × n matrix such that in entry (i, j) there is the transition probability from node i to node j), 1 hence M = d A. Thus the largest eigenvalue is 1 and every other eigenvalue λi satisfies |λi|≤λ/5. DRAFT -- DRAFT -- DRAFT -- DRAFT -- DRAFT -- A.3. EIGENVALUES AND EXPANDERS 411 1 Then M is ergodic, and the stationary distribution is n 1. Let X be some distribution on the vertices of G.ThenX can be written as: 1 n X = 1+ α e n i i i=2 Where the ei’s are the eigenvectors of M; they form an orthogonal basis 1 since M is symmetric. The coefficient of the first eigenvector is n is because 1 1 DRAFT -- DRAFT -- DRAFT -- DRAFT -- DRAFT -- 412 APPENDIX A. APPENDIX DRAFT -- DRAFT -- DRAFT -- DRAFT -- DRAFT -- Bibliography [AA02] M.Vyalyi A.Kitaev and A.Shen. Classical and Quantum compu- tation. AMS Press, 2002. [AD97] M. Ajtai and C. Dwork. A public-key cryptosystem with worst- case/average-case equivalence. In Proceedings of the 29th An- nual ACM Symposium on the Theory of Computing (STOC ’97), pages 284–293, New York, 1997. Association for Computing Ma- chinery. [Adl78] Leonard Adleman. Two theorems on random polynomial time. In 19th Annual Symposium on Foundations of Computer Sci- ence, pages 75–83, Ann Arbor, Michigan, 1978. IEEE. 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