On the correlation between variance in individual susceptibilities and infection prevalence in populations

Abstract

The hypothesis that infection prevalence in a population correlates negatively with variance in the susceptibility of its individuals has support from experimental, field, and theoretical studies. However, its generality has never been formally demonstrated. Here we formulate an endemic SIS model with individual susceptibility distributed according to a discrete or continuous probability function to assess the generality of such hypothesis. We introduce an ordering among susceptibility distributions with the same mean, analogous to that considered in Katriel (J Math Biol 65:237–262, 2012) to order the attack rates in an epidemic SIR model with heterogeneity. It turns out that if one distribution dominates another in this order then it has greater variance and corresponds to a lower infection prevalence for $R_0$ varying in a suitable maximal interval of the form $]1, R_0^*].$ We show that in both the discrete and continuous frameworks $R_0^*$ can be finite, so that the expected correlation among variance and prevalence does not always hold. For discrete distributions this fact is demonstrated analytically, and the proof introduces a constructive procedure to find ordered pairs for which $R_0^*$ is arbitrarily close to $1.$ For continuous distributions our conclusion is based on numerical studies with the beta distribution. Finally, we present explicit partial orderings among discrete susceptibility distributions and among symmetric beta distributions which guarantee that $R_0^*=+\infty$.

Appendix

Proof (Proposition 1)

The fact that $\varOmega _{\bar{s}}$ is an open subset of the manifold defined by (3) and Eq. (5) is a direct consequence of its definition. To see that the set $\varOmega _{\bar{s}}$ is not empty, we argue as follows. Fix any $\alpha _0$ satisfying (3) and consider the continuous function $\varPsi _{\alpha _0}(a)=a\cdot \alpha _0,$ where the dot denotes the standard inner product in ${\mathbb {R}}^n.$ Observe that it is possible to satisfy (4) choosing every $a_i,\,\,i=1,\ldots ,n,$ either arbitrarily close to zero or arbitrarily close to one. We conclude that the image through $\varPsi _{\alpha _0}$ of the set $\{ a\in {\mathbb {R}}^{n} : a\quad \text{ satisfies }\,\, (4)\}$ is the whole interval $]0,1[$ and therefore $\varOmega _{\bar{s}}\ne \emptyset .$

Finally, we show that the set $\varOmega _{\bar{s}}$ is path-wise connected. The discussion above implies that, fixed $\alpha _0$ satisfying (3), the set $\{(a,\alpha _0) : a\in {\mathbb {R}}^n \}\cap \varOmega _{\bar{s}}\ne \emptyset .$ Actually, this slice of $\varOmega _{\bar{s}},$ obtained intersecting $\varOmega _{\bar{s}}$ with the affine space $\alpha =\alpha _0,$ is by definition of the form $P_{\alpha _0}\times \{ \alpha _0\},$ where $P_{\alpha _0}$ is a convex polytope that depends continuously on $\alpha _0.$ We recall that a convex polytope in ${\mathbb {R}}^{n}$ is the intersection of a finite number of half-spaces (it can be thought of as the higher-dimensional analogous of a polygon in ${\mathbb {R}}^2).$

Then, fixed two points $(a_1,\alpha _1), (a_2,\alpha _2)\in \varOmega _{\bar{s}},$ if we take the segment $\alpha (t)=t\alpha _1+(1-t)\alpha _2,\quad t\in [0,1],$ we can define a continuous function $t\rightarrow a(t)\in P_{\alpha (t)},\quad t\in [0,1]$ such that $a(0)=a_1, \,\,a(1)=a_2$ and $(a(t),\alpha (t))\in \varOmega _{\bar{s}}.$ $\square$

Proof (Lemma 2)

To prove that $([\hat{a'}, \hat{a}],\hat{\alpha })\in \hat{\varOmega }_{\bar{s}}$ it suffices to show that

$$\begin{aligned} \phi (t)= & {} ta'_{n-1}+(1-t)a_{n-1}<\psi (t)\nonumber \\= & {} \frac{\bar{s}-\sum _{i=1}^{n-1} \alpha _i(ta_i'+(1-t)a_i)}{1-\sum _{i=1}^{n-1} \alpha _i}<1,\,\, t\in [0,1]. \end{aligned}$$

(19)

In fact, the remaining inequalities

$$\begin{aligned} 0<ta'_{i}+(1-t)a_{i}<ta'_{i+1}+(1-t)a_{i+1}, \,\,i=1,\ldots , n-2, \quad t\in [0,1] \end{aligned}$$

follow immediately from the assumption $(a,\alpha ),(a',\alpha ')\in \varOmega _{\bar{s}}.$ To prove (19) we start by observing that the functions $\phi$ and $\psi$ are linear. Therefore, it suffices to check that (19) holds for $t=0$ and $t=1.$ Since $(\hat{a}, \hat{\alpha })\in \hat{\varOmega }_{\bar{s}},$ we get directly the inequalities

$$\begin{aligned} \phi (0)= a_{n-1}<\psi (0)=\displaystyle \frac{\bar{s}-\sum _{i=1}^{n-1} \alpha _i a_i}{1-\sum _{i=1}^{n-1} \alpha _i}<1, \end{aligned}$$

so that (19) is true for $t=0.$ To show that

$$\begin{aligned} \psi (1)=\displaystyle \frac{\bar{s}-\sum _{i=1}^{n-1} \alpha _i a_i'}{1-\sum _{i=1}^{n-1} \alpha _i}<1, \end{aligned}$$

we note that this inequality is equivalent to the following

$$\begin{aligned} \bar{s}+\sum _{i=1}^{n-1} \alpha _i (1-a_i')<1. \end{aligned}$$

Now, by assumption we have $\alpha _i'\ge \alpha _i,\,i=1,\ldots ,n-1,\,$ with strict inequality for at least one index $i,$ and we conclude that

$$\begin{aligned} \bar{s}+\sum _{i=1}^{n-1} \alpha _i (1-a_i')<\bar{s}+\sum _{i=1}^{n-1} \alpha _i' (1-a_i')<1. \end{aligned}$$

In fact, the second inequality in the chain above is equivalent to

$$\begin{aligned} \frac{\bar{s}-\sum _{i=1}^{n-1} \alpha _i' a_i'}{1-\sum _{i=1}^{n-1} \alpha _i'}<1, \end{aligned}$$

which holds since $(\hat{a'},\hat{\alpha }')\in \hat{\varOmega }_{\bar{s}}.$ We have proven that $\psi (1)<1.$ It remains to see that $\phi (1)=a_{n-1}'<\psi (1).$ By assumption, $a_i\ge a_i',\quad i=1,\ldots , n-1,$ so that

$$\begin{aligned} \psi (1)>\displaystyle \frac{\bar{s}-\sum _{i=1}^{n-1} \alpha _i a_i}{1-\sum _{i=1}^{n-1} \alpha _i}>a_{n-1}\ge a_{n-1}'. \end{aligned}$$

Notice that the second inequality in the chain above holds since $(\hat{a},\hat{\alpha })\in \hat{\varOmega }_{\bar{s}}.$ Then, we see that (19) hold also for $t=1,$ and the proof that $([\hat{a'}, \hat{a}],\hat{\alpha })\in \hat{\varOmega }_{\bar{s}}$ is concluded.

To show that $(\hat{a'},[\hat{\alpha }, \hat{\alpha }'])\in \hat{\varOmega }_{\bar{s}}$ we have to prove the following:

$$\begin{aligned} a'_{n-1}<\eta (t):=\frac{\bar{s}-\sum _{i=1}^{n-1} (t\alpha _i+(1-t)\alpha _i')a_i}{1-\sum _{i=1}^{n-1} t\alpha _i+(1-t)\alpha _i'}<1,\quad t\in [0,1]. \end{aligned}$$

Since the function $\eta (t)$ is monotone, we must check the last chain of inequalities only in $t=0,1.$ The arguments used to this aim are similar to those given above, and we skip the details. $\square$