Janson s Inequality and Poisson Heuristic
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1 Janson s Inequality and Poisson Heuristic Dinesh K CS11M019 IIT Madras April 30, 2012 Dinesh (IITM) Janson s Inequality April 30, / 11
2 Outline 1 Motivation Dinesh (IITM) Janson s Inequality April 30, / 11
3 Outline 1 Motivation 2 Janson s Inequality Dinesh (IITM) Janson s Inequality April 30, / 11
4 Outline 1 Motivation 2 Janson s Inequality 3 Proof of Lower bound Dinesh (IITM) Janson s Inequality April 30, / 11
5 Outline 1 Motivation 2 Janson s Inequality 3 Proof of Lower bound 4 Proof of Upper Bound Dinesh (IITM) Janson s Inequality April 30, / 11
6 Outline 1 Motivation 2 Janson s Inequality 3 Proof of Lower bound 4 Proof of Upper Bound Dinesh (IITM) Janson s Inequality April 30, / 11
7 Motivation Given Ω, events {B i } i [n] defined on Ω. Find Pr[ n i=1b i ] Events Mutually independent = Can be easily handled. But cannot expect mutual independence among events always. Question Can the probability of occurrence of bad events be bounded even if there is no mutual independence among events? Poisson Paradigm When X is a sum of mostly independent random variables with µ = E[X ], Pr[X = 0] e µ Dinesh (IITM) Janson s Inequality April 30, / 11
8 Outline 1 Motivation 2 Janson s Inequality 3 Proof of Lower bound 4 Proof of Upper Bound Dinesh (IITM) Janson s Inequality April 30, / 11
9 Janson s Inequality Sample space : Ω. Create a random set R as follows. At each element r in sample space, toss coin with bias p r independently and add it to set R. Dinesh (IITM) Janson s Inequality April 30, / 11
10 Janson s Inequality Sample space : Ω. Create a random set R as follows. At each element r in sample space, toss coin with bias p r independently and add it to set R. Let {A i } i I be subsets of Ω. Dinesh (IITM) Janson s Inequality April 30, / 11
11 Janson s Inequality Sample space : Ω. Create a random set R as follows. At each element r in sample space, toss coin with bias p r independently and add it to set R. Let {A i } i I be subsets of Ω. B i : [A i R]. Dinesh (IITM) Janson s Inequality April 30, / 11
12 Janson s Inequality Sample space : Ω. Create a random set R as follows. At each element r in sample space, toss coin with bias p r independently and add it to set R. Let {A i } i I be subsets of Ω. B i : [A i R]. X i the indicator random variable for B i. Let X = i I X i. We want to bound Same as Pr[X = 0]. Pr[ i I B i ] Dinesh (IITM) Janson s Inequality April 30, / 11
13 Janson s Inequality Sample space : Ω. Create a random set R as follows. At each element r in sample space, toss coin with bias p r independently and add it to set R. Let {A i } i I be subsets of Ω. B i : [A i R]. X i the indicator random variable for B i. Let X = i I X i. We want to bound Same as Pr[X = 0]. Denote i j if A i A j. Observation Pr[ i I B i ] Events B i and B j are independent when A i A j =. Reason : Coin flips involved are separate. Dinesh (IITM) Janson s Inequality April 30, / 11
14 Janson s Inequality {B i } i I be the set of events as defined before. = i j Pr[B i B j ] µ = E[X ] = i I Pr[B i ] M = i I Pr[B i ] Janson s Inequality M Pr [ i I B i ] exp( µ + /2) Dinesh (IITM) Janson s Inequality April 30, / 11
15 Proof of Janson s Inequality Results used Pr[A B C] Pr[A B C], for any events A, B, C. Dinesh (IITM) Janson s Inequality April 30, / 11
16 Proof of Janson s Inequality Results used Pr[A B C] Pr[A B C], for any events A, B, C. (FGK Inequality) Pr [ ] B i 1 j<i B j Pr[Bi ] Dinesh (IITM) Janson s Inequality April 30, / 11
17 Proof of Janson s Inequality Results used Pr[A B C] Pr[A B C], for any events A, B, C. (FGK Inequality) Pr [ ] B i 1 j<i B j Pr[Bi ] Proof Idea : Let I = {1, 2,..., m}. Pr [ ] m i I B i = Pr [ ] B i 1 j<i B j Use FGK to lower bound. Upper bounding requires work. i=1 Dinesh (IITM) Janson s Inequality April 30, / 11
18 Outline 1 Motivation 2 Janson s Inequality 3 Proof of Lower bound 4 Proof of Upper Bound Dinesh (IITM) Janson s Inequality April 30, / 11
19 Proof of Lower bound Index set I = {1, 2,..., m}. Start with FKG inequality Pr [ ] B i 1 j<i B j Pr[Bi ] Inverting on both sides. Pr [ ] B i 1 j<i B j Pr[Bi ] Apply definition of conditional probabilities. Pr [ ] m i I B i = Pr [ ] B i 1 j<i B j i=1 m Pr[B i ] = M i=1 Dinesh (IITM) Janson s Inequality April 30, / 11
20 Outline 1 Motivation 2 Janson s Inequality 3 Proof of Lower bound 4 Proof of Upper Bound Dinesh (IITM) Janson s Inequality April 30, / 11
21 Proof of Upper Bound - I Index set I = {1, 2,..., m}. Need to upper bound Pr [ m i=1b i ] Enough to find lower bound for Pr [ ] B i 1 j<i B j P(i) = Pr [ ] B i B 1 B 2... B i 1 Re-number events so that i j for 1 j d and not for d + 1 j < i. Dinesh (IITM) Janson s Inequality April 30, / 11
22 Proof of Upper Bound - I Index set I = {1, 2,..., m}. Need to upper bound Pr [ m i=1b i ] Enough to find lower bound for Pr [ B i 1 j<i B j ] P(i) = Pr [ B i B 1 B 2... B i 1 ] Re-number events so that i j for 1 j d and not for d + 1 j < i. Split! B = B 1... B d C = B d+1... B i 1 Dinesh (IITM) Janson s Inequality April 30, / 11
23 Proof of Upper Bound - II P(i) = Pr[B i B C] Pr[B i B C] (By independence of B i and C) P(i) Pr[B i C] Pr[B B i C] P(i) Pr[B i ] Pr[B B i C] Dinesh (IITM) Janson s Inequality April 30, / 11
24 Proof of Upper Bound - II P(i) = Pr[B i B C] Pr[B i B C] (By independence of B i and C) P(i) Pr[B i C] Pr[B B i C] P(i) Pr[B i ] Pr[B B i C] How to lower bound Pr[B i ] Pr[B B i C]? P(i) Pr[B i ] Pr[ d j=1b j B i C] Dinesh (IITM) Janson s Inequality April 30, / 11
25 Proof of Upper Bound - II P(i) = Pr[B i B C] Pr[B i B C] (By independence of B i and C) P(i) Pr[B i C] Pr[B B i C] P(i) Pr[B i ] Pr[B B i C] How to lower bound Pr[B i ] Pr[B B i C]? P(i) Pr[B i ] Pr[ d j=1b j B i C] (Taking complementary event) P(i) Pr[B i ] (1 Pr[ d j=1b j B i C]) Dinesh (IITM) Janson s Inequality April 30, / 11
26 Proof of Upper Bound - III (By union bound) d P(i) Pr[B i ] 1 Pr[B j B i C] j=1 Dinesh (IITM) Janson s Inequality April 30, / 11
27 Proof of Upper Bound - III (By union bound) d P(i) Pr[B i ] 1 Pr[B j B i C] j=1 (FKG Inequality) Pr[B j B i C] Pr[B j B i ] d P(i) Pr[B i ] Pr[B i ] Pr[B j B i ] j=1 Dinesh (IITM) Janson s Inequality April 30, / 11
28 Proof of Upper Bound - III (By union bound) d P(i) Pr[B i ] 1 Pr[B j B i C] j=1 (FKG Inequality) Pr[B j B i C] Pr[B j B i ] d P(i) Pr[B i ] Pr[B i ] Pr[B j B i ] j=1 d P(i) Pr[B i ] Pr[B j B i ] j=1 Dinesh (IITM) Janson s Inequality April 30, / 11
29 Proof of Upper Bound - III (By union bound) d P(i) Pr[B i ] 1 Pr[B j B i C] j=1 (FKG Inequality) Pr[B j B i C] Pr[B j B i ] d P(i) Pr[B i ] Pr[B i ] Pr[B j B i ] j=1 d P(i) Pr[B i ] Pr[B j B i ] Pr [ ] d B i 1 j<i B j 1 P(i) exp( Pr[Bi ] + Pr[B j B i )] j=1 j=1 Dinesh (IITM) Janson s Inequality April 30, / 11
30 Summary Janson s Inequality provides exponential upper bound on sum of dependent Bernoulli random variables being zero. Quantity is a measure of pairwise dependence. For small we can get good bounds. If 2µ result is not of much use. Dinesh (IITM) Janson s Inequality April 30, / 11
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