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> [!proposition] Proposition. ([[SIR example]])
> ~
Consider an epidemic SIR outbreak (i.e., an outbreak that starts in the giant cluster of the corresponding percolation process) on a configuration model network with exponential degree distribution $p_{k}=(1-a)a^{k}$ with $a<1$. You can assume that the network is large.
**a. Using the results of Section 15.2.1 write down an expression for the probability $u$ appearing in Eq. (16.29) in terms of $\phi$ and $a$.**
The solution as shown in equation 15.17 is $u=a^{-1} -\frac{1}{2} \phi - \sqrt{ \frac{1}{4} \phi^{2} + \phi(a^{-1} - 1) }.$
**b. Hence find an expression for the probability that a node is infected by the disease if it has degree $k$.
If a node has degree $k$ then it does not belong to the [[giant cluster]] with probability $u^{k}$. So it does belong to the [[giant cluster]] with probability $1-u^{k}$. And the outbreak starts in the [[giant cluster]] with [[probability]] $S$. If the outbreak does start in the [[giant cluster]] then the node will get infected almost surely. So the node gets infected with probability $S(1-u^{k})$. Equation 16.30 then tells us giant cluster size is ![[CleanShot 2023-12-05 at 17.49.05.jpg]]
(note that if the node is not in the [[giant cluster]] then it almost surely will not be infected)
**c. Evaluate this probability for the case $a=0.4$ and $\phi=0.9$, for $k=0$, 1, and 10.**
For $a=0.4$ and $\phi=0.9$ we get $u=.804$ and $S=.116$. Then the probability for infection for $k=0$ is $0$, for $k=1$ is $.023$, and for $k=10$ is $.103$.
> [!proof]- Proof. ([[SIR example]])
> ~
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#### References
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