We defined the Erdôs- Rényi graph as follows:
$ER_n(P)$ is the random graph with vertex set $[n]$ where each pair $\{u,v\}$ of vertices is added to the edges set $E$ independently with probability $p$. The law of $ER_n(p)$ is $\mathbb {P}_\lambda$ with probability $\lambda=np$.
We have then a slightly different construction: that gives a coupling of $(\mathbb{P}_{n,p})_{p\in [0,1]}$. Take an i.i.d. family of $(U_e)_{e\in E_2}$ with $U_e\overset{d}{=}\text{Unif}([0,1])$, where $E_2$ is the complete graphe (all edges on $[n]$). Then define the random graph $ER_n(p)$ as having the edge $e\iff U_e\leq p$.
I found that the distribution of $X="\text{the number of edges in }E"=\mid E\mid\sim\text{Binomial}(k,2^n,p)$. Now I am trying to solve the following exercise but I don't understand what am I supposed to prove..
"Formulate and prove a CLT version for $\mid E\mid$ of $ER_n(p)$ for
- $p$ fixed
- $\lambda$ fixed
(better $p=p_n$, find for which $p_n$ a CLT holds)."
Thanks for any help
This may be a bit of overkill but the Lindeberg-Feller Central Limit Theorem (for triangular array) gives you the answer rather immediately. The condition that $Var[|E|]={n \choose 2} p_n (1-p_n) \to \infty$ is both a necessary and sufficiently condition that $\frac{|E|-{n \choose 2}p_n}{\sqrt{{n \choose 2}p_n (1-p_n)}}$ converges in distribution to a standard normal.