更改

\mathbb{E}[k_\text{in}^2]  

\mathbb{E}[k_\text{in}^2]  

+\mathbb{E}[k_\text{in}^2] \mathbb{E}[k_\text{out}^2]  

+\mathbb{E}[k_\text{in}^2] \mathbb{E}[k_\text{out}^2]  

  - \mathbb{E}[k_\text{in} k_\text{out}]^2 >0.</math>注意 <math display="inline">\mathbb E [k_\text{in}]</math> 和 <math display="inline">\mathbb E [k_\text{out}]</math> 是相等的，因此在后面的不等式中二者可交换。对于图中任意一个节点，它属于大小为<math>n</math>的子图的概率由下式给出：（入分量） <ref>{{Cite journal|last=Kryven|first=Ivan|date=2017-11-02|title=Finite connected components in infinite directed and multiplex networks with arbitrary degree distributions|journal=Physical Review E|volume=96|issue=5|pages=052304|doi=10.1103/PhysRevE.96.052304|pmid=29347790|arxiv=1709.04283|bibcode=2017PhRvE..96e2304K}}</ref><math display="block">h_\text{in}(n)=\frac{\mathbb E[k_{in}]}{n-1} \tilde u_\text{in}^{*n}(n-2),  \;n>1, \; \tilde u_\text{in}=\frac{k_\text{in}+1}{\mathbb E[k_\text{in}]}\sum\limits_{k_\text{out}\geq 0}u(k_\text{in}+1,k_\text{out}) </math>，（出分量） <math>h_\text{out}(n)=\frac{\mathbb E[k_\text{out}]}{n-1} \tilde u_\text{out}^{*n}(n-2),  \;n>1, \;\tilde u_\text{out}=\frac{k_\text{out}+1}{\mathbbE[k_\text{out}]}\sum\limits_{k_\text{in}\geq0}u(k_\text{in},k_\text{out}+1) </math>。

  - \mathbb{E}[k_\text{in} k_\text{out}]^2 >0.</math>注意 <math display="inline">\mathbb E [k_\text{in}]</math> 和 <math display="inline">\mathbb E [k_\text{out}]</math> 是相等的，因此在后面的不等式中二者可交换。对于图中任意一个节点，它属于大小为<math>n</math>的子图的概率由下式给出：

*入分量: <ref>{{Cite journal|last=Kryven|first=Ivan|date=2017-11-02|title=Finite connected components in infinite directed and multiplex networks with arbitrary degree distributions|journal=Physical Review E|volume=96|issue=5|pages=052304|doi=10.1103/PhysRevE.96.052304|pmid=29347790|arxiv=1709.04283|bibcode=2017PhRvE..96e2304K}}</ref><math display="block">h_\text{in}(n)=\frac{\mathbb E[k_{in}]}{n-1} \tilde u_\text{in}^{*n}(n-2),  \;n>1, \; \tilde u_\text{in}=\frac{k_\text{in}+1}{\mathbb E[k_\text{in}]}\sum\limits_{k_\text{out}\geq 0}u(k_\text{in}+1,k_\text{out}) </math>

*出分量: <math>h_\text{out}(n)=\frac{\mathbb E[k_\text{out}]}{n-1} \tilde u_\text{out}^{*n}(n-2),  \;n>1, \;\tilde u_\text{out}=\frac{k_\text{out}+1}{\mathbb E[k_\text{out}]}\sum\limits_{k_\text{in}\geq0}u(k_\text{in},k_\text{out}+1).</math>