Probability that we choose even edges for a vertex












0















We choose each edge in a graph $G$ (with $m$ edges) at random independently with probability $pin[0,1]$. What is a probability that we chose an even number of edges at the vertex $v$ which has degree $k$.




Let $X$ be a number of chosen edges at $v$ and let $q=1-p$.
$$begin{eqnarray}
P(X=;{is; even}) &=& P(X=0)+P(X=2)+P(X=4)+P(X=6)...\
&=& {kchoose 0}p^0q^k +{kchoose 2}p^2q^{k-2}+{kchoose 4}p^4q^{k-4}+...\
&=& {1over 2}Big( (q+p)^k+(q-p)^kBig)\
&=& {1over 2}Big( 1+(1-2p)^kBig)
end{eqnarray}$$
Edit: Thanks to Henry I will continue with my second question right here.




Prove that probability that we choose at each vertex odd number of edges if a graph $G$ is connected and with $2n$ vertices is nonzero.




$P(G ;is ;''odd'') = 1-P(G;is ;not "odd") $



$$begin{eqnarray}
P(G;is;not;"odd") &=& P(X_1 = even;or;X_2=even ;or; X_3=even...)\
&leq & P(X_1 = even)+P(X_2=even) + P(X_3=even)+...\
&=& {1over 2}sum_{i=1}^{2n} Big( 1+(1-2p)^{d_i}Big)
end{eqnarray}$$
where $d_i$ is a degree of vertex $v_i$. I'm stuck here, don't know how to show that this is (if it is) smaller than $1$ for some sutabile $p$. I know, that since $G$ is connected we have $mgeq 2n-1$.






probability graph-theory probabilistic-method







share|cite|improve this question




















  • 2




    Apart from a missing $P(X=0)$ at the start (now edited in), this looks sensible
    – Henry
    Dec 27 '18 at 23:19












  • What range is $p$ in? If $p le 1/2$ then the sum is at least $n$...
    – Sandeep Silwal
    Dec 28 '18 at 1:31










  • Your second question appears to be equivalent to your earlier question from a few days ago: math.stackexchange.com/questions/3052416 (That is, assuming $p ne 0,1$. If $p=0$ or $p=1$ the probability might be $0$.)
    – Misha Lavrov
    Dec 28 '18 at 3:02


















0















We choose each edge in a graph $G$ (with $m$ edges) at random independently with probability $pin[0,1]$. What is a probability that we chose an even number of edges at the vertex $v$ which has degree $k$.




Let $X$ be a number of chosen edges at $v$ and let $q=1-p$.
$$begin{eqnarray}
P(X=;{is; even}) &=& P(X=0)+P(X=2)+P(X=4)+P(X=6)...\
&=& {kchoose 0}p^0q^k +{kchoose 2}p^2q^{k-2}+{kchoose 4}p^4q^{k-4}+...\
&=& {1over 2}Big( (q+p)^k+(q-p)^kBig)\
&=& {1over 2}Big( 1+(1-2p)^kBig)
end{eqnarray}$$
Edit: Thanks to Henry I will continue with my second question right here.




Prove that probability that we choose at each vertex odd number of edges if a graph $G$ is connected and with $2n$ vertices is nonzero.




$P(G ;is ;''odd'') = 1-P(G;is ;not "odd") $



$$begin{eqnarray}
P(G;is;not;"odd") &=& P(X_1 = even;or;X_2=even ;or; X_3=even...)\
&leq & P(X_1 = even)+P(X_2=even) + P(X_3=even)+...\
&=& {1over 2}sum_{i=1}^{2n} Big( 1+(1-2p)^{d_i}Big)
end{eqnarray}$$
where $d_i$ is a degree of vertex $v_i$. I'm stuck here, don't know how to show that this is (if it is) smaller than $1$ for some sutabile $p$. I know, that since $G$ is connected we have $mgeq 2n-1$.






probability graph-theory probabilistic-method







share|cite|improve this question




















  • 2




    Apart from a missing $P(X=0)$ at the start (now edited in), this looks sensible
    – Henry
    Dec 27 '18 at 23:19












  • What range is $p$ in? If $p le 1/2$ then the sum is at least $n$...
    – Sandeep Silwal
    Dec 28 '18 at 1:31










  • Your second question appears to be equivalent to your earlier question from a few days ago: math.stackexchange.com/questions/3052416 (That is, assuming $p ne 0,1$. If $p=0$ or $p=1$ the probability might be $0$.)
    – Misha Lavrov
    Dec 28 '18 at 3:02
















0












0








0


1






We choose each edge in a graph $G$ (with $m$ edges) at random independently with probability $pin[0,1]$. What is a probability that we chose an even number of edges at the vertex $v$ which has degree $k$.




Let $X$ be a number of chosen edges at $v$ and let $q=1-p$.
$$begin{eqnarray}
P(X=;{is; even}) &=& P(X=0)+P(X=2)+P(X=4)+P(X=6)...\
&=& {kchoose 0}p^0q^k +{kchoose 2}p^2q^{k-2}+{kchoose 4}p^4q^{k-4}+...\
&=& {1over 2}Big( (q+p)^k+(q-p)^kBig)\
&=& {1over 2}Big( 1+(1-2p)^kBig)
end{eqnarray}$$
Edit: Thanks to Henry I will continue with my second question right here.




Prove that probability that we choose at each vertex odd number of edges if a graph $G$ is connected and with $2n$ vertices is nonzero.




$P(G ;is ;''odd'') = 1-P(G;is ;not "odd") $



$$begin{eqnarray}
P(G;is;not;"odd") &=& P(X_1 = even;or;X_2=even ;or; X_3=even...)\
&leq & P(X_1 = even)+P(X_2=even) + P(X_3=even)+...\
&=& {1over 2}sum_{i=1}^{2n} Big( 1+(1-2p)^{d_i}Big)
end{eqnarray}$$
where $d_i$ is a degree of vertex $v_i$. I'm stuck here, don't know how to show that this is (if it is) smaller than $1$ for some sutabile $p$. I know, that since $G$ is connected we have $mgeq 2n-1$.






probability graph-theory probabilistic-method







share|cite|improve this question
















We choose each edge in a graph $G$ (with $m$ edges) at random independently with probability $pin[0,1]$. What is a probability that we chose an even number of edges at the vertex $v$ which has degree $k$.




Let $X$ be a number of chosen edges at $v$ and let $q=1-p$.
$$begin{eqnarray}
P(X=;{is; even}) &=& P(X=0)+P(X=2)+P(X=4)+P(X=6)...\
&=& {kchoose 0}p^0q^k +{kchoose 2}p^2q^{k-2}+{kchoose 4}p^4q^{k-4}+...\
&=& {1over 2}Big( (q+p)^k+(q-p)^kBig)\
&=& {1over 2}Big( 1+(1-2p)^kBig)
end{eqnarray}$$
Edit: Thanks to Henry I will continue with my second question right here.




Prove that probability that we choose at each vertex odd number of edges if a graph $G$ is connected and with $2n$ vertices is nonzero.




$P(G ;is ;''odd'') = 1-P(G;is ;not "odd") $



$$begin{eqnarray}
P(G;is;not;"odd") &=& P(X_1 = even;or;X_2=even ;or; X_3=even...)\
&leq & P(X_1 = even)+P(X_2=even) + P(X_3=even)+...\
&=& {1over 2}sum_{i=1}^{2n} Big( 1+(1-2p)^{d_i}Big)
end{eqnarray}$$
where $d_i$ is a degree of vertex $v_i$. I'm stuck here, don't know how to show that this is (if it is) smaller than $1$ for some sutabile $p$. I know, that since $G$ is connected we have $mgeq 2n-1$.






probability graph-theory probabilistic-method




probability graph-theory probabilistic-method






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edited Dec 28 '18 at 12:34

























asked Dec 27 '18 at 23:11









greedoid

38.2k114797




38.2k114797








  • 2




    Apart from a missing $P(X=0)$ at the start (now edited in), this looks sensible
    – Henry
    Dec 27 '18 at 23:19












  • What range is $p$ in? If $p le 1/2$ then the sum is at least $n$...
    – Sandeep Silwal
    Dec 28 '18 at 1:31










  • Your second question appears to be equivalent to your earlier question from a few days ago: math.stackexchange.com/questions/3052416 (That is, assuming $p ne 0,1$. If $p=0$ or $p=1$ the probability might be $0$.)
    – Misha Lavrov
    Dec 28 '18 at 3:02
















  • 2




    Apart from a missing $P(X=0)$ at the start (now edited in), this looks sensible
    – Henry
    Dec 27 '18 at 23:19












  • What range is $p$ in? If $p le 1/2$ then the sum is at least $n$...
    – Sandeep Silwal
    Dec 28 '18 at 1:31










  • Your second question appears to be equivalent to your earlier question from a few days ago: math.stackexchange.com/questions/3052416 (That is, assuming $p ne 0,1$. If $p=0$ or $p=1$ the probability might be $0$.)
    – Misha Lavrov
    Dec 28 '18 at 3:02










2




2




Apart from a missing $P(X=0)$ at the start (now edited in), this looks sensible
– Henry
Dec 27 '18 at 23:19






Apart from a missing $P(X=0)$ at the start (now edited in), this looks sensible
– Henry
Dec 27 '18 at 23:19














What range is $p$ in? If $p le 1/2$ then the sum is at least $n$...
– Sandeep Silwal
Dec 28 '18 at 1:31




What range is $p$ in? If $p le 1/2$ then the sum is at least $n$...
– Sandeep Silwal
Dec 28 '18 at 1:31












Your second question appears to be equivalent to your earlier question from a few days ago: math.stackexchange.com/questions/3052416 (That is, assuming $p ne 0,1$. If $p=0$ or $p=1$ the probability might be $0$.)
– Misha Lavrov
Dec 28 '18 at 3:02






Your second question appears to be equivalent to your earlier question from a few days ago: math.stackexchange.com/questions/3052416 (That is, assuming $p ne 0,1$. If $p=0$ or $p=1$ the probability might be $0$.)
– Misha Lavrov
Dec 28 '18 at 3:02












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Unfortunately, the last sum is not necessarily less than $1$.



Say we have the line graph with $2n$ vertices. Then two vertices has a degree $1$ and the other $2$. So we have: $${1over 2}sum_{i=1}^{2n} Big( 1+(1-2p)^{d_i}Big) = (n-1)x^2+x+n =:f(x)$$



where $x = 1-2p$. But $f(x)$ can not be $<1$ since the inequality $ f(x)<1 $ is equivalent to $$ (n-1)x^2+x+(n-1)<0$$ and the discirminat is $1-4(n-1)^2<0$.






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    Unfortunately, the last sum is not necessarily less than $1$.



    Say we have the line graph with $2n$ vertices. Then two vertices has a degree $1$ and the other $2$. So we have: $${1over 2}sum_{i=1}^{2n} Big( 1+(1-2p)^{d_i}Big) = (n-1)x^2+x+n =:f(x)$$



    where $x = 1-2p$. But $f(x)$ can not be $<1$ since the inequality $ f(x)<1 $ is equivalent to $$ (n-1)x^2+x+(n-1)<0$$ and the discirminat is $1-4(n-1)^2<0$.






    share|cite|improve this answer


























      0














      Unfortunately, the last sum is not necessarily less than $1$.



      Say we have the line graph with $2n$ vertices. Then two vertices has a degree $1$ and the other $2$. So we have: $${1over 2}sum_{i=1}^{2n} Big( 1+(1-2p)^{d_i}Big) = (n-1)x^2+x+n =:f(x)$$



      where $x = 1-2p$. But $f(x)$ can not be $<1$ since the inequality $ f(x)<1 $ is equivalent to $$ (n-1)x^2+x+(n-1)<0$$ and the discirminat is $1-4(n-1)^2<0$.






      share|cite|improve this answer
























        0












        0








        0






        Unfortunately, the last sum is not necessarily less than $1$.



        Say we have the line graph with $2n$ vertices. Then two vertices has a degree $1$ and the other $2$. So we have: $${1over 2}sum_{i=1}^{2n} Big( 1+(1-2p)^{d_i}Big) = (n-1)x^2+x+n =:f(x)$$



        where $x = 1-2p$. But $f(x)$ can not be $<1$ since the inequality $ f(x)<1 $ is equivalent to $$ (n-1)x^2+x+(n-1)<0$$ and the discirminat is $1-4(n-1)^2<0$.






        share|cite|improve this answer












        Unfortunately, the last sum is not necessarily less than $1$.



        Say we have the line graph with $2n$ vertices. Then two vertices has a degree $1$ and the other $2$. So we have: $${1over 2}sum_{i=1}^{2n} Big( 1+(1-2p)^{d_i}Big) = (n-1)x^2+x+n =:f(x)$$



        where $x = 1-2p$. But $f(x)$ can not be $<1$ since the inequality $ f(x)<1 $ is equivalent to $$ (n-1)x^2+x+(n-1)<0$$ and the discirminat is $1-4(n-1)^2<0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '18 at 18:52









        greedoid

        38.2k114797




        38.2k114797






























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