Probability of subset of a graph being stable












2














Let $G=(V,E)$ be a graph with $|E|=m$. Let $Ssubseteq V$ such that $Pr(vin S)=frac{1}{2}$ for all $vin V$ and the events are independent for all $v in V$. Show $Pr(S text{ is stable})geq left( frac{3}{4}right)^{m}$. A set is called stable if it has no edges between any of its vertices.



$Pr(S text{ is stable}) =1-Pr(E_S neq emptyset)=1-Pr(exists v_1,v_2in S:{v_1,v_2}in E)=1-Pr(text{choosing }v_1,v_2)=1-frac{1}{4}cdot N$



Where $N$ is the number of possible choices for $v_1,v_2$.



The probability of an independent set is minimized when $G$ is a complete Graph. A complete graph with $n$ vertices has $frac{n(n-1)}{2}$edges. Suppose $G$ is complete and has $m$ edges, then possible number of vertices:
$$frac{1+sqrt{1+8m}}{2},quad frac{1-sqrt{1+8m}}{2}.$$



Not sure how to continue






probability graph-theory







share|cite|improve this question
























  • What is a definiton of stable?
    – greedoid
    Dec 1 at 18:46










  • I have added it and it is in the computation
    – orange
    Dec 1 at 18:47
















2














Let $G=(V,E)$ be a graph with $|E|=m$. Let $Ssubseteq V$ such that $Pr(vin S)=frac{1}{2}$ for all $vin V$ and the events are independent for all $v in V$. Show $Pr(S text{ is stable})geq left( frac{3}{4}right)^{m}$. A set is called stable if it has no edges between any of its vertices.



$Pr(S text{ is stable}) =1-Pr(E_S neq emptyset)=1-Pr(exists v_1,v_2in S:{v_1,v_2}in E)=1-Pr(text{choosing }v_1,v_2)=1-frac{1}{4}cdot N$



Where $N$ is the number of possible choices for $v_1,v_2$.



The probability of an independent set is minimized when $G$ is a complete Graph. A complete graph with $n$ vertices has $frac{n(n-1)}{2}$edges. Suppose $G$ is complete and has $m$ edges, then possible number of vertices:
$$frac{1+sqrt{1+8m}}{2},quad frac{1-sqrt{1+8m}}{2}.$$



Not sure how to continue






probability graph-theory







share|cite|improve this question
























  • What is a definiton of stable?
    – greedoid
    Dec 1 at 18:46










  • I have added it and it is in the computation
    – orange
    Dec 1 at 18:47














2












2








2


1





Let $G=(V,E)$ be a graph with $|E|=m$. Let $Ssubseteq V$ such that $Pr(vin S)=frac{1}{2}$ for all $vin V$ and the events are independent for all $v in V$. Show $Pr(S text{ is stable})geq left( frac{3}{4}right)^{m}$. A set is called stable if it has no edges between any of its vertices.



$Pr(S text{ is stable}) =1-Pr(E_S neq emptyset)=1-Pr(exists v_1,v_2in S:{v_1,v_2}in E)=1-Pr(text{choosing }v_1,v_2)=1-frac{1}{4}cdot N$



Where $N$ is the number of possible choices for $v_1,v_2$.



The probability of an independent set is minimized when $G$ is a complete Graph. A complete graph with $n$ vertices has $frac{n(n-1)}{2}$edges. Suppose $G$ is complete and has $m$ edges, then possible number of vertices:
$$frac{1+sqrt{1+8m}}{2},quad frac{1-sqrt{1+8m}}{2}.$$



Not sure how to continue






probability graph-theory







share|cite|improve this question















Let $G=(V,E)$ be a graph with $|E|=m$. Let $Ssubseteq V$ such that $Pr(vin S)=frac{1}{2}$ for all $vin V$ and the events are independent for all $v in V$. Show $Pr(S text{ is stable})geq left( frac{3}{4}right)^{m}$. A set is called stable if it has no edges between any of its vertices.



$Pr(S text{ is stable}) =1-Pr(E_S neq emptyset)=1-Pr(exists v_1,v_2in S:{v_1,v_2}in E)=1-Pr(text{choosing }v_1,v_2)=1-frac{1}{4}cdot N$



Where $N$ is the number of possible choices for $v_1,v_2$.



The probability of an independent set is minimized when $G$ is a complete Graph. A complete graph with $n$ vertices has $frac{n(n-1)}{2}$edges. Suppose $G$ is complete and has $m$ edges, then possible number of vertices:
$$frac{1+sqrt{1+8m}}{2},quad frac{1-sqrt{1+8m}}{2}.$$



Not sure how to continue






probability graph-theory




probability graph-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 at 18:53

























asked Dec 1 at 18:46









orange

615215




615215












  • What is a definiton of stable?
    – greedoid
    Dec 1 at 18:46










  • I have added it and it is in the computation
    – orange
    Dec 1 at 18:47


















  • What is a definiton of stable?
    – greedoid
    Dec 1 at 18:46










  • I have added it and it is in the computation
    – orange
    Dec 1 at 18:47
















What is a definiton of stable?
– greedoid
Dec 1 at 18:46




What is a definiton of stable?
– greedoid
Dec 1 at 18:46












I have added it and it is in the computation
– orange
Dec 1 at 18:47




I have added it and it is in the computation
– orange
Dec 1 at 18:47










1 Answer
1






active

oldest

votes


















2














Your goal is to prove that
$$
Prleft[bigwedge_{vw in E} {v,w}notsubseteq Sright] ge prod_{vw in E} Pr[{v,w} notsubseteq S]
$$
since $Pr[{v,w} notsubseteq S]$ is just $frac34$ so the right-hand side simplifies to $(frac34)^m$. (The $bigwedge$ in the left-hand side denotes the logical AND of all the statements ${v,w} notsubseteq S$ over $vw in E$: that is, no edge has both endpoints in $S$, making $S$ a stable set.)



One way to do so is with some variant of the FKG inequality. Each event ${v,w} notsubseteq S$ is a decreasing property of $S$: if it is true for $S$, it's true for all subsets of $S$. Decreasing events are positively correlated with each other: the probability that they hold simultaneously is at least as large as it would be if they were independent. This gives us the inequality above.



(Regarding which correlation inequality to use - of the ones mentioned in the Wikipedia article, the Harris inequality seems the most appropriate, and the example given there is very similar. If you have Alon and Spencer's Probabilistic Method as a reference, then Kleitman's Lemma (Proposition 6.3.1 in Chapter 6) is the simplest result there that gives us what we want.)






share|cite|improve this answer























  • Very simple and clear. Thank you.
    – orange
    Dec 1 at 19:07










  • Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
    – orange
    Dec 1 at 19:24










  • I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
    – Misha Lavrov
    Dec 1 at 20:38












  • Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
    – orange
    Dec 1 at 20:40












  • You have the right idea, but personally I would not want to mix notations like that. Formally, of course, all events are sets of atoms in a probability space, and so we can use $bigcap$ when discussing the joint probability. However, if we are using the fiction that a logical statement such as "${v,w} notsubseteq S$" is an event, I would also use logical connectors to combine them. Alternatively, we could define $A_{vw}$ to be the event ${Ssubseteq V: {v,w} notsubseteq S}$ and then take $bigcap_{vw in E} A_{vw}$.
    – Misha Lavrov
    Dec 1 at 22:22











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2














Your goal is to prove that
$$
Prleft[bigwedge_{vw in E} {v,w}notsubseteq Sright] ge prod_{vw in E} Pr[{v,w} notsubseteq S]
$$
since $Pr[{v,w} notsubseteq S]$ is just $frac34$ so the right-hand side simplifies to $(frac34)^m$. (The $bigwedge$ in the left-hand side denotes the logical AND of all the statements ${v,w} notsubseteq S$ over $vw in E$: that is, no edge has both endpoints in $S$, making $S$ a stable set.)



One way to do so is with some variant of the FKG inequality. Each event ${v,w} notsubseteq S$ is a decreasing property of $S$: if it is true for $S$, it's true for all subsets of $S$. Decreasing events are positively correlated with each other: the probability that they hold simultaneously is at least as large as it would be if they were independent. This gives us the inequality above.



(Regarding which correlation inequality to use - of the ones mentioned in the Wikipedia article, the Harris inequality seems the most appropriate, and the example given there is very similar. If you have Alon and Spencer's Probabilistic Method as a reference, then Kleitman's Lemma (Proposition 6.3.1 in Chapter 6) is the simplest result there that gives us what we want.)






share|cite|improve this answer























  • Very simple and clear. Thank you.
    – orange
    Dec 1 at 19:07










  • Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
    – orange
    Dec 1 at 19:24










  • I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
    – Misha Lavrov
    Dec 1 at 20:38












  • Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
    – orange
    Dec 1 at 20:40












  • You have the right idea, but personally I would not want to mix notations like that. Formally, of course, all events are sets of atoms in a probability space, and so we can use $bigcap$ when discussing the joint probability. However, if we are using the fiction that a logical statement such as "${v,w} notsubseteq S$" is an event, I would also use logical connectors to combine them. Alternatively, we could define $A_{vw}$ to be the event ${Ssubseteq V: {v,w} notsubseteq S}$ and then take $bigcap_{vw in E} A_{vw}$.
    – Misha Lavrov
    Dec 1 at 22:22
















2














Your goal is to prove that
$$
Prleft[bigwedge_{vw in E} {v,w}notsubseteq Sright] ge prod_{vw in E} Pr[{v,w} notsubseteq S]
$$
since $Pr[{v,w} notsubseteq S]$ is just $frac34$ so the right-hand side simplifies to $(frac34)^m$. (The $bigwedge$ in the left-hand side denotes the logical AND of all the statements ${v,w} notsubseteq S$ over $vw in E$: that is, no edge has both endpoints in $S$, making $S$ a stable set.)



One way to do so is with some variant of the FKG inequality. Each event ${v,w} notsubseteq S$ is a decreasing property of $S$: if it is true for $S$, it's true for all subsets of $S$. Decreasing events are positively correlated with each other: the probability that they hold simultaneously is at least as large as it would be if they were independent. This gives us the inequality above.



(Regarding which correlation inequality to use - of the ones mentioned in the Wikipedia article, the Harris inequality seems the most appropriate, and the example given there is very similar. If you have Alon and Spencer's Probabilistic Method as a reference, then Kleitman's Lemma (Proposition 6.3.1 in Chapter 6) is the simplest result there that gives us what we want.)






share|cite|improve this answer























  • Very simple and clear. Thank you.
    – orange
    Dec 1 at 19:07










  • Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
    – orange
    Dec 1 at 19:24










  • I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
    – Misha Lavrov
    Dec 1 at 20:38












  • Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
    – orange
    Dec 1 at 20:40












  • You have the right idea, but personally I would not want to mix notations like that. Formally, of course, all events are sets of atoms in a probability space, and so we can use $bigcap$ when discussing the joint probability. However, if we are using the fiction that a logical statement such as "${v,w} notsubseteq S$" is an event, I would also use logical connectors to combine them. Alternatively, we could define $A_{vw}$ to be the event ${Ssubseteq V: {v,w} notsubseteq S}$ and then take $bigcap_{vw in E} A_{vw}$.
    – Misha Lavrov
    Dec 1 at 22:22














2












2








2






Your goal is to prove that
$$
Prleft[bigwedge_{vw in E} {v,w}notsubseteq Sright] ge prod_{vw in E} Pr[{v,w} notsubseteq S]
$$
since $Pr[{v,w} notsubseteq S]$ is just $frac34$ so the right-hand side simplifies to $(frac34)^m$. (The $bigwedge$ in the left-hand side denotes the logical AND of all the statements ${v,w} notsubseteq S$ over $vw in E$: that is, no edge has both endpoints in $S$, making $S$ a stable set.)



One way to do so is with some variant of the FKG inequality. Each event ${v,w} notsubseteq S$ is a decreasing property of $S$: if it is true for $S$, it's true for all subsets of $S$. Decreasing events are positively correlated with each other: the probability that they hold simultaneously is at least as large as it would be if they were independent. This gives us the inequality above.



(Regarding which correlation inequality to use - of the ones mentioned in the Wikipedia article, the Harris inequality seems the most appropriate, and the example given there is very similar. If you have Alon and Spencer's Probabilistic Method as a reference, then Kleitman's Lemma (Proposition 6.3.1 in Chapter 6) is the simplest result there that gives us what we want.)






share|cite|improve this answer














Your goal is to prove that
$$
Prleft[bigwedge_{vw in E} {v,w}notsubseteq Sright] ge prod_{vw in E} Pr[{v,w} notsubseteq S]
$$
since $Pr[{v,w} notsubseteq S]$ is just $frac34$ so the right-hand side simplifies to $(frac34)^m$. (The $bigwedge$ in the left-hand side denotes the logical AND of all the statements ${v,w} notsubseteq S$ over $vw in E$: that is, no edge has both endpoints in $S$, making $S$ a stable set.)



One way to do so is with some variant of the FKG inequality. Each event ${v,w} notsubseteq S$ is a decreasing property of $S$: if it is true for $S$, it's true for all subsets of $S$. Decreasing events are positively correlated with each other: the probability that they hold simultaneously is at least as large as it would be if they were independent. This gives us the inequality above.



(Regarding which correlation inequality to use - of the ones mentioned in the Wikipedia article, the Harris inequality seems the most appropriate, and the example given there is very similar. If you have Alon and Spencer's Probabilistic Method as a reference, then Kleitman's Lemma (Proposition 6.3.1 in Chapter 6) is the simplest result there that gives us what we want.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 1 at 20:40

























answered Dec 1 at 19:01









Misha Lavrov

43.6k555104




43.6k555104












  • Very simple and clear. Thank you.
    – orange
    Dec 1 at 19:07










  • Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
    – orange
    Dec 1 at 19:24










  • I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
    – Misha Lavrov
    Dec 1 at 20:38












  • Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
    – orange
    Dec 1 at 20:40












  • You have the right idea, but personally I would not want to mix notations like that. Formally, of course, all events are sets of atoms in a probability space, and so we can use $bigcap$ when discussing the joint probability. However, if we are using the fiction that a logical statement such as "${v,w} notsubseteq S$" is an event, I would also use logical connectors to combine them. Alternatively, we could define $A_{vw}$ to be the event ${Ssubseteq V: {v,w} notsubseteq S}$ and then take $bigcap_{vw in E} A_{vw}$.
    – Misha Lavrov
    Dec 1 at 22:22


















  • Very simple and clear. Thank you.
    – orange
    Dec 1 at 19:07










  • Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
    – orange
    Dec 1 at 19:24










  • I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
    – Misha Lavrov
    Dec 1 at 20:38












  • Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
    – orange
    Dec 1 at 20:40












  • You have the right idea, but personally I would not want to mix notations like that. Formally, of course, all events are sets of atoms in a probability space, and so we can use $bigcap$ when discussing the joint probability. However, if we are using the fiction that a logical statement such as "${v,w} notsubseteq S$" is an event, I would also use logical connectors to combine them. Alternatively, we could define $A_{vw}$ to be the event ${Ssubseteq V: {v,w} notsubseteq S}$ and then take $bigcap_{vw in E} A_{vw}$.
    – Misha Lavrov
    Dec 1 at 22:22
















Very simple and clear. Thank you.
– orange
Dec 1 at 19:07




Very simple and clear. Thank you.
– orange
Dec 1 at 19:07












Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
– orange
Dec 1 at 19:24




Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
– orange
Dec 1 at 19:24












I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
– Misha Lavrov
Dec 1 at 20:38






I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
– Misha Lavrov
Dec 1 at 20:38














Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
– orange
Dec 1 at 20:40






Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
– orange
Dec 1 at 20:40














You have the right idea, but personally I would not want to mix notations like that. Formally, of course, all events are sets of atoms in a probability space, and so we can use $bigcap$ when discussing the joint probability. However, if we are using the fiction that a logical statement such as "${v,w} notsubseteq S$" is an event, I would also use logical connectors to combine them. Alternatively, we could define $A_{vw}$ to be the event ${Ssubseteq V: {v,w} notsubseteq S}$ and then take $bigcap_{vw in E} A_{vw}$.
– Misha Lavrov
Dec 1 at 22:22




You have the right idea, but personally I would not want to mix notations like that. Formally, of course, all events are sets of atoms in a probability space, and so we can use $bigcap$ when discussing the joint probability. However, if we are using the fiction that a logical statement such as "${v,w} notsubseteq S$" is an event, I would also use logical connectors to combine them. Alternatively, we could define $A_{vw}$ to be the event ${Ssubseteq V: {v,w} notsubseteq S}$ and then take $bigcap_{vw in E} A_{vw}$.
– Misha Lavrov
Dec 1 at 22:22


















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