Early epidemic spread, percolation and Covid-19.

Human to human transmissible infectious diseases spread in a population using human interactions as its transmission vector. The early stages of such an outbreak can be modeled by a graph whose edges encode these interactions between individuals, the vertices. This article attempts to account for the case when each individual entails in different kinds of interactions which have therefore different probabilities of transmitting the disease. The majority of these results can be also stated in the language of percolation theory. The main contributions of the article are: (1) Extend to this setting some results which were previously known in the case when each individual has only one kind of interactions. (2) Find an explicit formula for the basic reproduction number [Formula: see text] which depends only on the probabilities of transmitting the disease along the different edges and the first two moments of the degree distributions of the associated graphs. (3) Motivated by the recent Covid-19 pandemic, we use the framework developed to compute the [Formula: see text] of a model disease spreading in populations whose trees and degree distributions are adjusted to several different countries. In this setting, we shall also compute the probability that the outbreak will not lead to an epidemic. In all cases we find such probability to be very low if no interventions are put in place.

Introduction

The vector by which several infectious diseases propagate in a population is the human to human interaction. It is therefore natural to model their spread using such interactions as the “basic mechanism”. In general, it gives rise to a dynamic process which evolves in time with its early stages being reasonably well approximated by considering the patient zero as the root (apex) of a tree whose branches encode the interactions through which the disease can propagate. In such epidemiological and percolation problems it is commonly assumed that each individual has an equal probability to transmit the disease to any of its contacts. However, this is an over-simplification of the actual situation as the same person can have several classes of interactions. Namely, it is conceivable that an infected individual will more likely transmit the disease to someone with whom it maintains a close familiar relation than to another person who it sporadically meets. In this setting we shall encode these different probabilities of transmitting the disease by distinct trees all with the same root which is the patient zero. To each tree we associated a probability of transmitting the disease along an interaction modeled by the corresponding tree. In the case when the are all the same for each vertex (individual) it is known that the basic reproduction number controls the possibility of almost surely avoiding an epidemic, see Brauer (2008), Harris (1963), Jagers (1975), Kimmel and Axelrod (2002), Schinazi (1999) and Callaway et al. (2000) which phrases these results in terms of the percolation interpretation. When there are different a similar framework can be used to prove the following result.

Theorem 1

Suppose the basic reproduction number , then the outbreak will almost surely, i.e. with probability one, not lead to an epidemic. On the other hand, if there is still a nonzero probability that the outbreak will be contained.

In Sect. 2 we shall develop the framework of multivariate generating functions on which this work will be based. Section 3 will show how to use this framework to effectively compute and finally we will prove the main abstract results in Sect. 4. We will then exemplify the theory with a few examples. These are instructive in order to unravel an explicit formula for which does not depend on any “abstract” generating function. Such a formula is deduced in Sect. 6 where we show the following result.

Theorem 2

For each let denote the average degree of the tree and the standard deviation of the associated degree distribution. Suppose that a fraction Q of all infected individuals is completely isolated and does not transmit the disease to anyone. Then, if the probability of transmitting the disease along an arm of the tree is , the basic reproduction number isor

A somewhat interesting feature of the previous formula is that it together with the transmission probabilities , it depends solely on the first two moments of the degree distribution of the trees encoding the interactions.

As a final application of the theory, and motivated by the recent outbreak of Covid-19, we reserve the last section to do some specific country analysis. We have attempted to make the parameters of the theory be somewhat adequate to model the initial spread of Covid-19 but the results should be regarded as an “academic” toy example. A more robust analysis using our framework is possible, but would require detailed knowledge on the habits, family ties/interactions, attendace of public gatherings and other features of the analyzed populations, which are not uniform in each country. That last section computes the relevant values of for the countries considered and the probabilities that the outbreak will be contained. As we shall see, these are very small and in order to increase it we will investigate the effect of quarantining part of the infected individuals.

Generating functions

From the generating vertex

Let be a graph having a tree structure and whose edges are divided into n groups. Each of groups gives rise to subgraphs having the same vertices and the edges of the corresponding group. We will also assume that has a tree structure. For each of these trees , for , we shall denote by the corresponding degree distribution, i.e. for a randomly chosen vertex its degree is k with probability . Using these we can construct the corresponding generating functionsand the joint generating function

Remark 1

One can readily check that , and converges for . Furthermore, we compute that and thus the average degree of the tree isSimilarly, we find that from the total average degree is

Each of these trees corresponds to different classes of contacts which have unequal probability of transmitting the disease. For instance, people that live in the same house are more likely to transmit the disease to each other than those which occasionally meet on public transport. Thus, to each tree, i.e., to each class of interactions, we associate a probability of transmission and assume with no loss of generality that . Fix , then the probability that a randomly picked first infected individual transmits the disease to other individuals along the tree isFurther suppose there is a probability Q that an infected individual is detected and quarantined in complete isolation. Then, if , the probability above must be multiplied by a factor of which accounts for the possibility that it is not quarantined. At this point, it is convenient to define the generating function

By following a random infection

Suppose we place ourselves at a vertex which is obtained from following a randomly chosen transmission. As an element of the tree , the ramification of this vertex is the number of remaining edges emanating from it. The probability that such a vertex has ramification is therefore proportional towith each term accounting from the probability of arriving at the chosen vertex via a given tree. Normalizing this we find that such probability is obtained from the previous formula by dividing byi.e. the average total ramification. We have thus concluded thatAssociated with this we define the generating functionFor future reference, it is convenient to have this written in terms of the simpler generating functions for . For this, we insert the formula for previously obtained. This yieldswhich may be written in the following more explicit form

Remark 2

Consider of a sole tree , the corresponding ramification distribution iswhich the individual generating functionThen, we have which upon inserting in Eq. 2.3 yields

Let with . Using the distribution for the ramification, we conclude that by following the contacts of the trees up to a randomly infected individual infects up to other ones isif any of the is nonzero. Based on this, we define the generating function

The basic reproduction number

The basic reproduction number, usually denoted by , is defined as the average number of individuals which are infected by each previously infected one. In our setting this can be immediately computed as follows. First, suppose we stand at a randomly chosen individual which has ramification . The average number of individuals it infects iswhere we have used the binomial formula to deduce . For the average vertex we must weight this with the ramification distribution, i.e.Having in mind the formula for we findInserting into the above equation for and using again the binomial formula givesComparing this with the formula for we conclude the following result.

Proposition 1

The basic reproduction number can be obtained from the generating function for the distribution of individuals infected by following a randomly chosen infected individual, viawhich may also be written as

Containing an outbreak

In this section we shall compute the probability that the infection as it propagates eventually dies out. In the case of a unique tree, i.e. assuming all interactions have the same probability of transmitting the disease, such a computation have been carried out in Brauer (2008). See also Harris (1963), Jagers (1975), Kimmel and Axelrod (2002), Schinazi (1999) for related results and Callaway et al. (2000) for the same setup in the context of the theory of percolation.

Consider an individual which has been infected by another one, i.e. a vertex of the tree which is not its root and consider the probability that the infections generated by that vertex disappear within N generations. Denote such a probability by , thenwhich upon comparing with the definition of in Eq. 2.4 can equally be read asBy construction we must havewith Eq. 4.1 yielding all the following iterations. From inspection we find that and as is increasing we find assuming thatwhich inductively proves that the sequence is increasing. Hence, the numberis well defined and encodes the probability that the infection starting from any such individual eventually dies out. We can then conclude the following.

Proposition 2

The probability that the chain of infections generated from a randomly infected individual eventually disappears in a finite number of generations satisfiesFurthermore, there ate most two fixed points of in [0, 1].

Proof

The fact that satisfies follows immediately from the preceding discussion. Hence, the only remaining item to be shown is that there are at most two fixed points of in the interval [0, 1]. The fact that there is at least one is obvious as . We must now show that there is at most one other.

We argue by contradiction and assume there are at least two other different fixed points both in (0, 1).1 First, notice that both , are positive in (0, 1). Secondly, consider the function whose zeros correspond to the fixed points of , we haveHence, by the intermediate value theorem there must be two critical points and of f. As in (0, 1) we have we have that each of these must a maximum. Again, by the intermediate value theorem, between the two maxima must be a minimum contradicting .

Placing ourselves at the tip of the tree which originated the infection chain, the so called patient zero, the probability that the infection eventually dies out iswhere the last equality follows from comparison with the formula 2.1 for the generating function for infections starting from the patient zero.

The case when

We shall now prove that when the basic reproduction number is smaller than one, the chain of infections will almost surely extinguish.

Theorem 3

If , then both and equal 1.

Proof

By Proposition 2 we know that it must be a fixed point of . Furthermore, we know that and so the statement follows if we can show that there is no other fixed point of in the interval [0, 1]. In that direction we shall show that under the hypothesis that the mapis a contraction and so has a unique fixed point which must therefore the 1. This follows immediately from realizing that is nonnegative and so by 1. Hence, for we havewhich shows that is a contraction if .

Finally, the fact that also is then a consequence of .

Remark 3

(The minimum required quarantined) At the beginning of an outbreak the question arises of what is the minimum number of infected individuals that must be detected and subsequently quarantined in order to contain the possible epidemic outbreak.

If the disease is already well known, such as flu, measles or any other standard disease, not Covid-19, then its “free” basic reproduction number is known. Of course, this may depend on local conditions of where the outbreak takes place. By “free” we intend to emphasize that this is the basic reproduction number when the disease is free to propagate without taking in account any non-pharmaceutical intervention directed to slow its spread.

Now, suppose an aggressive testing capacity can be put in place in order to detect those which have been infected. We would like to know the minimal fraction Q of infected individuals which must be completely isolated so that the outbreak is almost surely controlled without having to take any other measures. The answer, as we shall now see is that .

The generating functions and can both be written as in Eq. 2.4 with the exception that for . Hence, and by Proposition 1 we findThen, the condition required to apply Theorem 3 turns intowhich corroborates the common intuition behind the basic reproduction number. For example, suppose there is an outbreak of disease for which each infected individual is expected to infect 3 others if nothing is done to prevent it, i.e. , then we expect that in order to cut the chain of transmission less than a third of the infections can be allowed to transmit the disease. Indeed, from the above computation, at least of the whole infected must be detected and isolated.

Finally, in the case when we shall now prove that there is still a positive, but not certain, probability that the infection disappears.

Theorem 4

If , then and .

In this setup we consider the function used in the proof of Proposition 2. This satisfies , and by Proposition 1 and so for sufficiently small nonzero . Thus, again the intermediate value theorem shows the existence of a zero of f which we shall denote by . Recalling that zeros of f correspond to fixed points of which by Proposition 2 has only and 1 as fixed points. Thus, in this case we can also have .

Remark 4

(Lower bounds for and ) From the fixed point equation and writing the generating function as in Remark 3, i.e. , we find thatand so and soFurthermore, we can equally write . Said in other words, we find that the probability of the infection eventually dying out is at least the fraction of infected individuals which are completely isolated.

We shall now consider the case when . We go back to the setup in the proof of Proposition 2 and Theorem 3, namely we consider the function whose zeros correspond to the fixed points of the map . We have seen that in (0, 1) and . Under the hypothesis that we have and as for we find that f is negative immediately before . Hence, if there was another zero of f, between the function f would have a minimum which contradicts in (0, 1). We then conclude that also in this case

Lower bounds on

In this section we shall elaborate on the question raised in Remark 4, namely: Whether it is possible to find lower bounds on the probability that the chain of infections eventually dies out not leading to an epidemic.

To answer the question raised we proceed by direct inspection of the fixed point equation in Proposition 2. Start by noticing that all the terms in the Taylor series for are positive as one can check from its definition in Eq. 2.4. Thus, as we find that is larger than the zeroth order term of , i.e.and from the monotonicity of Z(x) we then havewhich is itself grater than Z(0).

Examples

In the simplest nontrivial example we can consider a population in which individuals have to kinds of interactions: a close and continuous interaction with their family and friends, and a a more distant sporadic interaction with not so close friends and other people which cross their path, by chance, in their daily lives as they commute to work and so on. Of course, we expect the probability of transmitting the disease to be larger in the first case and so assign to a it a larger transmissibility than to the second interactions , i.e. .

Delta and Poisson

In this first example we assume for simplicity that all person have the same number, , of close contacts and their sporadic contacts follow a Poisson distribution with intensity . In formulas, we haveThen, we find the generating function for the degree distributionfrom which we compute and soIn order to compute we may first find the generating function for the ramification distribution. This can be done using Eq. 2.3 which yieldsWe can finally use this to compute the generating function using the formula 2.4. This givesand using it we can compute the basic reproduction number , which by Proposition 1 isor perhaps in a somewhat more suggestive mannerNotice in particular that scales homogeneously with degree 1 as a function of .

Remark 5

Notice that in the case , i.e. if only the sporadic contacts occur, we have , while if meaning that all sporadic contacts are cut out, then .

Also, by evaluating the generating function at we findIt then follows from the discussion in Sect. 4.4 thatand

Polynomial and Poisson I

In this second model we will elaborate slightly on the first example, in the sense that we still assume everyone to establishes sporadic contacts following a Poisson distribution with intensity . On the other hand, we shall encode the distribution of close contacts by a polynomial of degree . Furthermore, let be the average degree of the distribution of close contacts, i.e. . In formulas, we haveAs in the previous case, we can now find the generating function for these degree distributionsand compute . Then,and to compute we start by obtaining the generating function for the ramification distribution. Equation 2.3 yieldsFinally using Eq. 2.4 as before, we findfrom which we can compute using Proposition 1, i.e. using the formula . This requires computing the derivative of which readsand so

Remark 6

In order to evaluate more clearly, we must understand what is . This can be computed for any distribution with generating function . Indeed, from direct differentiationas the terms vanish an the terms cancel.

Inserting this into the previous formula for yieldsNotice in particular that scales homogeneously with degree 1 as a function of .

We turn now to compute lower bounds on and following the strategy of Sect. 4.4, from which we infer thatwhile

Polynomial and Poisson II

We shall now continue with a population organized as in the previous example but we assume that a fraction f of the population decides to social isolate and cut its sporadic contacts. One can imagine this can be done by not going into public transport, reducing contact with unknown people and working from home. To implement this we modify the previous example by writingAs in the previous case, we can now find the generating function for these degree distributionsand compute . Then,andfrom which we can computefrom which we can compute using Proposition 1, i.e. using the formula . This requires computing the derivative of which readsand evaluating this at yieldsAs before, if , we find the lower boundsand

Poisson and Poisson

In this final example we split the contacts again in two groups, the familiar close interactions and distant sporadic ones. In contrast to the previous examples we shall assume a Poisson distribution for both of these classes of contacts having intensity and respectively. As in Example 5.3 we will be assuming that a fractions of the population are isolating by cutting their contacts. To work with some generality we will assume that fraction of the population cuts its close contacts and another fraction 2 cuts its sporadic contacts. Then, the degree distributions of the relevant trees and arewith the corresponding generating functions beingand compute . Then,whilewhere in the first sum . We now compute using Proposition 1 which requires computing . This yieldsIf , we have the lower boundsand

An alternative formula for

This section is motivated by the examples in previous and in finding a more amenable general formula to compute . We start with Proposition 1, namely the equationand the Eq. 2.3 which we rewrite here for simplicityUsing this, we computeand from the discussion in Remark 6 we further findwhere we have used to denote the standard deviation of the degree distribution of the tree . Then, inserting this into the formula for yields the formulae in Theorem 2 which we shall restate here for convenience.

Theorem 5

For each let denote the mean degree of the tree and its standard deviation. Suppose that a fraction Q of all infected individuals is completely isolated and does not transmit the disease to anyone. Then, if the probability of transmitting the disease along an arm of the tree is , the basic reproduction number isor

Applications

We shall now apply these results to a realistic scenario where a disease spreads through a population. Our goal is to investigate the possibility of an effective combination of isolation of infectious individuals, and practicing of social distancing, which together are capable of bringing below the threshold of 1. When that is not possible we will compute the probability of an epidemic developing.

Random contacts occurring with two different constant rates

We shall use the setup of Example 5.4 where the contacts established by the population in two groups. These are the close and distant contacts encoded in the trees and respectively. As in that example we assume both degree distributions to be Poisson, having intensities and . The fractions of the population which are cutting their close contacts is and that cutting its sporadic contacts . We will also be assuming that as it seems reasonable to assume that everyone which cuts its close contacts also cuts its sporadic ones.

As computed in Example 5.4, the basic reproduction number isWhen no intervention is made all , and Q vanish the disease is free to propagate and the corresponding basic reproduction number will be denoted by

Example 1

Suppose for the sake of simplicity that . Then, can be reqritten in terms of as followsThen, the condition that which is sufficient to contain the outbreak turns intoFor example, suppose that , , and , . Then, which is actually a reasonable assumption a disease such as Covid-19 and the computation above yields which seems extremely difficult to achieve. On the other hand, if while all other parameters remain the same which seems a much more achievable goal.

The previous example assumes that one can also cut the close contacts and that is not a reasonable assumption in most situations. For instance, many people may leave in the same house in which way it is not possible to cut such contacts. In the next example we address this and suppose only the sporadic contacts are cut.

Example 2

Only cutting the sporadic contacts corresponds to having . Then, can be isand the condition that becomesWe shall now use the same numerical values as in the previous example, , and , . Then, we find from the previous computation that for any value of . Hence, there is no way of surely containing the disease simply from cutting out the sporadic contacts and not increasing Q above 1/2. Suppose then the borderline case when and consider the setting where also of the population is capable of cutting their sporadic contacts. Then, we havefrom which we findi.e. the probability of an epidemic forming, which is given by , is approximately 0.406.

Some country based analysis

We shall now due some country based analysis using the household composition from the United Nations database United Nations (2019). This analysis is motivated by the current pandemic of Coronavirus. However, given the existence of insufficient knowledge to estimate the and the rude data from the beginning of the outbreak, this section should be regarded as within the realm of academic exercise. We shall use a simple model with two trees and with the first modeling the close interactions and given by the household composition of the respective country with the degree distribution being encoded in a generating function which is then a polynomial. The second tree is intended to model the sporadic interactions in public gatherings. Assuming these to occur as a constant rate, the most suitable degree distribution is the Poisson of some intensity, say , so that . Having this in mind, we shall use the model of Sects. 5.2 and 5.3 above. In all cases we shall consider, we have from those examplesso that writing and

Germany

We start with the case of Germany, which according to the United Nations database United Nations (2019) has a household distribution in which of the population live alone, with one or two other individuals, with three or four other and with more than five other. The reader may note that we have not stated what is the exact percentage that live with only one or two other individuals. This is because such data is hard to locate for a large number of countries and we believe it will make little different in the qualitative, but also quantitative errors such as estimating the transmission probabilities and . Indeed, there are more serious issues contributing to quantitative deviations. Thus, we will assume that half of the corresponding live with one other individual and the other half with two other ones. Similar remarks hold for the of the population that lives with three or four other individuals.

Remark 7

Of course, if all households remain isolated there is no way for the disease to spread from one household to another. However, we know this is not a realistic situation as in general there are close family ties connecting individuals in living in different houses. Given the difficulty in quantifying this we shall simply use the household composition to model the degree distribution of the tree .

Based on the data mentioned above we shall assume thatAs for the transmission rates, we will set , i.e. the probability of infecting a close family member is 2/3. In a similar way, we shall assume that the probability of infecting one of the sporadic contacts is and the average number of sporadic contacts is . This includes everyone that an infected person stays next to in public transport, markets, restaurants, work and other common areas. Of course, these numbers are debatable and we have chosen these simply to illustrate the theory. Using them, and assuming that both and we compute that andThen, iterating six times the sequence appears to stabilize around which givesSee Fig. 1 where the intersection point can be visualized graphically.

Fig. 1

The function for Germany

Hence, according to the model there is a very slight chance, of approximately that the outbreak will not lead to an epidemic. Of course, this assumes that no non-pharmaceutical interventions have been put in place to control the outbreak. Suppose for instance that strict social distancing outside the household is imposed so that everyone adheres to it, which corresponds to . Then, a computation shows that and so is not yet enough to almost surely guarantee that the outbreak will not lead to an epidemic. Also, getting all the population to cut its sporadic contacts seems very difficult to achieve in practice. A more efficient and easier to achieve strategy seems to be that of identifying and isolating infected individuals and we shall now analyze it. For example, motivated by the case of Covid-19 let us consider a disease for which around or of individuals are asymptomatic. These estimates where obtained for Covid-19 in airport screening and the data from the Diamond Princess cruise ship, see Quilty et al. (2020), Sun et al. (2019). However, the validity of extrapolating thezse estimates to the remaining population is debatable as other studies seem to have quite disparate estimates for the fraction of asymptomatic carriers, see for instance Mizumoto et al. (2020). Let us say then, that at least 0.18 of all infected individuals will not be detected, i.e . We choose, for simplicity and which yields and the outbreak will almost surely be contained. As another example, suppose that which is easier to achieve practically, then even with we have and so it would be needed which, again, is very difficult to achieve in practice. The conclusion is that, to contain the spread of the disease, it is much easier and effective to quarantine a sufficient fraction of infected individuals effectively than to simply cut the sporadic contacts of a large fraction of the entire population.

Italy

We shall now consider the Italian case, and based on United Nations (2019) we shall assume thatThen, using the same parameters as in the previous example we compute and so thati.e. there is chance of that an outbreak can be avoided without taking any precautions. Still, as a matter of comparison we find that if then which even though below 1 is visibly higher than that of the previous example.

France

For modeling France, we setwhich with the same parameters of the previous two examples yields and which givesyielding a probability of to avoid an outbreak. When we compute which being below 1 is larger than that of the previous two examples.

Portugal

For Portugal, the United Nations database United Nations (2019) suggests usingAgain, with the same parameters used in the previous examples we find and . This giveswhich yields a probability of to avoid an outbreak. In this case, when we find which in contrast with the previous examples is already above 1. Thus, by Theorem 3 the probability of avoiding an epidemic is below 1. Nevertheless, a computation shows that which is still quite high.

Spain

In the case of Spain we assume, using the same reference, thatAgain, with the same parameters used in the previous examples we find and . This giveswhich yields a probability of to avoid an outbreak. In this case, when we find as in the previous example. Indeed, the whole situation is very much parallel as that of the previous example.

Brazil

For Brazil we haveand with the same parameters of the previous examples we find which is the highest so far. Using this distributions we compute which haswhich yields a very small probability of to avoid an outbreak. In this case, even when we find which is still high. Indeed, the associated probability of avoiding an outbreak is , i.e. there is a chance of of containing the outbreak.