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Consider a graph having n vertices labeled 1, 2, . . . , n, and suppose that, between each of the $\binom{n}{2}$ pairs of distinct vertices, an edge is independently present with probability p. The degree of vertex i, designated as $D_i$, is the number of edges that have vertex i as one of their vertices.

(a) What is the distribution of $D_i$?

(b) Find $ρ(D_i,D_j)$, the correlation between $D_i$ and $D_j$.

My attempt is:

Let $X_{i,k} = \begin{cases}1 & \text{ if there is an edge from } i \text{ to } k\\ 0 & \text{ otherwise }\end{cases}$

$D_i=\sum_{k=1}^{n-1} x_{i,k}$=number of edges from vertex i.

$D_i$ has a binomial distribution with parameters $n-1$ and p.

$E[D_i]=E[\sum_{k=1}^{n-1} x_{i,k}]=\sum_{k=1}^{n-1} E[x_{i,k}]=(n-1)*p$

Let $X_{j,k} = \begin{cases}1 & \text{ if there is an edge from} j \text{ to }k\\ 0 & \text{ otherwise}\end{cases}$

$D_j=\sum_{k=1}^{n-1} x_{j,k}$=number of edges from vertex j.

$D_j$ has a binomial distribution with parameters $n-1$ and p.

$E[D_j]=E[\sum_{k=1}^{n-1} x_{j,k}]=\sum_{k=1}^{n-1} E[x_{j,k}]=(n-1)*p$

then I try to calculate $cov(D_i,D_j)$

$$cov(D_i,D_j)= cov (\sum_{k=1}^{n-1} x_{i,k},\sum_{w=1}^{n-1} x_{j,w})=\binom{n}{2}*(-p^2)$$ with $cov(x_{i,k},x_{j,w})= E[x_{i,k}*x_{j,w}]-E[x_{i,k}]*E[x_{j,w}]$ (and here i'm not sure) $$cov(x_{i,k},x_{j,w}) = \begin{cases}0-p^2 & if k=w \\ 0 & if k \ne w \end{cases}$$

$$Var(D_i)=Var(D_j)=(n-1)*p*(1-p)$$

The final result in the book is $\frac{1}{n-1} $

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Notice that

$$cov(D_i, D_j)=cov\left(\sum_{k \neq i} X_{ik}, \sum_{w \neq j} X_{jw}\right)=\sum_{k \neq i} \sum_{w \neq j}cov\left( X_{ik}, X_{jw}\right)$$

$$cov(X_{ik}, X_{jw}) = \begin{cases} p-p^2 & ,k=j, w=i \\ 0 & \text{otherwise} \end{cases}$$

Hence,

$$cov(D_i, D_j)=\sum_{k \neq i} \sum_{w \neq j}cov\left( X_{ik}, X_{jw}\right)=cov(X_{ij}, X_{ji})=p-p^2$$

Since $Var(D_i)=(n-1)p(1-p)$,

$$\rho(D_i, D_j)=\frac{p-p^2}{\left(\sqrt{(n-1)p(1-p)}\right)^2}=\frac{1}{n-1}$$

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