Before determining the network structures under which dynamical partitioning occurs more generally, we consider some examples. For further examples in the context of undirected networks the reader is directed to Kiss et. al. (2014).
Example 1
Consider the network in Figure 3.5a. Let us suppose that all individuals recover at rate γ when infected, become infected at rate α when exposed, and that the contact rate across all network links is unity. For simplicity we shall also make this assumption through the remainder of the explicit examples in this chapter. We can apply the closure in corollary 3.4.1, which is a special case of the exact closure theorem, to build up the moment-induced system ME. Let us consider the probability of individual 1
being exposed to see how this works. We have: ˙
hE1i=hS1I2i −αhE1i. (3.28)
Here and throughout this chapter we order individuals according to the numerical order of their labels; the relevant subsystem configurations need to be understood with reference to the full network. Now, individual 2 is dynamically partitioning with respect to individuals 1 and 3, and it is also dynamically partitioning with respect to individuals 1 and 4. Hence: ˙ hS1I2i = −(1 +γ)hS1I2i+αhS1E2i, and for hS1E2i: ˙ hS1E2i = hS1S2I3i+hS1S2I4i −αhS1E2i = hS1S2ihS2I3i hS2i +hS1S2ihS2I4i hS2i − αhS1E2i. (3.29)
Rather than a complete analysis of all moment-induced subsystem configurations that arise, we take a single pair configuration S2I3 from this equation as an example. Here,
individual 3 is not dynamically partitioning with respect to individuals 2 and 4 but individual 2 is dynamically partitioning with respect to 1 and 3 so:
˙
hS2I3i=−h
I1S2ihS2I3i
hS2i − h
S2I3I4i −(1 +γ)hS2I3i+αhS2E3i. (3.30)
Then, for hS2I3I4i, individual 2 is dynamically partitioning with respect to individual
1 and individuals 3 and 4 so: ˙ hS2I3I4i = −h I1S2ihS2I3I4i hS2i − 2(1 +γ)hS2I3I4i +αhS2E3I4i+αhS2I3E4i. (3.31)
We see that here, ME represents a significant dimensional reduction compared to the
full system M.
Example 2
For the undirected network in Figure 3.5b there is dynamical partitioning about indi- vidual 1. Starting with (for example) the probability of individual 1 being exposed, we have:
˙
hE1i=hS1I2i+hS1I4i+hS1I5i+hS1I6i −αhE1i. (3.32)
Now, choosing the first of these pairs to develop one part of the moment-induced system
ME gives: ˙ hS1I2i = −hS1I2I4i −h S1I2ihS1I5i hS1i − hS1I2ihS1I6i hS1i −(1 +γ)hS1I2i+αhS1E2i, (3.33) and ˙ hS1E2i = hS1S2I3i − hS1E2I4i −h S1E2ihS1I5i hS1i − hS1E2ihS1I6i hS1i −αhS1E2i. (3.34)
Then, for the first of these triples: ˙ hS1S2I3i = −hS1S2I3I4i −h S1S2I3ihS1I5i hS1i − hS1S2I3ihS1I6i hS1i −(1 +γ)hS1S2I3i+αhS1S2E3i, (3.35)
and for the quad we have: ˙ hS1S2I3I4i = −h S1S2I3I4ihS1I5i hS1i − hS1S2I3I4ihS1I6i hS1i −2(1 +γ)hS1S2I3I4i+αhS1S2E3I4i+αhS1S2I3E4i. (3.36)
Here, the maximum size of a subsystem configuration is four individuals. We note that this is equal to the size of the largest cycle and that this was also true for example 1. However, this is not always the case as shown by the next example.
Example 3
Figure 3.5c shows a network where the maximum simple cycle size is 4 but the maximum size of a subsystem configuration in ME is 5. Starting with the the probability of
individual 1 being exposed we have: ˙
hE1i=hS1I2i+hS1I4i+hS1I5i −αhE1i. (3.37)
Then, taking just the subsystem configuration in the first term: ˙
hS1I2i=−hS1I2I4i − hS1I2I5i −(1 +γ)hS1I2i+αhS1E2i, (3.38)
and again taking the first term gives: ˙
hS1I2I4i = −hS1I2I4I5i −2(1 +γ)hS1I2I4i
+αhS1E2I4i+αhS1I2E4i, (3.39)
and for the quad: ˙
hS1I2I4I5i = −3(1 +γ)hS1I2I4I5i+αhS1E2I4I5i
+αhS1I2E4I5i+αhS1I2I4E5i. (3.40)
Finally, taking the last of these terms gives: ˙
hS1I2I4E5i = −(2 + 2γ+α)hS1I2I4E5i
+hS1I2I3I4S5i+αhS1E2I4E5i+αhS1I2E4E5i. (3.41)
In this case we see that the maximum size of a subsystem configuration is at the size of the system (5 individuals) and is not constrained by the largest cycle (4 individuals). This leads to the question of what aspect of a network specifies the largest subsystem configuration that appears inME. We answer this question in the following section.