Probability of subset of a graph being stable
$begingroup$
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
asked Dec 1 '18 at 18:46
orangeorange
620215
$endgroup$
add a comment |
$begingroup$
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
asked Dec 1 '18 at 18:46
orangeorange
620215
$endgroup$
$begingroup$
What is a definiton of stable?
$endgroup$
– greedoid
Dec 1 '18 at 18:46
$begingroup$
I have added it and it is in the computation
$endgroup$
– orange
Dec 1 '18 at 18:47
add a comment |
$begingroup$
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
asked Dec 1 '18 at 18:46
orangeorange
620215
$endgroup$
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
asked Dec 1 '18 at 18:46
orangeorange
620215
asked Dec 1 '18 at 18:46
orangeorange
620215
edited Dec 1 '18 at 18:53
orange
asked Dec 1 '18 at 18:46
orangeorange
620215
asked Dec 1 '18 at 18:46
orangeorange
620215
asked Dec 1 '18 at 18:46
orangeorange
620215
620215
$begingroup$
What is a definiton of stable?
$endgroup$
– greedoid
Dec 1 '18 at 18:46
$begingroup$
I have added it and it is in the computation
$endgroup$
– orange
Dec 1 '18 at 18:47
add a comment |
$begingroup$
What is a definiton of stable?
$endgroup$
– greedoid
Dec 1 '18 at 18:46
$begingroup$
I have added it and it is in the computation
$endgroup$
– orange
Dec 1 '18 at 18:47
$begingroup$
What is a definiton of stable?
$endgroup$
– greedoid
Dec 1 '18 at 18:46
$begingroup$
What is a definiton of stable?
$endgroup$
– greedoid
Dec 1 '18 at 18:46
$begingroup$
I have added it and it is in the computation
$endgroup$
– orange
Dec 1 '18 at 18:47
$begingroup$
I have added it and it is in the computation
$endgroup$
– orange
Dec 1 '18 at 18:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.)
answered Dec 1 '18 at 19:01
Misha LavrovMisha Lavrov
44.7k556107
$endgroup$
$begingroup$
Very simple and clear. Thank you.
$endgroup$
– orange
Dec 1 '18 at 19:07
$begingroup$
Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
$endgroup$
– orange
Dec 1 '18 at 19:24
$begingroup$
I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 20:38
$begingroup$
Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
$endgroup$
– orange
Dec 1 '18 at 20:40
$begingroup$
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 $bigcap_{vw in E} A_{vw}$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 22:22
|
show 2 more comments
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
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active
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$begingroup$
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.)
answered Dec 1 '18 at 19:01
Misha LavrovMisha Lavrov
44.7k556107
$endgroup$
$begingroup$
Very simple and clear. Thank you.
$endgroup$
– orange
Dec 1 '18 at 19:07
$begingroup$
Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
$endgroup$
– orange
Dec 1 '18 at 19:24
$begingroup$
I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 20:38
$begingroup$
Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
$endgroup$
– orange
Dec 1 '18 at 20:40
$begingroup$
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 $bigcap_{vw in E} A_{vw}$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 22:22
|
show 2 more comments
$begingroup$
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.)
answered Dec 1 '18 at 19:01
Misha LavrovMisha Lavrov
44.7k556107
$endgroup$
$begingroup$
Very simple and clear. Thank you.
$endgroup$
– orange
Dec 1 '18 at 19:07
$begingroup$
Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
$endgroup$
– orange
Dec 1 '18 at 19:24
$begingroup$
I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 20:38
$begingroup$
Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
$endgroup$
– orange
Dec 1 '18 at 20:40
$begingroup$
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 $bigcap_{vw in E} A_{vw}$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 22:22
|
show 2 more comments
$begingroup$
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.)
answered Dec 1 '18 at 19:01
Misha LavrovMisha Lavrov
44.7k556107
$endgroup$
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.)
answered Dec 1 '18 at 19:01
Misha LavrovMisha Lavrov
44.7k556107
edited Dec 1 '18 at 20:40
answered Dec 1 '18 at 19:01
Misha LavrovMisha Lavrov
44.7k556107
answered Dec 1 '18 at 19:01
Misha LavrovMisha Lavrov
44.7k556107
answered Dec 1 '18 at 19:01
Misha LavrovMisha Lavrov
44.7k556107
44.7k556107
$begingroup$
Very simple and clear. Thank you.
$endgroup$
– orange
Dec 1 '18 at 19:07
$begingroup$
Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
$endgroup$
– orange
Dec 1 '18 at 19:24
$begingroup$
I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 20:38
$begingroup$
Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
$endgroup$
– orange
Dec 1 '18 at 20:40
$begingroup$
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 $bigcap_{vw in E} A_{vw}$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 22:22
|
show 2 more comments
$begingroup$
Very simple and clear. Thank you.
$endgroup$
– orange
Dec 1 '18 at 19:07
$begingroup$
Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
$endgroup$
– orange
Dec 1 '18 at 19:24
$begingroup$
I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 20:38
$begingroup$
Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
$endgroup$
– orange
Dec 1 '18 at 20:40
$begingroup$
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 $bigcap_{vw in E} A_{vw}$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 22:22
$begingroup$
Very simple and clear. Thank you.
$endgroup$
– orange
Dec 1 '18 at 19:07
$begingroup$
Very simple and clear. Thank you.
$endgroup$
– orange
Dec 1 '18 at 19:07
$begingroup$
Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
$endgroup$
– orange
Dec 1 '18 at 19:24
$begingroup$
Quick clarification, does $bigwedge_{vw in E}$ mean $bigcap_{vw in E}$ or something else in the language of lattice theory or something?
$endgroup$
– orange
Dec 1 '18 at 19:24
$begingroup$
I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 20:38
$begingroup$
I mean a logical "and" ($land$) of the statements ${v,w} notsubseteq S$ over all $vw in E$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 20:38
$begingroup$
Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
$endgroup$
– orange
Dec 1 '18 at 20:40
$begingroup$
Equivalently $bigcap_{vwin E}{{v,w}notsubseteq S}$
$endgroup$
– orange
Dec 1 '18 at 20:40
$begingroup$
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 $bigcap_{vw in E} A_{vw}$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 22:22
$begingroup$
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 $bigcap_{vw in E} A_{vw}$.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 22:22
|
show 2 more comments
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$begingroup$
What is a definiton of stable?
$endgroup$
– greedoid
Dec 1 '18 at 18:46
$begingroup$
I have added it and it is in the computation
$endgroup$
– orange
Dec 1 '18 at 18:47