I was reading the proof of crossing number inequality and there was one step in the proof which I cannot prove rigorously. Firstly, let me remind the definition of the crossing number and then I state the lemma and its proof. I will not write the whole proof since I understood it pretty well except one moment.
Definition The crossing number of a graph $G=(V,E)$, denoted $\text{cr}(G)$, is the smallest integer $k$ such that we can draw $G$ in the plane with $k$ edge crossings.
Lemma (the crossing lemma) Let $G=(V,E)$ be a graph with $|E|\geq 4|V|$. Then $\text{cr}(G)\geq \frac{|E|^3}{64|V|^2}.$
Proof. Consider a drawing of $G$ with $\text{cr}(G)$ crossings. Set $p=\frac{4|V|}{|E|}$. The assumption of the lemma implies that $0<p\leq 1$. We remove every vertex of $V$ from the drawing with probability $1-p$ (together with the edges adjacent to the vertex).
More precisely, we consider the probability space $\Omega:=\{V': V'\subset V\}$ and we define probability as follows: $\mathbb{P}(V'):=p^{|V'|}(1-p)^{n-|V'|}$, where $n=|V|$.
We consider the following three random variables: $$\xi_1:\Omega \to \mathbb{R} \quad \text{defined as} \quad \xi_1(V'):=|V'|,$$ $$\xi_2:\Omega \to \mathbb{R} \quad \text{defined as} \quad \xi_2(V'):=|E'|,$$$$\xi_3:\Omega \to \mathbb{R} \quad \text{defined as} \quad \xi_3(V'):=\text{cr}(G'),$$ where $|E'|$ is the number of edges in the induced subgraph $G[V']$ and $\text{cr}(G')$ is crossing number of the induced subgraph $G':=G[V']$.
It is not difficult to compute that $\mathbb{E}[\xi_1]=p|V|$ and $\mathbb{E}[\xi_2]=p^2|E|$. I computed them by the definition of expectation.
Can anyone show rigorously why $\mathbb{E}[\xi_3]\leq p^4\text{cr}(G)$? The rest of the proof makes sense to me.
I'd be very happy if someone can show the detailed proof of this inequality.
EDIT: For example, this is how I computed $\mathbb{E}[\xi_1]$: since $\xi_1(V')=|V'|=\sum_{k=1}^{n}\mathbb{1}_{v_k}(V')$, where $$\mathbb{1}_{x}(V')=\begin{cases} 1, & \text{if } x\in V' \\ 0, & \text{if }x\notin V' \end{cases}$$ By linearity of expectation we have $\mathbb{E}[\xi_1]=\sum_{k=1}^n \mathbb{E}[1_{v_k}]=\sum_{k=1}^n p=p|V|.$ I wonder if it can be shown that $\mathbb{E}[\xi_3]\leq p^4\text{cr}(G)$ rigorously as I did?
Here is my reasoning: so we have a r.v. $\xi_3(V')=\text{cr}(G[V'])$. Consider the drawing of $G$ which has minimal number of edge crossings. In this drawing, we consider the induced subgraph $G[V']$. Obviously, $$\xi_3(V')\leq \#\left\{\{\{x,y\},\{z,w\}\}:x,y,z,w\in V',\ xy,zw\in E,\ xy\cap zw=\varnothing,\ xy \ \text{and} \ zw \ \text{have crossing}\right\}$$ It is not difficult to see that RHS is $$\frac{1}{8}\#\{(x,y,z,w)\in (V')^4:xy,zw\in E,\ xy\cap zw=\varnothing,\ xy \ \text{and} \ zw \ \text{have crossing}\}.$$ Of course, here we assume that no edge crossing comes from three vertices (but I don't know how to prove it yet) and the notation $xy\cap zw=\varnothing$ means that $\{x,y\}\cap \{z,w\}=\varnothing$.
Taking expectation, have
$$\mathbb{E}[\xi_3]\leq \frac{1}{8}\sum_{\cdots}\mathbb{E}[\mathbf{1}_x]\mathbb{E}[\mathbf{1}_y]\mathbb{E}[\mathbf{1}_z]\mathbb{E}[\mathbf{1}_w]=\frac{p^4}{8}\sum_{\cdots}1=p^4\text{cr}(G).$$
UPD. If we consider an optimal drawing of $G$ which has minimal number of edge crossings, then why we cannot have the following situation? More precisely, why edge crossing cannot arise from three points?
Suppose we have such a situation (see the picture below). So we have a crossing of blue and red edges. One can untangle it (see the picture on the right). Even in that case blue and red edges touch each other in a point which I denoted as $i$. But they should not touch each other at all. That is my concern.