Expectation of degree in Bernoulli graphs
$begingroup$
Let $mathcal{G}(n,p)$ be the Bernoulli graph ditribution of $n$ verteces with edge probability $p$. It is known that the degree distribution of such graphs is the binomial distribution.
My question is the following: Let $G$ be a graph drawn from $mathcal{G}(n,p)$ and let $m(G)$ be the mean degree of the graph. Then $m(cdot)$ is a random variable on $mathcal{G}(n,p)$. Is it known what is its distribution?
random-graphs
asked Jan 21 at 22:51
tsttst
722412
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add a comment |
$begingroup$
Let $mathcal{G}(n,p)$ be the Bernoulli graph ditribution of $n$ verteces with edge probability $p$. It is known that the degree distribution of such graphs is the binomial distribution.
My question is the following: Let $G$ be a graph drawn from $mathcal{G}(n,p)$ and let $m(G)$ be the mean degree of the graph. Then $m(cdot)$ is a random variable on $mathcal{G}(n,p)$. Is it known what is its distribution?
random-graphs
asked Jan 21 at 22:51
tsttst
722412
$endgroup$
add a comment |
$begingroup$
Let $mathcal{G}(n,p)$ be the Bernoulli graph ditribution of $n$ verteces with edge probability $p$. It is known that the degree distribution of such graphs is the binomial distribution.
My question is the following: Let $G$ be a graph drawn from $mathcal{G}(n,p)$ and let $m(G)$ be the mean degree of the graph. Then $m(cdot)$ is a random variable on $mathcal{G}(n,p)$. Is it known what is its distribution?
random-graphs
asked Jan 21 at 22:51
tsttst
722412
$endgroup$
Let $mathcal{G}(n,p)$ be the Bernoulli graph ditribution of $n$ verteces with edge probability $p$. It is known that the degree distribution of such graphs is the binomial distribution.
My question is the following: Let $G$ be a graph drawn from $mathcal{G}(n,p)$ and let $m(G)$ be the mean degree of the graph. Then $m(cdot)$ is a random variable on $mathcal{G}(n,p)$. Is it known what is its distribution?
random-graphs
random-graphs
asked Jan 21 at 22:51
tsttst
722412
asked Jan 21 at 22:51
tsttst
722412
asked Jan 21 at 22:51
tsttst
722412
asked Jan 21 at 22:51
tsttst
722412
asked Jan 21 at 22:51
tsttst
722412
722412
add a comment |
add a comment |
1 Answer
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It's $frac{2}{n}$ times the binomial distribution with probability $p$ on $binom{n}{2}$ trials. Each possible edge appears with probability $p$. If it does, it contributes $2$ to the sum of degrees ($1$ for each endpoint) or $frac2n$ to the average degree.
Mean $(n-1)p$, variance $frac{2(n-1)}{n}p(1-p)$.
answered Jan 21 at 23:06
jmerryjmerry
10.6k1225
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
It's $frac{2}{n}$ times the binomial distribution with probability $p$ on $binom{n}{2}$ trials. Each possible edge appears with probability $p$. If it does, it contributes $2$ to the sum of degrees ($1$ for each endpoint) or $frac2n$ to the average degree.
Mean $(n-1)p$, variance $frac{2(n-1)}{n}p(1-p)$.
answered Jan 21 at 23:06
jmerryjmerry
10.6k1225
$endgroup$
add a comment |
$begingroup$
It's $frac{2}{n}$ times the binomial distribution with probability $p$ on $binom{n}{2}$ trials. Each possible edge appears with probability $p$. If it does, it contributes $2$ to the sum of degrees ($1$ for each endpoint) or $frac2n$ to the average degree.
Mean $(n-1)p$, variance $frac{2(n-1)}{n}p(1-p)$.
answered Jan 21 at 23:06
jmerryjmerry
10.6k1225
$endgroup$
add a comment |
$begingroup$
It's $frac{2}{n}$ times the binomial distribution with probability $p$ on $binom{n}{2}$ trials. Each possible edge appears with probability $p$. If it does, it contributes $2$ to the sum of degrees ($1$ for each endpoint) or $frac2n$ to the average degree.
Mean $(n-1)p$, variance $frac{2(n-1)}{n}p(1-p)$.
answered Jan 21 at 23:06
jmerryjmerry
10.6k1225
$endgroup$
It's $frac{2}{n}$ times the binomial distribution with probability $p$ on $binom{n}{2}$ trials. Each possible edge appears with probability $p$. If it does, it contributes $2$ to the sum of degrees ($1$ for each endpoint) or $frac2n$ to the average degree.
Mean $(n-1)p$, variance $frac{2(n-1)}{n}p(1-p)$.
answered Jan 21 at 23:06
jmerryjmerry
10.6k1225
answered Jan 21 at 23:06
jmerryjmerry
10.6k1225
answered Jan 21 at 23:06
jmerryjmerry
10.6k1225
answered Jan 21 at 23:06
jmerryjmerry
10.6k1225
10.6k1225
add a comment |
add a comment |
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