Expectation of independent sets

Question

Let G be a undirected graph with vertices $V = \lbrace1,..., n\rbrace$ and edges $E \subseteq {V \choose 2}$ and $|E| = m$. If we choose a set $R$ of $r$ vertices at random (every set in ${V \choose r}$ with same probability) , then the probability for an edge $\lbrace i, j \rbrace$ to have both endpoints in $R$ is:

$$\frac{n-2 \choose r-2}{n \choose r} = \frac{r (r -1)}{n(n-1)}$$

Let $X$ be a random variable such that X = #$(\frac{R}{2} \cap E)$. We compute: $$E(X) = m. \frac{r (r -1)}{n(n-1)}$$.

I understand how to calculate $E(X)$. So, it should be for every value of $X$ being $x$, probability $Pr[X=x]$ should get multiplied with $X$ and all such values should get summed up. In this case, it should be $$E(X) = \sum i * Pr[X = i]$$

But, I am not being able to figure out how the $E(X) = m. \frac{r (r -1)}{n(n-1)}$ is being calculated here. Any clue on this is welcome.

Answer 1

In this case the formula $E(X)=\sum_iiP(X=i)$ is not helpful, since it is not possible to work out the individual probabilities $P(X=i)$ (they depend on the exact graph $G$, not just the number of edges it has).

However, it is possible to deduce the expectation of X, using the fact that expectation is linear. $X=\sum_{e\in E}X_e$, where $X_e$ is $1$ if the edge $e$ has both ends in $R$ and $0$ otherwise (this is just going through edges one by one and counting the ones that we are interested in). Now $$E(X_e)=P(X_e=1)=\frac{r(r-1)}{n(n-1)},$$ and so $$E(X)=E\Big(\sum_eX_e\Big)=\sum_eE(X_e)=m\frac{r(r-1)}{n(n-1)}.$$

Crucially, you can just add expectations together like this even if the variables are not independent.