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| In mathematics, the '''Fortuin–Kasteleyn–Ginibre (FKG) inequality''' is a [[correlation]] inequality, a fundamental tool in [[statistical mechanics]] and [[Combinatorics#Probabilistic_combinatorics|probabilistic combinatorics]] (especially [[random graph]]s and the [[probabilistic method]]), due to {{harvs | last1=[[Cees M. Fortuin|Fortuin]] | first1=Cees M. | last2=[[Pieter Kasteleyn|Kasteleyn]] | first2=Pieter W. | last3=[[Jean Ginibre|Ginibre]] | first3=Jean | title=Correlation inequalities on some partially ordered sets | url=http://projecteuclid.org/euclid.cmp/1103857443 | mr=0309498 | year=1971 | journal=Communications in Mathematical Physics | volume=22 | pages=89–103|txt}}. Informally, it says that in many random systems, increasing events are positively correlated, while an increasing and a decreasing event are negatively correlated.
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| An earlier version, for the special case of [[i.i.d.]] variables, called '''Harris inequality''', is due to {{harvs|last=[[Ted Harris (mathematician)|Harris]] |first=Theodore Edward|year=1960|txt}}, see [[#A special case: the Harris inequality|below]]. One generalization of the FKG inequality is the [[#A generalization: the Holley inequality|Holley inequality (1974)]] below, and an even further generalization is the [[Ahlswede–Daykin inequality|Ahlswede–Daykin "four functions" theorem (1978)]]. Furthermore, it has the same conclusion as the [[Griffiths inequalities]], but the hypotheses are different.
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| ==The inequality== | |
| Let <math>X</math> be a finite [[distributive lattice]], and ''μ'' a nonnegative function on it, that is assumed to satisfy the ('''FKG''') '''lattice condition''' (sometimes a function satisfying this condition is called '''log supermodular''') i.e.,
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| :<math>\mu(x\wedge y)\mu(x\vee y) \ge \mu(x)\mu(y)</math>
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| for all ''x'', ''y'' in the lattice <math>X</math>.
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| The FKG inequality then says that for any two monotonically increasing functions ''ƒ'' and ''g'' on <math>X</math>, the following positive correlation inequality holds:
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| :<math> \Big(\sum _{x\in X}f(x)g(x)\mu(x)\Big)\Big(\sum _{x\in X}\mu(x)\Big) \ge \Big(\sum _{x\in X}f(x)\mu(x)\Big)\Big(\sum _{x\in X}g(x)\mu(x)\Big).</math>
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| The same inequality (positive correlation) is true when both ''ƒ'' and ''g'' are decreasing. If one is increasing and the other is decreasing, then they are negatively correlated and the above inequality is reversed.
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| Similar statements hold more generally, when <math>X</math> is not necessarily finite, not even countable. In that case, ''μ'' has to be a finite measure, and the lattice condition has to be defined using [[cylinder (algebra)|cylinder]] events; see, e.g., Section 2.2 of {{harvtxt|Grimmett|1999}}.
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| For proofs, see the original {{harvtxt|Fortuin|Kasteleyn|Ginibre|1971}} or the [[Ahlswede–Daykin inequality|Ahlswede–Daykin inequality (1978)]]. Also, a rough sketch is given below, due to {{harvtxt|Holley|1974}}, using a [[Markov chain]] [[coupling (probability)|coupling]] argument.
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| ==Variations on terminology==
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| The lattice condition for ''μ'' is also called '''multivariate total positivity''', and sometimes the '''strong FKG condition'''; the term ('''multiplicative''') '''FKG condition''' is also used in older literature.
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| The property of ''μ'' that increasing functions are positively correlated is also called having '''positive associations''', or the '''weak FKG condition'''.
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| Thus, the FKG theorem can be rephrased as "the strong FKG condition implies the weak FKG condition".
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| ==A special case: the Harris inequality==
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| If the lattice <math>X</math> is [[totally ordered]], then the lattice condition is satisfied trivially for any measure ''μ''. For this case, the FKG inequality is [[Chebyshev's sum inequality]]: if the two increasing functions take on values <math>a_1\leq a_2 \leq \cdots \leq a_n</math> and <math>b_1\leq b_2 \leq \cdots \leq b_n</math>, then (we may assume that the measure ''μ'' is uniform)
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| :<math>\frac{a_1b_1+\cdots+a_nb_n}{n} \geq \frac{a_1+\cdots+a_n}{n} \; \frac{b_1+\cdots+b_n}{n}.</math>
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| More generally, for any probability measure ''μ'' on <math>\R</math> and increasing functions ''ƒ'' and ''g'',
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| :<math> \int_\R f(x)g(x) \,d\mu(x) \geq \int_\R f(x)\,d\mu(x) \, \int_\R g(x)\,d\mu(x),</math>
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| which follows immediately from
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| :<math>\int_\R\int_\R [f(x)-f(y)][g(x)-g(y)]\,d\mu(x)\,d\mu(y) \geq 0.</math>
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| The lattice condition is trivially satisfied also when the lattice is the product of totally ordered lattices, <math>X=X_1\times\cdots\times X_n</math>, and <math>\mu=\mu_1\otimes\cdots\otimes\mu_n</math> is a product measure. Often all the factors (both the lattices and the measures) are identical, i.e., ''μ'' is the probability distribution of [[i.i.d.]] random variables.
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| The FKG inequality for the case of a product measure is known also as the '''Harris inequality''' after [[Ted Harris (mathematician)|Harris]] {{harv|Harris|1960}}, who found and used it in his study of [[percolation theory|percolation]] in the plane. A proof of the Harris inequality that uses the above double integral trick on <math>\R</math> can be found, e.g., in Section 2.2 of {{harvtxt|Grimmett|1999}}.
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| ===Simple examples===
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| A typical example is the following. Color each hexagon of the infinite [[honeycomb lattice]] black with probability <math>p</math> and white with probability <math>1-p</math>, independently of each other. Let ''a, b, c, d'' be four hexagons, not necessarily distinct. Let <math>a \leftrightarrow b</math> and <math>c\leftrightarrow d</math> be the events that there is a black path from ''a'' to ''b'', and a black path from ''c'' to ''d'', respectively. Then the Harris inequality says that these events are positively correlated: <math>\Pr(a \leftrightarrow b,\ c\leftrightarrow d) \geq \Pr(a \leftrightarrow b)\Pr(c\leftrightarrow d)</math>. In other words, assuming the presence of one path can only increase the probability of the other.
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| Similarly, if we randomly color the hexagons inside an <math>n\times n</math> rhombus-shaped [[hex (board game)|hex board]], then the events that there is black crossing from the left side of the board to the right side is positively correlated with having a black crossing from the top side to the bottom. On the other hand, having a left-to-right black crossing is negatively correlated with having a top-to-bottom white crossing, since the first is an increasing event (in the amount of blackness), while the second is decreasing. In fact, in any coloring of the hex board exactly one of these two events happen — this is why hex is a well-defined game.
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| In the [[Erdos–Renyi model|Erdős–Rényi random graph]], the existence of a [[Hamiltonian cycle]] is negatively correlated with the [[graph coloring|3-colorability of the graph]], since the first is an increasing event, while the latter is decreasing.
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| ==Beyond product measures==
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| In statistical mechanics, the usual source of measures that satisfy the lattice condition (and hence the FKG inequality) is the following:
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| If <math>S</math> is an ordered set (such as <math>\{-1,+1\}</math>), and <math>\Gamma</math> is a finite or infinite [[graph (mathematics)|graph]], then the set <math>S^\Gamma</math> of <math>S</math>-valued configurations is a [[poset]] that is a distributive lattice.
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| Now, if <math>\Phi</math> is a '''submodular [[Gibbs measure|potential]]''' (i.e., a family of functions
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| :<math>\Phi_\Lambda: S^\Lambda \longrightarrow \R\cup\{\infty\},</math> | |
| one for each finite <math>\Lambda \subset \Gamma</math>, such that each <math>\Phi_\Lambda</math> is [[submodular]]), then one defines the corresponding [[Gibbs measure|Hamiltonian]]s as
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| :<math>H_\Lambda(\phi):=\sum_{\Delta\cap\Lambda\not=\emptyset} \Phi_\Delta(\phi).</math>
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| If ''μ'' is an [[Gibbs measure|extremal Gibbs measure]] for this Hamiltonian on the set of configurations <math>\phi</math>, then it is easy to show that ''μ'' satisfies the lattice condition, see {{harvtxt|Sheffield|2005}}.
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| A key example is the [[Ising model]] on a graph <math>\Gamma</math>. Let <math>S=\{-1,+1\}</math>, called spins, and <math>\beta\in [0,\infty]</math>. Take the following potential:
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| :<math>\Phi_\Lambda(\phi)=\begin{cases}
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| \beta 1_{\{\phi(x)\not=\phi(y)\}} & \text{ if }\Lambda=\{x,y\}\text{ is a pair of adjacent vertices of }\Gamma;\\
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| 0 & \text{ otherwise.}\end{cases}
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| </math>
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| Submodularity is easy to check; intuitively, taking the min or the max of two configurations tends to decrease the number of disagreeing spins. Then, depending on the graph <math>\Gamma</math> and the value of <math>\beta</math>, there could be one or more extremal Gibbs measures, see, e.g., {{harvtxt|Georgii|Häggström|Maes|2001}} and {{harvtxt|Lyons|2000}}.
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| ==A generalization: the Holley inequality==
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| The '''Holley inequality''', due to {{harvs|last=Holley|first=Richard|year=1974|txt}}, states that the [[expected value|expectations]]
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| :<math> \langle f\rangle_i = \frac{\sum _{x\in X}f(x)\mu_i(x)}{\sum_{x\in X}\mu_i(x)} </math> | |
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| of a monotonically increasing function ''ƒ'' on a finite [[distributive lattice]] <math>X</math> with respect to two positive functions ''μ''<sub>1</sub>, ''μ''<sub>2</sub> on the lattice satisfy the condition
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| :<math> \langle f\rangle_1 \ge \langle f\rangle_2, </math>
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| provided the functions satisfy the '''Holley condition''' ('''criterion''')
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| :<math>\mu_2(x\wedge y)\mu_1(x\vee y) \ge \mu_1(x)\mu_2(y)</math>
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| for all ''x'', ''y'' in the lattice.
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| To recover the [[#The inequality|FKG inequality]]: If ''μ'' satisfies the lattice condition and ''ƒ'' and ''g'' are increasing functions on <math>X</math>, then ''μ''<sub>1</sub>(''x'')=''g''(''x'')''μ''(''x'') and ''μ''<sub>2</sub>(''x'')=''μ''(''x'') will satisfy the lattice-type condition of the Holley inequality. Then the Holley inequality states that
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| :<math> \frac{ \langle fg\rangle_\mu }{\langle g\rangle_\mu} = \langle f\rangle_1 \ge \langle f\rangle_2 =\langle f\rangle_\mu, </math> | |
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| which is just the FKG inequality.
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| As for FKG, the Holley inequality follows from the [[Ahlswede–Daykin inequality]].
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| ==Weakening the lattice condition: monotonicity== | |
| Consider the usual case of <math>X</math> being a product <math>\R^V</math> for some finite set <math>V</math>. The lattice condition on ''μ'' is easily seen to imply the following '''monotonicity''', which has the virtue that it is often easier to check than the lattice condition:
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| Whenever one fixes a vertex <math>v \in V</math> and two configurations ''φ'' and ''ψ'' outside ''v'' such that <math>\phi(w) \geq \psi(w)</math> for all <math>w\not=v</math>, the ''μ''-conditional distribution of ''φ(v)'' given <math>\{\phi(w) : w\not=v\}</math> [[stochastic ordering|stochastically dominates]] the ''μ''-conditional distribution of ''ψ(v)'' given <math>\{\psi(w) : w\not=v\}</math>.
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| Now, if ''μ'' satisfies this monotonicity property, that is already enough for the FKG inequality (positive associations) to hold.
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| Here is a rough sketch of the proof, due to {{harvtxt|Holley|1974}}: starting from any initial configuration on <math>V</math>, one can run a simple [[Markov chain]] (the [[Metropolis algorithm]]) that uses independent Uniform[0,1] random variables to update the configuration in each step, such that the chain has a unique stationary measure, the given ''μ''. The monotonicity of ''μ'' implies that the configuration at each step is a monotone function of independent variables, hence the [[#A special case: the Harris inequality|product measure version of Harris]] implies that it has positive associations. Therefore, the limiting stationary measure ''μ'' also has this property.
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| The monotonicity property has a natural version for two measures, saying that ''μ''<sub>1</sub> conditionally pointwise dominates ''μ''<sub>2</sub>. It is again easy to see that if ''μ''<sub>1</sub> and ''μ''<sub>2</sub> satisfy the lattice-type condition of the [[#A generalization: the Holley inequality|Holley inequality]], then ''μ''<sub>1</sub> conditionally pointwise dominates ''μ''<sub>2</sub>. On the other hand, a Markov chain [[coupling (probability)|coupling]] argument similar to the above, but now without invoking the Harris inequality, shows that conditional pointwise domination, in fact, implies [[stochastic ordering|stochastically domination]]. Stochastic domination is equivalent to saying that <math> \langle f\rangle_1 \ge \langle f\rangle_2</math> for all increasing ''ƒ'', thus we get a proof of the Holley inequality. (And thus also a proof of the FKG inequality, without using the Harris inequality.)
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| See {{harvtxt|Holley|1974}} and {{harvtxt|Georgii|Häggström|Maes|2001}} for details.
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| ==See also==
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| *[[Ahlswede–Daykin inequality]]
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| ==References==
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| *{{springer|id=f/f110120|first=P.C.|last= Fishburn}}
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| *{{Citation | last1=Fortuin | first1=C. M. | last2=Kasteleyn | first2=P. W. | last3=Ginibre | first3=J. | title=Correlation inequalities on some partially ordered sets | url=http://projecteuclid.org/euclid.cmp/1103857443 | mr=0309498 | year=1971 | journal=Communications in Mathematical Physics | volume=22 | pages=89–103 | doi=10.1007/BF01651330 | issue=2}}
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| *{{Citation | first1=H-O.|last1=Georgii| last2=Häggström| first2=O.| last3=Maes| first3=C.| chapter=The random geometry of equilibrium phases| arxiv=math/9905031| year=2001|title=[[Phase transitions and critical phenomena]] |volume=18|pages=1–142|publisher=Academic Press, San Diego, CA|mr=2014387}}
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| *{{Citation | last=Grimmett | first=G. R. | title=Percolation. Second edition| publisher=Springer-Verlag | year=1999|mr=1707339| isbn=3-540-64902-6}}
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| *{{Citation | last=Harris | first=T. E. | title=A lower bound for the critical probability in a certain percolation | year=1960 | journal=Proceedings of the Cambridge Philosophical Society | volume=56 | pages=13–20|mr=0115221 | doi=10.1017/S0305004100034241}}
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| *{{Citation | last1=Holley | first1=R. | title=Remarks on the FKG inequalities | url=http://projecteuclid.org/euclid.cmp/1103859732 | mr=0341552 | year=1974 | journal=Communications in Mathematical Physics | volume=36 | pages=227–231 | doi=10.1007/BF01645980 | issue=3}}
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| *{{citation|last=Lyons|first=R.|year=2000|title=Phase transitions on nonamenable graphs|arxiv=math/9908177|journal=J. Math. Phys.|volume=41|pages=1099–1126|mr=1757952|doi=10.1063/1.533179|issue=3}}
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| *{{Citation|last=Sheffield|first=S.|title=Random surfaces|arxiv=math/0304049|journal=Asterisque|volume=304|year=2005| mr=2251117}}
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| {{DEFAULTSORT:Fkg Inequality}}
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| [[Category:Inequalities]]
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| [[Category:Statistical mechanics]]
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| [[Category:Covariance and correlation]]
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Russell Wilson, the 5'11, 25 year old Seattle Seahawks quarterback who just won the Superbowl, loves to tell the story of how his dad pulled him aside when he was a kid struggling to make teams because he was so small.
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Wilson says those words became the driving force that kept him competing for starting jobs on college teams and kept him working hard to get drafted in the NFL. On draft day, he watched as other quarterbacks got picked in the first round, and then the second round.
The announcers who covered Wilson back then all said that if he were only 3 inches taller he would be a first round draft pick. Finally, with the 12th pick in the third round, the Seahawks drafted the undersized quarterback.
When Wilson showed up for training camp, it was almost certain he would sit on the bench behind highly prized free agent, Matt Flynn. Matt had been Aaron Rodgers' back up in Green Bay and had a franchise record setting, 6 touchdown performance at the end of the previous season. Millions of dollars later, Matt was projected to be the Seahawk's starter for years to come.
As Wilson started training camp, he heard his coaches say that the starting quarterback spot was open to the best player at that position.
"Why not you?"
These words rang in Wilson's head again as he practiced, competed and performed during off season workouts and throughout the preseason. By the final game, Wilson was named the starter, and he went on to have an amazing 2012 rookie season.
As the 2013 ugg boots outlet season began, Wilson had a message for his team: "Why not us?" Every player bought into those words and that became their team mantra as they went on to a dominating 13-3 season. In February as they headed out of the tunnel to start Superbowl XLVIII, they all believed that no matter discount ugg boots who their opponent was, there was no reason they couldn't be hoisting the Lombardi trophy at the end of the game.
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