Let $H(n,p)$ be a $r$-uniform random hypergraph on $n$ vertices, with vertex set $V$ and edge set $E$. Let $E' \subset E$ be a set of edges with some known characteristics, and $V' = V \setminus \{u_i: u_i \text{ is a vertex of } e, e \in E'\}$.
I have several questions about this:
- My note says that $$|V'| = \sum_{v \in V} 1_{\{v \in V'\}}, \text{ and } \mathbb{E}\left[ 1_{\{v \in V'\}} \right] = \mathbb{P}(v \in V') = (1 - p)^{\text{deg}(v)}.$$ But if we know deg$(v)$, shouldn't we already know the exact value of $1_{\{v \in V'\}}$, and hence do not need to estimate it by its expected value?
- Say we accept this line of reasoning, wouldn't a more appropriate statmement be $$\mathbb{P}(v \in V') \leq (1 - p)^{\text{deg}(v)},$$ with equality achieved if all edges adjacent to $v$ are in $E'$?
- Regardless of the above, why can't I get $\mathbb{P}(v \in V')$ in the following way? \begin{align} & \mathbb{P}(v \in \{u_i: u_i \text{ is a vertex of } e, e \in E'\}) = p^{\text{deg}(v)} \\ \Rightarrow & \ \mathbb{P}(v \in V') = \mathbb{P}(v \notin \{u_i: u_i \text{ is a vertex of } e, e \in E'\}) = 1 - p^{\text{deg}(v)}.\end{align}
Edit: Please have a look at slides 33 and 37 for the context of the question, if necessary.