T. Hagerup and C. R ub. A guided tour of chernoff bounds. Inform. Process. Lett., 33(6):305--308, Feb. 1990.  Home/Search   Document Not in Database   Summary   Related Articles   Check

....facts, known as Chernoff bounds [3] These bounds provide estimates of the tail probabilities of binomial distributions. Let X be the number of heads in n independent flips of a biased coin, the probability of a head in a single flip being p. Such an X has the binomial distribution B(n; p) In [7] it is shown, that for all 0 h n Delta p PrfX n Delta p hg e ; 1) PrfX n Delta p Gamma hg e : 2) The results of this paper hold with high probability : Definition 1 An event A happens with high probability, if PrfAg 1 Gamma n Gammaff , for some constant ff 0. ....

T. Hagerup and C. Rub. A Guided Tour of Chernoff Bounds. Inf. Proc. Lett., 33:305--308, 1990.

....which is the set composed of the first M j (n#p) members in SE j . However, since each element of SE j is independently chosen to be a sample with probability 1#p, the probability of this event occurring is given by s# (n#p) b s; M j , p . 6) Using the following Chernoff type bound [17] for estimating the head of a binomial distribution s#=rq (1 = 2 (rq#2) 7) where s# j (n#p ) r=M j (n#p) and q=1#p, it follows that the probability that a set SE j among the p sets of input elements has more than M j (n#p) members is bounded by p 1 (1 j #M ) 2 M n#2p ....

T. Hagerup and C. Ru# b, A guided tour of Chernoff bounds, Inform. Process. Lett. 33, No. 6 (Feb. 1990), 305#308.

....paths in (Dn, c) consist then of O( log n) 2) edges with high probability. 3. 2 The general case: Approximating the vertex potential differences We will use the following form of the well known Chernoff Hoeffding bound on the tail of the distribution of a sum of independent random variables; see [16, 14] for a proof. Lemma 3 Let X be the sum of independent, identically distributed random variables Xl, X m with values in [0, 1] let : Xl] Then, for any e, O e l, Pr(IX m l st) 2. e e2m 3 (4) We now show how to compute vertex potentials i, i 6 [n] so that, for any k, In] with high ....

T. Hagerup and C. Ri/b, A guided tour of Chernoff bounds, Inform. Process. 33 (1989/90) 305-308

....introduce the Chernoff bound and the Hoeffding inequality. These are well known bounds on the tail of the distribution of a sum of random variables. Both bounds can be found in many standard works on probability theory. Our basic formulation of the Ghernoff bounds were taken from Hagerup and Riib [3]. The Hoeffding inequality we took from Hofri [6] and will be derived in appendix B. Often we will have a pair of formulas of the form P(A f(1 ) B) C, P(A f(1 ) B) C, with A, B and C expressions and f a (simple) function. Such a pair of formulas will be presented concisely with the ....

....bounds on the tail probabilities are given by the Hoeffding inequalities. First we give some results for the case that 0 Xi 1 for all i. The strongest form of the Hoeffding inequality is [6] e(zn , t) t) 0 ) 0 t) n (3) Using an estimate from [3], we find from (3) P(Z (l ) n. l )l 1 ( 1 ) 1 )l J 1.2. e , 0 1. 4) The generalizations of (4) to X with arbitrary support [m, M] become particularly nice when the Xi are equally distributed. So, assume m Xi M, for some m and M. In this case P(Z (1 q e) n. ....

Hagerup, T., C. liib, A guided tour of Chernoff Bounds, Techn. Rep. A 88/08, Dep. of Comp. Sc., Universitit des Saarlandes, Saarbriicken, Germany, 1988. Also, Inf. Proc. ett., 33 (1990), 305-308.

....deviation is oe = n=2. The above coefficients increase up to the mean n=2 and then decrease. Because the ratio oe=n tends to zero for n tending to infinity, for large n values most strings are clustered in a narrow peak at Hamming distance H = n=2. As an example, one can use the Chernoff bound [19]: Pr[H (1 Gamma )pn] e Gamma np=2 (18) the probability to find a point at a distance less than np = n=2 decreases in the above exponential way ( 0) The distribution of Hamming distances for n = 500 is shown in Fig 3. Clearly, if better local optima are located in a cluster that is ....

T. Hagerup and C. Rub, "A guided tour of Chernoff bounds," Information Processing Letters 33 (1989/90), 305--308.

....all c 0, there is a C, such that for all x C, f(x) c Delta g(x) Tail Estimates. In the analysis of our randomized algorithms, we need bounds on the tails of functions of random variables. For a binomial distribution, we can use the Chernoff bounds as presented by Hagerup and Rub: Lemma 5 [8] Let X 1 ; Xm be independent Bernoulli trials, with P [X i = 1] p. Let Z = P i X i . Then for any t 0, P [Z p Delta n t] e Gammat 2 = 3 Deltap Deltam) This is a strong result, but we will have to deal with more general functions than independent sums. The following ....

Hagerup, T., C. Rub, `A Guided Tour of Chernoff Bounds,' Information Processing Letters, 33, 305--308, 1990.

....bounds for U k are easy to obtain now. U k can be expressed as the sum of independent random variables U ki ; i = 1; r, where U k;i : ae 1; if i k 0; otherwise; and it holds that prob(U k;i = 1) k n 1 Gamma i = p i : The following is one of the basic Chernoff bounds [19]. Lemma 4.7 With E(U k ) p 1 Delta Delta Delta p r ) r and t 0, prob(U k E(U k ) Gamma t) exp Gamma t 2 2E(U k ) Using t = E(U k ) Gamma d 1 (which is nonnegative for the values of k we will be interested in below) we obtain prob(U k d Gamma 1) exp Gamma ....

T. Hagerup and C. Rub. A guided tour of Chernoff bounds. Inform. Process. Lett., 33:305-- 308, 1990.

....that the probability of either of those events is bounded by (n=e) Gammac ln(c=e) We will just consider the case of R. The other case is symmetric. The crucial observation is that for i the 0 1 random variables A i; are independent and one can therefore apply so called Chernoff bounds [14, 9] which, in one form, state that if random variable Z is the sum of independent 0 1 variables, then Pr[Z c Delta Ex[Z] e Gammac ln(c=e)Ex[Z] In order to make the bound independent of we consider random variable R 0 = P i n A i; where we use additional independent 0 1 ....

T. Hagerup and C. Rub, A guided tour of Chernoff bounds. Inf. Proc. Letters 33 (1989/90) 305--308.

.... hypothesis in H with error greater than , is less than jHj Delta (1 Gamma ) m , the probability that such a hypothesis was found at least l times can be bounded above by R l Delta (jHj Delta (1 Gamma ) m ) l : 1) Another bound can be obtained using the Chernoff bound (see Hagerup and Rub (1989) for a proof of the Chernoff bound) The Chernoff bound states that for n independent random variables Z i 2 f0; 1g with E(Z i ) p i , p = P n i=1 p i ) n, and ffi 0, the probability that the sum of the Z i is at least (1 ffi ) Delta n Delta p can be bounded above in the following ....

Hagerup, T. and Rub, C. 1989. A Guided Tour of Chernoff Bounds. Information Processing Letters. Number 33. Pages 305-308.

.... We get prob(Y k 1) prob(U d Gamma 1) U can be expressed as the sum of independent random variables U i ; i = 1; r, where U i : ae 1; if i k 0; otherwise; and it holds that prob(U i = 1) k n 1 Gamma i = p i : The following is one of the basic Chernoff bounds [17]. Lemma 4.7. With E(U) p1 Delta Delta Delta pr ) r and t 0, prob(U E(U) Gamma t) exp Gamma t 2 2E(U) Using t = E(U) Gamma d 1 (which is nonnegative for the values of k we will be interested in below) we obtain prob(U d Gamma 1) exp Gamma (E(U) Gamma d ....

T. Hagerup and C. Rub. A guided tour of Chernoff bounds. Inform. Process. Lett., 33:305--308, 1990.

.... L = j S G j there is a good set A t for each large gate with probability at least 1 Gamma Le Gamma(kl) c 1 : 3) Let p = 8l 2 =k: Consider a random assignment ae of 0 s and 1 s to variables y; where 1 is assigned with probability p: We shall use the following Chernoff type bound, cf. [HR89]: Lemma 3.3 Let S = X 1 : XN ; where X i are independent 0 1 random variables with P r[X i = 1] p; let M = pN: Then P r[jS Gamma M j ffM ] 2e Gammaff 2 M=3 : 2 First we observe that the number of 1 s in ae is at most 2pkl = 16l 3 with probability 1 Gamma 2e Gammapkl=3 ....

Hagerup, T., Rub, C.: A guided tour of Chernoff bounds, Inf. Process. Letters 33, 1989/90, 305-308.

....c. We consider the set C of all colorings with (p) q) 0 for all fp; qg 2 M . Note that every element of C has discrepancy at most c. We show that there is a better coloring in C by considering colorings randomly chosen from C. We need the well known Chernoff bound (see e.g. Spe87] HR90] in the following form: If X is the sum of k independent random f Gamma1; 1g variables each variable attains Gamma1 and 1 with equal probability , then Prob(jXj p k) 2e Gamma 2 =2 . Let h be a nonvertical line disjoint from S with c h crossings in M , and let h Gamma be ....

Torben Hagerup and Christine Rub. A guided tour of Chernoff bounds. Inform. Process. Lett., 33:305--308, 1990.

....facts, known as Chernoff bounds [3] These bounds provide estimates of the tail probabilities of binomial distributions. Let X be the number of heads in n independent flips of a biased coin, the probability of a head in a single flip being p. Such an X has the binomial distribution B(n; p) In [7] it is shown, that for all 0 h n Delta p PrfX n Delta p hg e Gammah 2 = 3 Deltan Deltap) 1) PrfX n Delta p Gamma hg e Gammah 2 = 2 Deltan Deltap) 2) The results of this paper hold with high probability : Definition 1 An event A happens with high probability, if ....

T. Hagerup and C. Rub. A Guided Tour of Chernoff Bounds. Inf. Proc. Lett., 33:305--308, 1990.

....more than 10KB, the feeding time exceeds the start up time. Random Packet Distributions. If each PU holds k packets with random destinations, then the expected number of packets between a pair of PUs equals k=P . To estimate the maximum arising number, we can use the Chernoff bounds, as given in [5], to derive Lemma 1 On a processor network consisting of P PUs, we are performing a routings. In each routing every PU sends k packets to randomly selected destinations. We may assume that the number of packets send between any pair of PUs is bounded by k=P (3 Delta k=P Delta ln(10 7 ....

Hagerup, T., C. Rub, `A Guided Tour of Chernoff Bounds,' IPL, 33, 305--308, 1990.

....all c 0, there is a C, such that for all x C, f(x) c Delta g(x) Tail Estimates. In the analysis of our randomized algorithms, we need bounds on the tails of functions of random variables. For a binomial distribution, we can use the Chernoff bounds as presented by Hagerup and Rub: Lemma 4 [6] Let X 1 ; Xm be independent Bernoulli trials, with P [X i = 1] p. Let Z = P i X i . Then for any t 0, P [Z p Delta m t] e Gammat 2 = 3 Deltap Deltam) This is a strong result, but we will have to deal with more general functions than independent sums. The following ....

Hagerup, T., C. Rub, `A Guided Tour of Chernoff Bounds,' Information Processing Letters, 33, 305--308, 1990.

....Randomness. An event A happens with high probability if Pr(A) 1 Gamma n Gammaffl , for some ffl 0. All our results for randomized algorithms hold with high probability. We use Chernoff bounds to bound the tail probabilities of binomial distributions, B(n; p) Using the estimates in [6], it is easy to derive that Lemma 1 Let X 0 ; X t Gamma1 be random variables with t = poly (n) and X i = B(n; p) for 0 i t. Then jX i Gamma p Delta nj = O( p Delta n Delta log n) 1=2 ) for all 0 i t, with high probability. Definition 1 A k randomization is a distribution of ....

Hagerup, T., C. Rub, `A Guided Tour of Chernoff Bounds,' Inf. Proc. Lett. 33, 305--308, 1990.

....to bound the tail probabilities of binomial distributions. Let X be the random variable denoting the number of heads in n independent flips of a biased coin, where the probability of a head in each coin flip is p. Thus, X has binomial distribution B(n; p) Then for all 0 h n Delta p, we have [11] PrfX n Delta p hg e Gammah 2 = 3 Deltan Deltap) PrfX n Delta p Gamma hg e Gammah 2 = 2 Deltan Deltap) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 Figure 2: The semi layered order ....

Hagerup, T., C. Rub, `A Guided Tour of Chernoff Bounds,' Inf. Proc. Lett. 33, 305--308, 1990.

....more details about the architecture can be found in [11] Random Packet Distributions. If each PU holds k packets with random destinations, then the expected number of packets between a pair of PUs equals k=P . To estimate the maximum arising number, we can use the Chernoff bounds, as given in [5]. Lemma 1 On a processor network consisting of P PUs, we are performing a routings. In each routing every PU sends k packets to randomly selected destinations. With very high probability, the number of packets send between any pair of PUs is bounded by k=P (3 Delta k=P Delta ln(10 7 Delta ....

Hagerup, T., C. Rub, `A Guided Tour of Chernoff Bounds,' Information Processing Letters, 33, 305--308, 1990.

....are taken into account. Mathematical Issues. Let f and g be positive functions on R. f is of larger order than g, denoted f = g) if for all c 0, there is a C(c) such that for all x C(c) f(x) c Delta g(x) For estimates on the binomial distribution, we use Lemma 1 (Chernoff Bounds) [6] Let X 1 ; Xm be independent Bernoulli trials, with P [X i = 1] p. Let Z = P i X i . Then for any t 0, P [Z p Delta m t] e Gammat 2 = 3 Deltap Deltam) For more general estimates the following inequality by McDiarmid is very useful: Lemma 2 (Azuma Inequality) 11] Let X ....

Hagerup, T., C. Rub, `A Guided Tour of Chernoff Bounds,' Information Processing Letters, 33, 305--308, 1990.

....known as Chernoff bounds [1] These bounds provide strong estimates of the tail probabilities of binomial distributions. Let X be the number of heads in n independent flips of a biased coin, the probability of a head in a single flip being p. Such an X has the binomial distribution B(n; p) In [4] it is shown, that for all 0 h n Delta p PrfX n Delta p hg e Gammah 2 = 3 Deltan Deltap) 1) PrfX n Delta p Gamma hg e Gammah 2 = 2 Deltan Deltap) 2) The results of this paper hold with high probability . We formalize this notion: Definition 1 An event A happens ....

Hagerup, T., C. Rub, `A Guided Tour of Chernoff Bounds,' Information Processing Letters 33, 305--308, 1990.

....fl 1 flae ln(ffi Gamma1 ) C1) In Case B, we show that procedure MI estimates fairly well the absolute error of every hypothesis, so that it is not likely to output one with large error. To do this we can use the standard Chernoff bounds for sums of Bernoulli variables (see, e.g. [4] and the Appendix in [6] Let Y i be the observed error for h i ; note that Y i is the sum of p independent 0=1 variables with expectation i , so E(Y i ) i p. We may have bad hypotheses with i and we know for sure that there is at least one good hypothesis with j ae . For ....

T. Hagerup and C. Rub: "A guided tour of Chernoff bounds". Information Processing Letters 33 (1989/90), 305--308.

....that 0 Y i d 2 and applying theorem A. 1 (the former observation follows from the fact that f is d 2 perfect for B f i , for all 0 i r) Pr f; f h b h f; f j (1 u) n s i e u (1 u) 1 u) n= d2s) 0 j s: By the right hand side of the following inequality [22], Gamma(1=2) 2 Gamma (1 ) log (1 ) Gamma(1=3) 2 ; 0 1; we further conclude that Pr f; f h b h f; f j (1 ) n s i e Gamma(1=3) 2 n= d2s) 0 j s: It remains to establish that for suitably chosen d 1 , d 2 , r and s, a randomly chosen h f; f 2 H d1 ....

T. Hagerup and C. Rub. A guided tour of Chernoff bounds. IPL, 33:305-308, 1990.

....of (small) subsets of mesh data. A single permutation of all mesh data at the end of the partitioning procedure suffice to order submeshes into consecutive sub array locations. The basic idea is sampling. It is based on the following simple probabilistic fact. Lemma 3. 2 (Chernoff Hoeffding[6]) There is a constant c 1 such that the following is true: Suppose there are L red balls in a set of n balls. Then, for any sample of s(n) random balls from the set containing r red balls, Prob[r= 2s(n) L=n 2r=s(n) 1 Gamma e Gammacs(n)L=n : This lemma can be applied to estimate the ....

T. Hagerup and C. Rub. A guided tour of Chernoff bounds. Info. Proc. Let., 33:305-308, 1990.

....Pr[jY Gamma E(Y )j t] 2 exp( Gamma2t 2 = X c 2 k ) see [26] Definition: Chernoff bound Suppose that all the random variables Xk are Bernoulli trials with Pr[Xk = 1] p and Pr[Xk = 0] 1 Gamma p. Let Y = P r k=1 Xk . Then E(Y ) r Delta p = and the following Chernoff bound [16] holds, for 0 1: Pr[jY Gamma j Delta ] 2 exp( Gamma 2 =3) Definition: A property Phi depending on a natural number n is said to hold with a high probability (abbreviated to w.h.p. if, for some constant ffi 0, Phi holds with the probability at least 1 Gamma n Gammaffi ....

T. Hagerup, and Ch. Rub, A Guided Tour of Chernoff Bounds, Inf. Proc. Letters 33 (1989/90), pp. 305--308.

.... L = j S G j there is a good set A t for each large gate with probability at least 1 Gamma Le Gamma(kl) c 1 : 3) Let p = 8l 2 =k: Consider a random assignment ae of 0 s and 1 s to variables y; where 1 is assigned with probability p: We shall use the following Chernoff type bound, cf. [HR89]: Lemma 3.3 Let S = X 1 : XN ; where X i are independent 0 1 random variables with P r[X i = 1] p; let M = pN: Then P r[jS Gamma M j ffM ] 2e Gammaff 2 M=3 : 2 First we observe that the number of 1 s in ae is at most 2pkl = 16l 3 with probability 1 Gamma 2e Gamma4pkl=3 ....

Hagerup, T., Rub, C.: A guided tour of Chernoff bounds, Inf. Process. Letters, 33, 1989/90, 305-308.

....algorithm it is easy to see that Y r is 1 with probability exactly 1=r. Thus E[Y] P 1rn E[Y r ] P 1rn 1=r = H n and therefore E[X] the expected number of comparisons in our slowed down algorithm is 2nH n . To estimate Pr(Y c Delta E[Y] we can now use the well known Chernoff bound (see [47, 31]) which in one form states that if a random variable Z is the sum of n independent 0 1 random variables and the expectation of Z is E, then for c 1 Pr(Z c Delta E) e GammaE(1 Gammac c log c) In our case the Y i s can easily be proven to be independent. Thus we obtain Pr(Y c Delta ....

T. Hagerup and C. Rub. "A Guided Tour of Chernoff Bounds." Inform. Proc. Letters 33 (1989/90), pp 305--308.

....distribution. Call an n a Theta n a subarray a bin. For each white key the probability of being placed in a fixed bin is p = n 2(a Gamma1) Hence the expected number of white keys in a bin is = p Delta k. Let X be the number of white keys in a bin. We use the Chernoff bound (cf. [5]) which is the following estimate: Pr[jX Gamma j t ] 2 exp( Gammat 2 =3) Take t = n a b . Then t = n 2 b Gammaa =k and we obtain Pr[jX Gamma j n a b ] 2 exp Gamma n 4 2b Gamma2a Delta n 2a Gamma2 Delta k 3k 2 = 2 exp Gamma n 2 2b 3k 2 exp i ....

T. Hagerup, and C. Rub, A guided tour of Chernoff bounds, Inform. Process. Letters 33 (1989/90) 305--308.

....of e 3x=4 Gamma x Gamma 1 ( 3 4 e 3x=4 Gamma 1) is positive for x 1. Now let X 1 ; Xn be independent random variables with geometric distributions with parameter p, and let Z = X 1 Delta Delta Delta Xn . Let fi 0 and t 0. Then following along the lines of Hagerup and Rub [30] Pr(Z (1 fi)nq=p) e Gammat(1 fi)nq=p e t(1 fi)nq=p P (e tZ e t(1 fi)nq=p ) e Gammat(1 fi)nq=p E(e tZ ) Then since X 1 ; Xn are independent and identically distributed, we get E(e tZ ) E(e t(X1 Delta Delta Delta X n ) E(e tX1 Delta Delta Delta e tXn ....

T. Hagerup and C. R ub, A guided tour of Chernoff bounds, Inform. Process. Lett., 33 (1990), pp. 305--308.

....interval. The worst case buffer utilization occurs if all (1 ffl)N=k bursts in one interval arrive at the end, and all bursts in the following interval arrive at the beginning. To show that it is likely that all intervals will be light, we use the following theorem. Theorem 3. 1 (Angluin,Valiant [1, 3]) Let X i , 1 i n, be independent random variables, each of which has probability p of being 1 and probability 1 Gamma p of being 0. Then for any t, 0 t 1 Pr ( n X i=1 X i (1 t)np ) e Gammat 2 np=3 Choose one of the k intervals to examine. In our application of this theorem, ....

Torben Hagerup and Christine Rub. "A guided tour of Chernoff bounds." Information Processing Letters, 33(6):305--308, 1990.

....deviation is oe = p n=2. The above coefficients increase up to the mean n=2 and then decrease. Because the ratio oe=n tends to zero for n tending to infinity, for large n values most strings are clustered in a narrow peak at Hamming distance H = n=2. As an example, one can use the Chernoff bound [19]: Pr[H (1 Gamma )pn] e Gamma 2 np=2 (18) the probability to find a point at a distance less than np = n=2 decreases in the above exponential way ( 0) The distribution of Hamming distances for n = 500 is shown in Fig 3. Clearly, if better local optima are located in a cluster that is ....

T. Hagerup and C. Rub, "A guided tour of Chernoff bounds," Information Processing Letters 33 (1989/90), 305--308.

....random inputs r 1 ; r k independently, and compute the empirical mean 1 k P k i=1 Y (r i ) If the empirical mean exceeds j 2 decide on the first hypothesis, else decide on the second. The probability bound is an immediate consequence of the Chernoff inequality: Theorem 3. 1 [Che52, HR90] Let fX i g 1 i=1 be i.i.d (independent and identically distributed) 0; 1 random variables with a mean . Then: Pr 1 k k X i=1 X i Gamma a # ( ff fi fi 1 Gamma ff 1 Gamma fi 1 Gammafi ) k where ff = and fi = a Note that kjrj perfect random bits are ....

T. Hagerup, and C. Rub, A Guided tour of Chernoff bounds. Inf. Proc. Lett, 33, 305-308.

....an incident node. We know that P r(X j = 1) 1=ffi i . Let X = P X j . Note that E(X) 1. We want to bound the deviation from the expected value. Because the decisions of the adjacent nodes are independent, we can use the CONTENTION RESOLUTION IN SHARED MEMORY SIMULATIONS 35 Chernoff bound [15]: P r(X ff) P r(X ffE(X) 1 2 ff for ff 5 : In particular, this implies that P r(a node has degree smaller than ffi i 1 ) 1 2 ffi i Gammaffi i 1 Gamma1 : Let x l , 1 l ffi d i 1 i Gamma1 d i Gamma1 , be the sequence of independent random decisions made by the ffi ....

T. Hagerup and C. R ub, A guided tour of Chernoff bounds, Inform. Process. Lett., 33 (1989/90), pp. 305--308.

....v and C [ D exist independently with probability 0:5. Furthermore, Y s CD = P v2Tsn(C[D) X v , i.e. Y s CD is the number of ones in a Bernoulli trial of length jT s n (C [D)j = n=2 Gamma x with p = 2 GammajC[Dj , where x = jT s (C [ D)j. Note that x r(n) We can use Chernoff bounds [Hagerup and Rub 1989] to estimate Y s CD . max (C;D)2C r(n) P Y s CD fi n 2 2 GammajC[Dj max (C;D)2C r(n) n=2 Gamma x)p fi n 2 p fi(n=2)p e fi(n=2)p Gamma(n=2 Gammax)p max (C;D)2C r(n) 1 fi fi(n=2)p e fi(n=2)p Gamma(n=2)p n n 1 fi fi e 1 Gammafi (n=2)2 ....

Hagerup, T. and R ub, C. 1989. A Guided Tour of Chernoff Bounds. Information Processing Letters 33, 305 -- 308.

.... distribution is a satisfactory approximation [3] Moreover, X = P n i X i , and the expected value of X is given as = E[X] E[ P n i=1 X i ] P n i=1 E[X i ] n , since E[X i ] 0 Delta P (X = 0) 1 Delta P (X = 1) For any positive constant, 0 ffl 1, the Chernoff bounds [4] state that P (X (1 Gamma ffl)n ) e Gammaffl 2 n =2 (1) P (X (1 ffl)n ) e Gammaffl 2 n =3 (2) Chernoff bounds provide information on how close is the actual occurrence of an itemset in the sample, as compared to the expected count in the sample. This aspect, which we call as ....

T. Hagerup and C. Rub. A guided tour of chernoff bounds. In Information Processing Letters, pages 305--308. North-Holland, 1989/90.

....= X jk = 1) P (X i = 1) Lemma 2.6 Consider n 0 1 random variables X 1 ; X 2 ; Xn which are self weakening. Let X = P n i=1 X i and let E(X) for some . Then P (X (1 ffl) e ffl (1 ffl) 1 ffl) e ffl 2 =3 : Proof. For a proof see [17] and [12]. Second, we use the General Markov Inequality (see, e.g. 19] to obtain bounds for sums of s wise independent binary random variables. Lemma 2.7 (General Markov Inequality) Let X be an arbitrary random variable then, for every ffl 0 and k 0, it holds: Prob jX Gamma E(X)j ffl k q ....

T. Hagerup and C. Rub, A guided tour of Chernoff bounds, Inform. Proc. Letters 33 (1989/90) 305-308.

....Lemma 2.2 (Alon and Megiddo [2] Given n half space constraints in d , for fixed d, linear programming can be performed in O(1) time with n processors on a CRCW PRAM, with (nexponential) probability 1 Gamma 2 cn 1 3 , where c 1 is a constant. Lemma 2. 3 (Tail Estimation (Chernoff bound [8, 20]) Let X 1 ; X 2 ; X n 2 f0; 1g be n Bernoulli trials: independent trials with probability p i that X i = 1, where p i 2 (0; 1) Define X = Sigma n i=1 X i , Sigma n i=1 p i . Then, for all ffi 0, P rob(X (1 ffi) e ffi (1 ffi) 1 ffi # ; and, for all ffi ....

....writes the coordinates of its point into that location. 5. Each processor that did suffer a collision repeats steps 2 4 for a total of up to d attempts, where d is a constant. Note that all these steps can be performed in O(1) time with m processors, on the CRCW PRAM. By a Chernoff bound [8, 20] (Lemma 2.3) fewer than 4k processors will attempt a write, with probability 1 Gamma 4 e 2k ; and more than k processors, with probability 1 Gamma e 2 Gammak : Thus, given that m 0 processors attempt to write, with k m 0 4k, the probability of such a processor ....

T. Hagerup and C. Rub, "A guided tour of Chernoff bounds," Information Processing Letters, 33, 305--308, 1989/90.

....and only if the jth incident edge of v will be removed by an incident node. We know that P r(X j = 1) 1=ffi i . Let X = P X j . Note that E(X) 1. We want to bound the deviation from the expected value. Because the decisions of the adjacent nodes are independent, we can use the Chernoff bound [17]: P r(X ff) P r(X ffE(X) 1 2 ff for ff 5 Particularly, this implies that P r(a node has degree smaller than ffi i 1 ) 1 2 ffi i Gammaffi i 1 Gamma1 Let x l , 1 l ffi d i 1 i Gamma1 d Gamma1 , be the sequence of independent random decisions made by the ffi d i 1 i ....

T. Hagerup and C. R¨ ub, A guided tour of Chernoff bounds, Inform. Process. Lett., 33 (1989/90), pp. 305--308.

....of n Bernouli trials where in the i th trial, for i = 1; 2; n, success occurs with probability p i = p and failure occurs with probability q i = 1 Gamma p. If X is a random variable that describes the total number of successes in this sequence of n trials, then for 0 a np we have [5]: P rfX np ag e Gammaa 2 3np 2 3 1 4 5 6 7 8 2 3 1 4 5 6 7 8 A B C D E F G H A B C D E F G H (a) Before routing (b) After routing Figure 5: Colouring B is an (n; n; n 2 ; n 2 ) colouring. In our case, success is considered to be the event of colouring a packet black and failure the ....

T. Hagerup, C. Rub, "A Guided Tour of Chernoff Bounds", Information Processing Letters, Vol. 33, pp. 305-308, 1990.

....the event that the kth task of size log i N does not depart; the x k i are independent random variables. x k i = ae 1 with probability 1= log N 0 with probability 1 Gamma (1= log N) The size s(oer ) can be expressed as a weighted sum of the variables, x k i . We now use Chernoff bounds [16] in a standard fashion to derive the high probability bound. 2 For any j, let l(T 0 j ; i) denote the load of a (log j N) PE submachine T 0 j of T at the beginning of phase i. Define the following potential functions. ffl The potential of each (log i N) PE submachine at the beginning ....

T. Hagerup and C. Rub (1989): A Guided Tour of Chernoff Bounds. Inf. Proc. Let., 305-308.

.... distribution is a satisfactory approximation [4] Moreover, X = P n i X i , and the expected value of X is given as = E[X] E[ P n i=1 X i ] P n i=1 E[X i ] n , since E[X i ] 0 Delta P (X = 0) 1 Delta P (X = 1) For any positive constant, 0 ffl 1, the Chernoff bounds [5] state that P (X (1 Gamma ffl)n ) e Gammaffl 2 n =2 (1) P (X (1 ffl)n ) e Gammaffl 2 n =3 (2) Chernoff bounds provide information on how close is the actual occurrence of an itemset in the sample, as compared to the expected count in the sample. This aspect, which we call as ....

T. Hagerup and C. Rub. A guided tour of chernoff bounds. In Information Processing Letters, pages 305--308. NorthHolland, 1989/90.

....11 t 21 t 31 t 41 t 12 t 22 t 32 t 42 t 13 t 23 t 33 t 43 t 14 t 24 t 34 t 44 E : T j W t 21 t 31 t 41 t 32 t 42 t 43 Figure 5: Removing unwanted copies. The probabilistic analysis of Lemma 7.5 below is based largely on the Chernoff bounds expressed in the following lemma. For proofs, see, e.g. [21]. Lemma 7.4 For every binomially distributed random variable Z and for all ffl with 0 ffl 1, a) For all z E(Z) Pr(Z (1 ffl)z) e Gammaffl 2 z=3 . b) Pr(Z (1 Gamma ffl)E(Z) e Gammaffl 2 E(z) 2 . Lemma 7.5 Suppose that we are given integers M;m 2 and f log M 2 with m ....

T. Hagerup and C. Rub. A guided tour of Chernoff bounds. Inform. Process. Lett., 33, pp. 305--308, 1990.

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T. Hagerup and C. R ub. A guided tour of chernoff bounds. Inform. Process. Lett., 33(6):305--308, Feb. 1990.

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T. Hagerup and C. Rub, A guided tour of the Chernoff bounds, Inf. Proc. Letters 33 (1989/90) 305-308. 3

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Hagerup, T., & Rub, C. (1990). A guided tour of Chernoff bounds. Information Processing Letters, 33, 305--308.

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T. Hagerup and C. Rub, A Guided Tour of Chernoff Bounds, Inf. Proc. Letters, 33, 305--308, (1989).

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T. Hagerup, C. Rub, A guided tour of Chernoff bounds, Inform. Proces. Lett. 33 (1990), 305-308.

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Hagerup, T., & Rub, C. (1990). A guided tour of Chernoff bounds. Information Processing Letters, 33, 305--308.

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Torben Hagerup, Christiane Rüb, A Guided Tour of Chern- off Bounds, Information Processing Letters 33, pp. 305-308, 1989. 10

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T. Hagerup and Ch. Rub, A Guided Tour of Chernoff Bounds, Inf. Proc. Letters 33 (1989/90), 305-308.

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