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Let $G(n,m)$ be the probability space of all graphs on $n$ vertices, having $m$ edges, where all graphs are equi-probable. Can the distribution of the degree of a vertex be approximated by the Poisson distribution ?

In $G(n,p)$ (assuming $p$ is proportional to $\frac{1}{n}$) it's simple. Since a vertex $v$ has $n-1$ potential incident edges, and since each edge is chosen independently with probability $p$, the degree is a Binomial r.v., $B(n-1,p)$, which is known to be approximated by Poisson with parameter $\lambda$, assuming that $(n-1)p \to \lambda$.

In $G(n,m)$, the probability for choosing each of the $n-1$ incident edges is $\frac{m}{n\choose 2}$, but they're not independent. Is it still the case that the distribution approaches Poisson, as $n\to \infty$ ?

Teddy
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2 Answers2

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The standard way of proving that a numerical random graph property $\mathbf X$ is asymptotically Poisson is the method of moments: show that for all constant $k$, $$ \mathbb E\left[\binom{\mathbf X}{k}\right] \sim \frac{\lambda^k}{k!} $$ as $n \to \infty$.

In this case, the calculation boils down to showing that any fixed set of $k$ edges out of $v$ is present with probability asymptotically $(m/\binom n2)^k$, as opposed to $p^k$ in the binomial random graph. This holds because the exact probability (taking $N = \binom n2$) is $$ \frac{\binom{N-k}{m-k}}{\binom{N}{m}} = \frac{(N-k)!/(m-k)!}{N!/m!} = \frac{m(m-1)\dotsb(m-k+1)}{N(N-1)\dotsb(N-k+1)} $$ which is $(1+o(1))\frac{m^k}{N^k}$ because $k$ is constant and $N, m \to \infty$ with $n$.

From there, everything is straightforward: $$ \mathbb E\left[\binom{\mathbf X}{k}\right] \sim \binom nk (m/N)^k \sim \frac{n^k}{k!} (m/N)^k = \frac{(mn/N)^k}{k!} $$ and in the case corresponding to $p = \frac cn$, we should have $mn/N$ approach a constant in the limit.

Misha Lavrov
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So your question boils down to approximating the hypergeometric distribution with the Poisson distribution. See this discussion.

kimchi lover
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