Random graph application

Question

Hi am looking at the theorem 1.1 but I can't come to that distribution.

Can someone tell me how does this factor comes in the distribution of $D_n^*$? That sounds trivial but I cannot get it.

Thanks for any help.

Answer 1

First we can make the following observations:

1. $\mathbb P(D_n=k)=\displaystyle\frac 1n\sum_{i=1}^n \mathbf{1} [d_i=k]$
2. $\displaystyle|E|=\frac 12\sum_{i=1}^n d_i$
3. $\displaystyle \sum_{e_{ij}} \mathbb{1}(d_i=k)+\mathbb{1}(d_j=k)=k\sum_{i=1}^n \mathbf{1} [d_i=k]$

(1) and (2) are straight forward, and (3) is because the sum on the left side counts each vertex of degree $k$ for $k$ times. .Now use simple Bayesian decomposition: $$ \mathbb P(D^*_n=k)=\frac{2}{\sum_{i=1}^n d_i}\sum_{e_{ij}} \Pr(d_i=k \text{ or } d_j=k | \mathbb{e}=e_{ij})\\ =\frac{2}{\sum_{i=1}^n d_i}\frac 12\sum_{e_{ij}} \mathbb{1}(d_i=k)+\mathbb{1}(d_j=k)\\ =\frac{k}{\sum_{i=1}^n d_i}\sum_{{i}} \mathbb{1}(d_i=k) $$ Now use $(1)$ and equality $(1.2.3)$ in the text to conclude the theorem.