Theoretical and Numerical Considerations of the Assumptions Behind Triple Closures in Epidemic Models on Networks

Abstract

Networks are widely used to model the contact structure within a population and in the resulting models of disease spread. While networks provide a high degree of realism, the analysis of the exact model is out of reach and even numerical methods fail for modest network size. Hence, mean-field models (e.g. pairwise) focusing on describing the evolution of some summary statistics from the exact model gained a lot of traction over the last few decades. In this paper we revisit the problem of deriving triple closures for pairwise models and we investigate in detail the assumptions behind some of the well-known closures as well as their validity. Using a top-down approach we start at the level of the entire graph and work down to the level of triples and combine this with information around nodes and pairs. We use our approach to derive many of the existing closures and propose new ones and theoretically connect the two well-studied models of multinomial link and Poisson link selection. The theoretical work is backed up by numerical examples to highlight where the commonly used assumptions may fail and provide some recommendations for how to choose the most appropriate closure when using graphs with no or modest degree heterogeneity.

Appendix: Facts of the Multinomial Distribution

A random vector

$$\displaystyle \begin{aligned}{\mathbf{X}}_{(r)} = (X_1, \ldots, X_r) \in \{0, 1, \ldots, n\}^r \end{aligned}$$

is multinomially distributed with n trials and r types (written as Mult(n, r;p ₁, p ₂, …, p _r)) if its mass function can be given by

$$\displaystyle \begin{aligned} f_{{\mathbf{X}}_{(r)}}(x_1, x_2, \ldots, x_r) =\begin{cases} \displaystyle {n \choose x_1, x_2, \ldots, x_r} p_1^{x_1}p_2^{x_2}\ldots p_r^{x_r}, & \mbox{ if } \displaystyle\sum_{i=1}^r x_i =n \\ 0, &\mbox{ otherwise }. \end{cases} \end{aligned} $$

(7.1)

This is provided that $\sum _{i=1}^r p_i = 1.$ Each p _i represents the probability of seeing type i in one independent experiment, and the vector X _(r) counts the number of outcomes of each type in n independent experiments. Note that the sum $\sum _{i=1}^r X_i = n$, so already there is some dependency between the coordinates. In fact, the last (or any one) coordinate can be given by $X_r = n - \sum _{i=1}^{r-1} X_i$, so in many sources the multinomial distribution has mass function

(7.2)

Throughout this article we used the following facts about multinomial distributions.

Lemma 7.1

Let X = (X ₁, …, X _r) ∼Mult(n, r;p ₁, p ₂, …, p _r) so that p ₁ + p ₂ + … + p _r = 1 where the success probability for type i is p _i.

• (1) The marginal distribution of X _i is binomial(n, p _i).

• (2) The joint distribution for (X _i, X _j) is the same as that of (X _i, X _j, n − X _i − X _j) ∼Mult(n, 3;p _i, p _j, 1 − p _i − p _j).

• (3) The $\mathbb {E}(X_iX_j) =n(n-1)p_ip_j$.

Proof

We show (1) for i = 1 (i.e. the X ₁ marginal.) First the domain of the p.m.f: Since X ₁ is a coordinate of a multinomial with n independent trials, its state space is $S_{X_1} = \{0,1, \ldots , n\}$. Now fix $x \in S_{X_1}$ and compute

$$\displaystyle \begin{aligned} f_{X_1}(x) &= \mathbb{P}\{ X_1 = x\} = \sum_{ (x_2, \ldots, x_r) : \sum x_i = n-x }\!\!\!\!\!\mathbb{P}\{ X_1 = x, X_2 = x_2, \ldots X_r=x_r \}\\ &= \sum_{ (x_2, \ldots, x_r) : \sum x_i = n-x }\!\!\!\!\! \frac{n!}{x! x_2!x_3!\ldots x_r!} p_1^x p_2^{x_2} \ldots p_r^{x_r}\\ &= \frac{n!}{x!(n-x)!}p_1^x\sum_{ (x_2, \ldots, x_r) : \sum x_i = n-x }\!\!\!\!\! \frac{(n-x)!}{x_2!x_3!\ldots x_r!} p_2^{x_2} \ldots p_r^{x_r}. \end{aligned} $$

The reason for the last manipulation is to reduce the expression in the sum to something that looks like the p.m.f. of a Mult(n − x, r − 1, q ₁, …, q _r−1). The expression is almost correct, except that the sum p ₂ + … + p _r = 1 − p ₁ instead of 1. To fix this little problem, multiply and divide by (1 − p ₁)^n−x and recall that n − x = x ₂ + … + x _r. Then

The sum is 1 since we add the p.m.f. of a Mult$\displaystyle \Big (n-x, r-1, \frac {p_2}{1-p_1},\ldots , \frac {p_r}{1-p_1}\Big )$. Observe that X ₁ ∼Bin(n, p ₁) = Mult(n, 2;p ₁, 1 − p ₁) and in particular each X _i ∼Bin(n, p _i).

For (2) we compute the joint marginal for (X ₁, X ₂). Similar calculations will work for any k-dimensional marginal, but we do not use them in this article. First the domain of the joint p.m.f: Since (X ₁, X ₂) are coordinates of a multinomial with n independent trials, its state space is $S_{X_1,X_2} = \{0,1, \ldots , n\}$ ×{0, 1, …, n}. Now fix $(x,y) \in S_{X_1, X_2}$. Assume x + y ≤ n otherwise the p.m.f. is 0 and compute as before

Therefore (X _i, X _j) ∼Mult(n, 3;p _i, p _j, 1 − p _i − p _j).

Now that we have the two-dimensional marginal distribution (X _i, X _j) ∼Mult(n, 3;p _i, p _j, 1 − p _i − p _j) and the one-dimensional marginals are X _k ∼ Bin(n, p _k) we can proceed with the expected value of the product.

$$\displaystyle \begin{aligned} &\mathbb{E}(X_iX_j)\\ &\quad = \sum_{(x_i, x_j): x_i+x_j \le n } x_ix_j \frac{n!}{x_i!x_j!(n-x_i-x_j)!} p_1^{x_1}p_2^{x_2}(1-p_1-p_2)^{n - x_1-x_2}\\ &\quad {=}p_1p_2 n(n-1)\!\!\!\!\!\sum_{(x_i, x_j): x_i+x_j \le n, x_ix_j \ge 1 } \frac{(n-2)! p_1^{x_1-1}p_2^{x_2-1}(1-p_1-p_2)^{n - x_1-x_2}}{(x_i - 1)!(x_j - 1)!(n-x_i-x_j)!} \\ &\quad {=}p_1p_2 n(n-1)\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum_{(x_i, x_j): 0\le (x_i -1)+(x_j-1) \le n-2 } \!\!\!\frac{(n-2)! p_1^{x_1-1}p_2^{x_2-1}(1-p_1-p_2)^{n - x_1-x_2}}{(x_i - 1)!(x_j - 1)!(n-x_i-x_j)!} \\ &\quad {=}p_1p_2 n(n-1). \end{aligned} $$

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