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> [!proposition] Proposition. ([[percolation in configuration model with exponential distribution]])
> [!proof]- Proof. ([[percolation in configuration model with exponential distribution]])
> **Uniform removal.**
> The [[network]] has a [[exponential random variable|geometric distribution]] $p_{k}=(1-a)a^{k}$, where $a<1$ and $1-a$ ensures normalization. Then, as discussed in [[exponential degree distribution configuration model]], we have $g_{0}(z)=\frac{1-a}{1-az}, \ \ g_{1}(z)=(\frac{1-a}{1-az})^{2}.$
> Then by [[percolation by uniform removal in the configuration model]] the average [[probability]] that a node $u$ is not connected to the [[giant cluster]] via a particular neighbor is $u=1-\phi + \phi g_{1}(u)=1-\phi + \phi (\frac{1-a}{1-az})^{2}$ and rearranging we find $u(1-au)^{2} - (1-\phi)(1-au)^{2} - \phi (1-a)^{2} = 0.$
> This is a cubic equation $u$; using [[synthetic division]] we factor as $(u-1)[a^{2}u^{2} + a(a\phi - 2)u + \phi - 2a \phi + 1] = 0.$
> Using the [[quadratic formula]] it is clear that one of the two remaining solutions is greater that 1 for $a<1$ and so cannot be a [[probability]]. The other must be $u$: $u=a^{-1} - \frac{1}{2}\phi - \sqrt{ \frac{1}{4}\phi^{2} + \phi(a^{-1} - 1) }.$
> Now we can plug this into [[percolation by uniform removal in the configuration model|the expression for]] $S$ thus: $\begin{align}
> S= & \phi \left[ 1- \frac{2(a^{-1} - 1)}{\phi + \sqrt{ \phi^{2} + 4\phi (a^{-1} - 1) }} \right] \\
> = & \phi\left[ 1- 2(a^{-1} - 1) \frac{\phi - \sqrt{ \phi^{2} + 4 \phi (a^{-1} - 1) }}{\phi^{2} -(\phi^{2} + 4 \phi (a^{-1} - 1))} \right] \\
> = & \frac{3}{2}\phi = \sqrt{ \frac{1}{4} \phi^{2} + \phi(a^{-1} - 1) }.
> \end{align}$
> The percolation threshold is $\phi_{c}=\frac{1-a}{2a}.$
>
>
>
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#### References
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