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Probabilistic Method
Presented by Jordon LynCourse COMP 4804
Seminar Oulinebull Short talk about its origin and definitionbull Explain the various techniques used in performing this
method (Proofs will be used to show how it works)
Origin and definition
Created by Paul Erdos in 1947
Used to help prove the existence of certain objects without having to construct them explicitly
Achieved by the construction of a sample space with all possible candidates given that the probability of finding the desired object is positive
Examples
Probabilistic Method Succeds
bull Getting an even number on a fair 6-sided diebull Only 6 number on the die
bull (1 2 3 4 5 6)
bull Pr(Even Number) = 36 = 12 gt 0 meaning an even number does exist
Probabilistic Method Failsbull Finding 5 Aces in a shuffled
deck of cardsbull 52 cards in a shuffled deck
with 13 different ranks and 4 suitsbull Each rank appears 5213 = 4
times (one for each suit) meaning there can only be a maximum of four Aces in a shuffled deck
bull Pr(5 Aces) = 0 meaning a fifth Ace does not exist
Techniques bull Basic Methodbull Linearity of Expectationbull Alterationsbull The Second Momentbull Lovasz Local Lemma
Basic Methodbull Best illustrated through many methods
bull Ramsey Numbersbull Any sufficiently large graph contains either a clique (a set of vertices that induce a
subgraph) or an independent set (a set of vertices that induce edgeless subgraph)
bull Hypergraph Coloringbull A hypergraph is c-colorable if the vertices of the graph can be colored
with c colors given that at least two colors appear in every edgebull Erdos-Ko-Rado Theorem
bull A family of sets F is intersecting if for every pair of sets within F the intersection set of both those sets is greater than zero
bull Overall goal prove a lower bound exists for each method when dealing with large integer valuesbull More sophisticated and simpler than counting arguments
Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set
of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of
vertices
bull Graphs are 2-colorablebull Generalized notion of graph coloring
bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound
Hypergraph Coloring Proofbull Theorem For any k ge 2
m(k)ge 2^k-1bull Consider a hypergraph H that has less than
2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1
bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is
at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1
bull Therefore there is a probability greater than 0 that every edge in H has at least two colors
Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao
Ko and Richard Radobull Lemma X = 0 1n-1 with
modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting
family F contains at most k of the sets of As
bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1
Erdos-Ko-Rado Theorem Proof
Lemma Proof
bull Assume the set Ai is an element of F bull Therefore any other set in As is
one of the followingbull Ai Ai+1 Ai+k-1
bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)
bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k
Theorem Proofbull Assume X = 0 1n-1 and
F (a subset of X choose k) is an intersecting familybull By following the lemma at
least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =
|F|(n choose k)bull |F| = (n choose k) kn= n-1
choose k-1
Linearity of Expectation bull Takes the expected values of individual indicators and uses
their summation as the expected value of a random variablebull Also knows as the first moment method
bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality
bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random
variable
bull Such examplesbull Tournamentsbull Hamiltonian Paths
Tournaments and Hamiltonian Paths
Tournaments
bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices
and n choose 2 edges
Hamiltonian Paths
bull A path that visits every vertex in the graph exactly oncebull Named after William
Rowan Hamilton inventor of Hamiltons puzzle
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Seminar Oulinebull Short talk about its origin and definitionbull Explain the various techniques used in performing this
method (Proofs will be used to show how it works)
Origin and definition
Created by Paul Erdos in 1947
Used to help prove the existence of certain objects without having to construct them explicitly
Achieved by the construction of a sample space with all possible candidates given that the probability of finding the desired object is positive
Examples
Probabilistic Method Succeds
bull Getting an even number on a fair 6-sided diebull Only 6 number on the die
bull (1 2 3 4 5 6)
bull Pr(Even Number) = 36 = 12 gt 0 meaning an even number does exist
Probabilistic Method Failsbull Finding 5 Aces in a shuffled
deck of cardsbull 52 cards in a shuffled deck
with 13 different ranks and 4 suitsbull Each rank appears 5213 = 4
times (one for each suit) meaning there can only be a maximum of four Aces in a shuffled deck
bull Pr(5 Aces) = 0 meaning a fifth Ace does not exist
Techniques bull Basic Methodbull Linearity of Expectationbull Alterationsbull The Second Momentbull Lovasz Local Lemma
Basic Methodbull Best illustrated through many methods
bull Ramsey Numbersbull Any sufficiently large graph contains either a clique (a set of vertices that induce a
subgraph) or an independent set (a set of vertices that induce edgeless subgraph)
bull Hypergraph Coloringbull A hypergraph is c-colorable if the vertices of the graph can be colored
with c colors given that at least two colors appear in every edgebull Erdos-Ko-Rado Theorem
bull A family of sets F is intersecting if for every pair of sets within F the intersection set of both those sets is greater than zero
bull Overall goal prove a lower bound exists for each method when dealing with large integer valuesbull More sophisticated and simpler than counting arguments
Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set
of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of
vertices
bull Graphs are 2-colorablebull Generalized notion of graph coloring
bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound
Hypergraph Coloring Proofbull Theorem For any k ge 2
m(k)ge 2^k-1bull Consider a hypergraph H that has less than
2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1
bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is
at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1
bull Therefore there is a probability greater than 0 that every edge in H has at least two colors
Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao
Ko and Richard Radobull Lemma X = 0 1n-1 with
modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting
family F contains at most k of the sets of As
bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1
Erdos-Ko-Rado Theorem Proof
Lemma Proof
bull Assume the set Ai is an element of F bull Therefore any other set in As is
one of the followingbull Ai Ai+1 Ai+k-1
bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)
bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k
Theorem Proofbull Assume X = 0 1n-1 and
F (a subset of X choose k) is an intersecting familybull By following the lemma at
least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =
|F|(n choose k)bull |F| = (n choose k) kn= n-1
choose k-1
Linearity of Expectation bull Takes the expected values of individual indicators and uses
their summation as the expected value of a random variablebull Also knows as the first moment method
bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality
bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random
variable
bull Such examplesbull Tournamentsbull Hamiltonian Paths
Tournaments and Hamiltonian Paths
Tournaments
bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices
and n choose 2 edges
Hamiltonian Paths
bull A path that visits every vertex in the graph exactly oncebull Named after William
Rowan Hamilton inventor of Hamiltons puzzle
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Origin and definition
Created by Paul Erdos in 1947
Used to help prove the existence of certain objects without having to construct them explicitly
Achieved by the construction of a sample space with all possible candidates given that the probability of finding the desired object is positive
Examples
Probabilistic Method Succeds
bull Getting an even number on a fair 6-sided diebull Only 6 number on the die
bull (1 2 3 4 5 6)
bull Pr(Even Number) = 36 = 12 gt 0 meaning an even number does exist
Probabilistic Method Failsbull Finding 5 Aces in a shuffled
deck of cardsbull 52 cards in a shuffled deck
with 13 different ranks and 4 suitsbull Each rank appears 5213 = 4
times (one for each suit) meaning there can only be a maximum of four Aces in a shuffled deck
bull Pr(5 Aces) = 0 meaning a fifth Ace does not exist
Techniques bull Basic Methodbull Linearity of Expectationbull Alterationsbull The Second Momentbull Lovasz Local Lemma
Basic Methodbull Best illustrated through many methods
bull Ramsey Numbersbull Any sufficiently large graph contains either a clique (a set of vertices that induce a
subgraph) or an independent set (a set of vertices that induce edgeless subgraph)
bull Hypergraph Coloringbull A hypergraph is c-colorable if the vertices of the graph can be colored
with c colors given that at least two colors appear in every edgebull Erdos-Ko-Rado Theorem
bull A family of sets F is intersecting if for every pair of sets within F the intersection set of both those sets is greater than zero
bull Overall goal prove a lower bound exists for each method when dealing with large integer valuesbull More sophisticated and simpler than counting arguments
Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set
of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of
vertices
bull Graphs are 2-colorablebull Generalized notion of graph coloring
bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound
Hypergraph Coloring Proofbull Theorem For any k ge 2
m(k)ge 2^k-1bull Consider a hypergraph H that has less than
2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1
bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is
at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1
bull Therefore there is a probability greater than 0 that every edge in H has at least two colors
Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao
Ko and Richard Radobull Lemma X = 0 1n-1 with
modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting
family F contains at most k of the sets of As
bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1
Erdos-Ko-Rado Theorem Proof
Lemma Proof
bull Assume the set Ai is an element of F bull Therefore any other set in As is
one of the followingbull Ai Ai+1 Ai+k-1
bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)
bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k
Theorem Proofbull Assume X = 0 1n-1 and
F (a subset of X choose k) is an intersecting familybull By following the lemma at
least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =
|F|(n choose k)bull |F| = (n choose k) kn= n-1
choose k-1
Linearity of Expectation bull Takes the expected values of individual indicators and uses
their summation as the expected value of a random variablebull Also knows as the first moment method
bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality
bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random
variable
bull Such examplesbull Tournamentsbull Hamiltonian Paths
Tournaments and Hamiltonian Paths
Tournaments
bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices
and n choose 2 edges
Hamiltonian Paths
bull A path that visits every vertex in the graph exactly oncebull Named after William
Rowan Hamilton inventor of Hamiltons puzzle
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Examples
Probabilistic Method Succeds
bull Getting an even number on a fair 6-sided diebull Only 6 number on the die
bull (1 2 3 4 5 6)
bull Pr(Even Number) = 36 = 12 gt 0 meaning an even number does exist
Probabilistic Method Failsbull Finding 5 Aces in a shuffled
deck of cardsbull 52 cards in a shuffled deck
with 13 different ranks and 4 suitsbull Each rank appears 5213 = 4
times (one for each suit) meaning there can only be a maximum of four Aces in a shuffled deck
bull Pr(5 Aces) = 0 meaning a fifth Ace does not exist
Techniques bull Basic Methodbull Linearity of Expectationbull Alterationsbull The Second Momentbull Lovasz Local Lemma
Basic Methodbull Best illustrated through many methods
bull Ramsey Numbersbull Any sufficiently large graph contains either a clique (a set of vertices that induce a
subgraph) or an independent set (a set of vertices that induce edgeless subgraph)
bull Hypergraph Coloringbull A hypergraph is c-colorable if the vertices of the graph can be colored
with c colors given that at least two colors appear in every edgebull Erdos-Ko-Rado Theorem
bull A family of sets F is intersecting if for every pair of sets within F the intersection set of both those sets is greater than zero
bull Overall goal prove a lower bound exists for each method when dealing with large integer valuesbull More sophisticated and simpler than counting arguments
Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set
of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of
vertices
bull Graphs are 2-colorablebull Generalized notion of graph coloring
bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound
Hypergraph Coloring Proofbull Theorem For any k ge 2
m(k)ge 2^k-1bull Consider a hypergraph H that has less than
2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1
bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is
at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1
bull Therefore there is a probability greater than 0 that every edge in H has at least two colors
Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao
Ko and Richard Radobull Lemma X = 0 1n-1 with
modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting
family F contains at most k of the sets of As
bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1
Erdos-Ko-Rado Theorem Proof
Lemma Proof
bull Assume the set Ai is an element of F bull Therefore any other set in As is
one of the followingbull Ai Ai+1 Ai+k-1
bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)
bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k
Theorem Proofbull Assume X = 0 1n-1 and
F (a subset of X choose k) is an intersecting familybull By following the lemma at
least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =
|F|(n choose k)bull |F| = (n choose k) kn= n-1
choose k-1
Linearity of Expectation bull Takes the expected values of individual indicators and uses
their summation as the expected value of a random variablebull Also knows as the first moment method
bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality
bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random
variable
bull Such examplesbull Tournamentsbull Hamiltonian Paths
Tournaments and Hamiltonian Paths
Tournaments
bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices
and n choose 2 edges
Hamiltonian Paths
bull A path that visits every vertex in the graph exactly oncebull Named after William
Rowan Hamilton inventor of Hamiltons puzzle
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Techniques bull Basic Methodbull Linearity of Expectationbull Alterationsbull The Second Momentbull Lovasz Local Lemma
Basic Methodbull Best illustrated through many methods
bull Ramsey Numbersbull Any sufficiently large graph contains either a clique (a set of vertices that induce a
subgraph) or an independent set (a set of vertices that induce edgeless subgraph)
bull Hypergraph Coloringbull A hypergraph is c-colorable if the vertices of the graph can be colored
with c colors given that at least two colors appear in every edgebull Erdos-Ko-Rado Theorem
bull A family of sets F is intersecting if for every pair of sets within F the intersection set of both those sets is greater than zero
bull Overall goal prove a lower bound exists for each method when dealing with large integer valuesbull More sophisticated and simpler than counting arguments
Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set
of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of
vertices
bull Graphs are 2-colorablebull Generalized notion of graph coloring
bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound
Hypergraph Coloring Proofbull Theorem For any k ge 2
m(k)ge 2^k-1bull Consider a hypergraph H that has less than
2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1
bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is
at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1
bull Therefore there is a probability greater than 0 that every edge in H has at least two colors
Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao
Ko and Richard Radobull Lemma X = 0 1n-1 with
modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting
family F contains at most k of the sets of As
bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1
Erdos-Ko-Rado Theorem Proof
Lemma Proof
bull Assume the set Ai is an element of F bull Therefore any other set in As is
one of the followingbull Ai Ai+1 Ai+k-1
bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)
bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k
Theorem Proofbull Assume X = 0 1n-1 and
F (a subset of X choose k) is an intersecting familybull By following the lemma at
least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =
|F|(n choose k)bull |F| = (n choose k) kn= n-1
choose k-1
Linearity of Expectation bull Takes the expected values of individual indicators and uses
their summation as the expected value of a random variablebull Also knows as the first moment method
bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality
bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random
variable
bull Such examplesbull Tournamentsbull Hamiltonian Paths
Tournaments and Hamiltonian Paths
Tournaments
bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices
and n choose 2 edges
Hamiltonian Paths
bull A path that visits every vertex in the graph exactly oncebull Named after William
Rowan Hamilton inventor of Hamiltons puzzle
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Basic Methodbull Best illustrated through many methods
bull Ramsey Numbersbull Any sufficiently large graph contains either a clique (a set of vertices that induce a
subgraph) or an independent set (a set of vertices that induce edgeless subgraph)
bull Hypergraph Coloringbull A hypergraph is c-colorable if the vertices of the graph can be colored
with c colors given that at least two colors appear in every edgebull Erdos-Ko-Rado Theorem
bull A family of sets F is intersecting if for every pair of sets within F the intersection set of both those sets is greater than zero
bull Overall goal prove a lower bound exists for each method when dealing with large integer valuesbull More sophisticated and simpler than counting arguments
Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set
of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of
vertices
bull Graphs are 2-colorablebull Generalized notion of graph coloring
bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound
Hypergraph Coloring Proofbull Theorem For any k ge 2
m(k)ge 2^k-1bull Consider a hypergraph H that has less than
2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1
bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is
at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1
bull Therefore there is a probability greater than 0 that every edge in H has at least two colors
Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao
Ko and Richard Radobull Lemma X = 0 1n-1 with
modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting
family F contains at most k of the sets of As
bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1
Erdos-Ko-Rado Theorem Proof
Lemma Proof
bull Assume the set Ai is an element of F bull Therefore any other set in As is
one of the followingbull Ai Ai+1 Ai+k-1
bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)
bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k
Theorem Proofbull Assume X = 0 1n-1 and
F (a subset of X choose k) is an intersecting familybull By following the lemma at
least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =
|F|(n choose k)bull |F| = (n choose k) kn= n-1
choose k-1
Linearity of Expectation bull Takes the expected values of individual indicators and uses
their summation as the expected value of a random variablebull Also knows as the first moment method
bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality
bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random
variable
bull Such examplesbull Tournamentsbull Hamiltonian Paths
Tournaments and Hamiltonian Paths
Tournaments
bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices
and n choose 2 edges
Hamiltonian Paths
bull A path that visits every vertex in the graph exactly oncebull Named after William
Rowan Hamilton inventor of Hamiltons puzzle
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set
of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of
vertices
bull Graphs are 2-colorablebull Generalized notion of graph coloring
bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound
Hypergraph Coloring Proofbull Theorem For any k ge 2
m(k)ge 2^k-1bull Consider a hypergraph H that has less than
2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1
bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is
at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1
bull Therefore there is a probability greater than 0 that every edge in H has at least two colors
Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao
Ko and Richard Radobull Lemma X = 0 1n-1 with
modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting
family F contains at most k of the sets of As
bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1
Erdos-Ko-Rado Theorem Proof
Lemma Proof
bull Assume the set Ai is an element of F bull Therefore any other set in As is
one of the followingbull Ai Ai+1 Ai+k-1
bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)
bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k
Theorem Proofbull Assume X = 0 1n-1 and
F (a subset of X choose k) is an intersecting familybull By following the lemma at
least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =
|F|(n choose k)bull |F| = (n choose k) kn= n-1
choose k-1
Linearity of Expectation bull Takes the expected values of individual indicators and uses
their summation as the expected value of a random variablebull Also knows as the first moment method
bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality
bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random
variable
bull Such examplesbull Tournamentsbull Hamiltonian Paths
Tournaments and Hamiltonian Paths
Tournaments
bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices
and n choose 2 edges
Hamiltonian Paths
bull A path that visits every vertex in the graph exactly oncebull Named after William
Rowan Hamilton inventor of Hamiltons puzzle
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Hypergraph Coloring Proofbull Theorem For any k ge 2
m(k)ge 2^k-1bull Consider a hypergraph H that has less than
2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1
bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is
at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1
bull Therefore there is a probability greater than 0 that every edge in H has at least two colors
Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao
Ko and Richard Radobull Lemma X = 0 1n-1 with
modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting
family F contains at most k of the sets of As
bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1
Erdos-Ko-Rado Theorem Proof
Lemma Proof
bull Assume the set Ai is an element of F bull Therefore any other set in As is
one of the followingbull Ai Ai+1 Ai+k-1
bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)
bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k
Theorem Proofbull Assume X = 0 1n-1 and
F (a subset of X choose k) is an intersecting familybull By following the lemma at
least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =
|F|(n choose k)bull |F| = (n choose k) kn= n-1
choose k-1
Linearity of Expectation bull Takes the expected values of individual indicators and uses
their summation as the expected value of a random variablebull Also knows as the first moment method
bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality
bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random
variable
bull Such examplesbull Tournamentsbull Hamiltonian Paths
Tournaments and Hamiltonian Paths
Tournaments
bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices
and n choose 2 edges
Hamiltonian Paths
bull A path that visits every vertex in the graph exactly oncebull Named after William
Rowan Hamilton inventor of Hamiltons puzzle
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao
Ko and Richard Radobull Lemma X = 0 1n-1 with
modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting
family F contains at most k of the sets of As
bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1
Erdos-Ko-Rado Theorem Proof
Lemma Proof
bull Assume the set Ai is an element of F bull Therefore any other set in As is
one of the followingbull Ai Ai+1 Ai+k-1
bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)
bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k
Theorem Proofbull Assume X = 0 1n-1 and
F (a subset of X choose k) is an intersecting familybull By following the lemma at
least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =
|F|(n choose k)bull |F| = (n choose k) kn= n-1
choose k-1
Linearity of Expectation bull Takes the expected values of individual indicators and uses
their summation as the expected value of a random variablebull Also knows as the first moment method
bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality
bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random
variable
bull Such examplesbull Tournamentsbull Hamiltonian Paths
Tournaments and Hamiltonian Paths
Tournaments
bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices
and n choose 2 edges
Hamiltonian Paths
bull A path that visits every vertex in the graph exactly oncebull Named after William
Rowan Hamilton inventor of Hamiltons puzzle
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Erdos-Ko-Rado Theorem Proof
Lemma Proof
bull Assume the set Ai is an element of F bull Therefore any other set in As is
one of the followingbull Ai Ai+1 Ai+k-1
bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)
bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k
Theorem Proofbull Assume X = 0 1n-1 and
F (a subset of X choose k) is an intersecting familybull By following the lemma at
least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =
|F|(n choose k)bull |F| = (n choose k) kn= n-1
choose k-1
Linearity of Expectation bull Takes the expected values of individual indicators and uses
their summation as the expected value of a random variablebull Also knows as the first moment method
bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality
bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random
variable
bull Such examplesbull Tournamentsbull Hamiltonian Paths
Tournaments and Hamiltonian Paths
Tournaments
bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices
and n choose 2 edges
Hamiltonian Paths
bull A path that visits every vertex in the graph exactly oncebull Named after William
Rowan Hamilton inventor of Hamiltons puzzle
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Linearity of Expectation bull Takes the expected values of individual indicators and uses
their summation as the expected value of a random variablebull Also knows as the first moment method
bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality
bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random
variable
bull Such examplesbull Tournamentsbull Hamiltonian Paths
Tournaments and Hamiltonian Paths
Tournaments
bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices
and n choose 2 edges
Hamiltonian Paths
bull A path that visits every vertex in the graph exactly oncebull Named after William
Rowan Hamilton inventor of Hamiltons puzzle
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Tournaments and Hamiltonian Paths
Tournaments
bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices
and n choose 2 edges
Hamiltonian Paths
bull A path that visits every vertex in the graph exactly oncebull Named after William
Rowan Hamilton inventor of Hamiltons puzzle
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that
has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament
bull Each edge has 12 chance of being chosen independently
bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =
12^n-1
bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Alterationsbull Sometimes the first attempt at finding the desired
object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired
properties in a deterministic time
bull Such examplesbull High Girthbull High Chromatic Number
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that
the graph can be colored with a small number of colorsbull Erdos proved this was false
bull There do exist some graphs that have large chromatic numbers despite having no short cycles
bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k
such that the graph has proper k-coloring
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
High Girth and High Chromatic Example
High Girth High Chromatic
This graph has a girth of 6
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Second Momentbull Similar to the first moment method as it deals with showing how
a random variable has positive probability of being positivebull Accomplishes this with a random variables variance
bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate
the variance of the sum of random variables we need to understand the pairwise independence or covariance
bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Equations and proofbull Variance equation
bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2
bull Standard deviation of X = (Var(X))^12
bull If X1Xn are independent the variance is the sum of all Xi
bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))
= E(XY) = E(X) E(Y)bull If X1Xn are independent then
the covariance is 0bull Cov(XY) = 0 does not imply
independence of X and Y
bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof
bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)
bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within
two standard deviations of the mean
bull 89 within three standard deviations of the mean
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Empirical Rule vs Chebyshevs Inequality
Empirical Rule Chebyshevs Inequality
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)
and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent
bull Still a positive probability none of the bad events happen
bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have
relatively small probability and their dependency digraph does not have too many edges
bull Two Ways Symmetric and Generalbull Symmetric is the most often used
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all
outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj
bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0
bull Proof bull If d = 0 the events are mutually independent and the result follows
easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so
bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
General Local Lemmabull Let A1A2An be events
and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total
product of (1-Xj)bull Pr(Complements of
A1An) ge the total product of (1-Xi) ge 0
bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on
the size of Sbull If S = 0 the statement
follows directly from the lemmabull Pr(Ai) le the total product of
Xi (1-Xj)le Xi
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|
bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all
complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)
bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all
complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)
bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)
bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem
References
For informationbull Matousek Jiri Jan Vondrak The
Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File
bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File
For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le
mma
bull httpenwikipediaorgwikiTournament_(graph_theory)
bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht
ml
bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem