Epidemic Models. Maaike Koninx July 12, Bachelor thesis Supervisor: dr. S.G. Cox

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1 Epidemic Models Maaike Koninx July 12, 2015 Bachelor thesis Supervisor: dr. S.G. Cox Korteweg-de Vries Instituut voor Wiskunde Faculteit der Natuurwetenschappen, Wiskunde en Informatica Universiteit van Amsterdam

2 Abstract In our everyday lives, we are challanged with many questions concerning our health. In this thesis, we will explore the spread of infectious diseases using a mathematical approach. Our primary goal is to examine multiple epidemic models and their related traits that contribute to the occurrence of an epidemic. We will consider discrete and continuous time epidemic models and argue that for all models the same general result emerges: in the supercritical regime an epidemic may occur while in the subcritical regime the disease will surely die out. For our discrete time models, we will compute the probability that an infectious disease will die out and discuss the probability that a certain proportion of the population is infected. For our continuous time models, we will explore the conditions that will lead to a decrease in the number of infected people. We will argue that if we view the number of infected people as a Markov chain on N, the number of infected people approximates a differential equation in large populations. Title: Epidemic Models Author: Maaike Koninx, maaike.koninx@student.uva.nl, Supervisor: dr. S.G. Cox Second grader: dhr. dr. B.J.K. Kleijn Date: July 12, 2015 Korteweg-de Vries Instituut voor Wiskunde Universiteit van Amsterdam Science Park 904, 1098 XH Amsterdam 2

3 Contents 1. Introduction 4 2. The Galton-Watson branching process 6 3. Dicrete time SIR epidemic models The Reed-Frost model The Erdős-Rényi model Simulation Continuous time epidemic models The deterministic model Kurtz s Theorem Proof of Lemma Simulation Conclusion Populaire samenvatting 33 Bibliografie 35 A. Appendix A 36 B. Appendix B 37 B.1. Supporting theorems for Chapter C. Appendix C 38 C.1. The number of infected people in an Reed-Frost epidemic C.2. The number of infected people on G(n,p) C.3. The mean number of infected people for 100 simulations on G(n,p) C.4. Number of infected people in a Markov chain model C.5. Accessory function for C

4 1. Introduction In our everyday lives, we are challanged with many questions concerning our health. New infectious diseases get discovered at a regular basis, while older, well-known diseases are still present in our world today. Some of the most well-known infectious diseases have played a major part in history, and even altered its course. For example in the 1300s a series of epidemics of the plague, also known as the Second plague pandemic, that started with the Black Death wiped out 30 to 70 percent of the population in Europe. Since then, many more outbreaks of the plague 1 have been well documented, as well as major outbreaks of cholera, influenza, measles, yellow fever, etc. The world is becoming increasingly interconnected and naturally a concern for our health arises. In this thesis, we will explore the spread of infectious diseases using a mathematical approach. Figure 1.1.: A depiction of the Black Death from a 15th century bible. Our main concern for this thesis is in which circumstances an epidemic occurs. We will define an epidemic as the presence of a certain amount of infected individuals of an infectious disease in a population for a substantial amount of time. In general, an epidemic may occur when a single individual or small proportion of the population is exposed to an infectious disease. If the infectious individuals make contact with other members of the population, the disease is spread throughout the population. Whether or not an epidemic occurs, is determined by certain traits of the disease as well as the population. When the population is exposed to the disease, an epidemic may or may not occur according to these traits. For example, high rates on the infectiousness of the disease and the rate at which people make contact in a population may lead to an 1 at the time of writing, an outbreak was documented in Madagascar 4

5 epidemic, while other traits like the rate at which sick people recover may lead to a decline in the number of infected people. The primary goal of this thesis is to examine multiple epidemic models and their related traits that contribute to the occurrence of an epidemic. Furthermore, if an epidemic occurs, we are interested in the number of people that become infected and the probability that the disease will die out eventually. We will support our theoretic findings with computer simulations. We argue that the same, general result emerges for all of our epidemic models, which is the following; If the expected number of people a single infected person infects is greater than one, an epidemic may occur. If the expected number of people he infects equals or is less than one, the disease will surely die out. In Chapter 2, we consider the spread of an infectious disease in a branching process and we will compute the probability that an infectious disease will die out. In Chapter 3 we discuss a discrete time epidemic-model, where we will argue that the Reed-Frost SIR model is equivalent to an E-R random graph given the same initial state. Here we will discuss the probability that a certain proportion of the population is infected. In Chapter 4, we introduce two continuous time epidemic models. We argue that if we view the number of infected people as a Markov chain on N, we find by applying Kurtz s Theorem, that the number of infected people approximates the solution to a differential equation in large populations. Acknowledgments. I would like to truly thank my supervisor Sonja Cox for offering me guidance throughout this project. Enjoy! 5

6 2. The Galton-Watson branching process For our first exploration of epidemics, we consider a simple branching process in which an epidemic starts at a so-called patient zero and spreads with a certain probability to a number of his contacts. In their turn, the infected individuals infect their contacts with a certain probability. This process continues as long as the disease is passed on between generations and is called the Galton-Watson branching process. 1 Definition 2.1. Let the random variable N represent the number of people each infected individual spreads the disease on to. The distribution of N is referred to as the offspring distribution, and its expected value R 0 = EN] is called the basic reproductive number. We are interested in the probability that the disease lives on for a certain amount of time. The probability p ext n that the disease has died out by the nth generation is a sequence in 0, 1] that converges to a point p ext, the probability that the disease dies out eventually. By Theorem 2.2, the probability that the epidemic either dies out or continues to exist indefinitely depends completely on the offspring distribution. If the number of new infections an infected individual causes is strictly less than one, the disease will surely die out. If this number is greater than one, the disease may continue to exist indefinitely. This is stated in the next theorem. Note that in Theorem 2.2 below we do not consider the case where P N = 1] = 1. In this case, every infected individual spreads his disease to exactly one individual. It is clear that p ext n = 0 for every n N, and therefore p ext = 0. Theorem 2.2. Conditional on P N = 1] < 1, the following regimes occur: (a) If R 0 1 then p ext = 1. (b) If R 0 > 1, then p ext < 1. Proof. Let X n be the number of infected people in the nth generation. We view an individual i from the 1 st generation and notice that if i is infected, he can be considered as patient zero in his own branching process. In particular, if Z n,i represents the number of infected people in the nth generation after i, (in)directly caused by i X 1, then Z n+1,i X n. Furthermore if i = 1, 2,... X 1, the {Z n,i } are i.i.d by definition and we 1 This model is best known not for its application in epidemics, but for its application in genealogy. The model was first introduced by Francis Galton to study the extinction of family names. The terminology used in this chapter is therefore closely related to common word usage in genealogy. 6

7 can deduce the following Es X n+1 : X 1 = j] = Es j Z n+1,i ] = Es Z n+1,1 s Z n+1,2 ] i.i.d = ( Es Z n+1,1 ] ) j = ( Es X n ] ) j. (2.1) Now we can introduce the probability generating function φ n (s) of X n, by φ n (s) = Es Xn ] = 2.1 = = Es Xn : X 1 = j]p X 1 = j] j=0 ( Es X n 1 ] ) j P N = j] j=0 (φ n 1 (s)) j P N = j]. j=0 Note that φ n (0) = P X n equation gives us = 0] = p ext n. Consequently, setting s = 0 in the previous p ext n = j=0 ( p ext n 1) j P N = j]. (2.2) Since p ext n converges to a unique point p ext in 0, 1], we know that p ext n p ext n 1 as n. We will now argue that by setting p ext n = p ext n 1, the smallest solution to the equation (2.2) in 0, 1] gives us p ext. This will finalize our proof. Let f : 0, 1] R be defined by f(x) = j=0 xj P N = j]. We deduce the following: (i) f is non-decreasing as f (x) = j=1 jxj 1 P (N = j) 0, for all x 0, 1]. (ii) f is strictly convex, indeed since P N = 1] < 1 and EN] = 1, we know that P N 2] > 0. Therefore f (x) = j=2 j(j 1)xj 2 P (N = j) > 0 for all x 0, 1]. (iii) f(1) = j=0 P N = j] = 1. We can see from Figure 2.1 that in the cases where (a) R 0 1 and (b) R 0 > 1 the following is true: (a) In the case where R 0 1, the strict convexity of f together with the condition that f (1) = EN] = R 0 1 force the graph of f to lie strictly above the function y = x on 0, 1). This implies that the sequence p ext n is increasing on 0, 1) and will converge to p ext = 1. (b) In the case where R 0 > 1, the condition that f (1) = R 0 > 1 and the above statements force f to lie under the function y = x on the interval (a, 1) for some a < 1 7

8 (a) (b) Figure 2.1.: A graph of f in the case where (a) R 0 1 and (b) R 0 > 1. and that a is the unique solution to f(x) = x in 0, 1). Indeed, if p ext n is a decreasing sequence and if p ext n the sequence p ext 0, a) then (p ext n n will converge to a = p ext < 1. (a, 1) then (p ext ) is increasing. We conclude that Definition 2.3. We will denote R 0 < 1 as the subcritical regime, R 0 = 1 the critical regime and R 0 > 1 as the supercritical regime. The previous theorem tells us that if we know the value of R 0, we can predict whether a disease will die out or persist indefinitely in a population. We refer the interested reader to Appendix A for an overview of well-known infectious diseases and their basic reproductive numbers. In particular, if we know the distribution of N, we can calculate the extinction probability of the disease by finding the minimal non-negative solution to 2.2. We conclude this chapter by examining the extinction probability for some distributions of N. Example 2.4. Let 0 < p < 1 and consider p k = P N = k] = p k (1 p) for k = 0, 1,... as the probability of infecting k people. Clearly p k (0, 1) for all k = 0, 1,..., and these probabilities add to one seeing as k=0 p k = (1 p) k=0 pk = (1 p)(1 p) 1 = 1. Therefore the set {p k } is a probability distribution 2. Suppose that p 1. Since 2 n ) EN] = kp k = (1 p) kp k p = (1 p) (1 p) = p 2 1 p k=0 k=0 we know that EN] 1. Seeing as p 1 < 1, Theorem 2.2 tells us that p ext = 1. In this case, the disease will almost surely die out at a certain point in time. Now suppose p > 1 2, then EN] > 1 and by Theorem 2.2 pext < 1. We are interested in finding the value of p ext. Therefore we have to find the minimal non-negative solution 2 In fact, N is Geometric(1 p) distributed 8

9 to φ(x) = x for φ(x) = x k p k k=0 = (1 p) x k p k k=0 1 = (1 p) 1 xp Some calculations are required to find that φ(x) = x on 0, 1] when x equals 1 or 1 p. p Seeing as p ext < 1, we find that p ext = 1 p. In this case, the disease will live on p indefinitely with probability 1 p ext = 2p 1. p Example 2.5. We can consider the spread of an infectious disease not only by counting infectious individuals but also by counting the healthy ones. Whenever a strand of DNA is exposed to a certain amount of radiation, its molecular composition can be permanently altered. These errors in genetic material are called mutations. Suppose that a healthy strand of DNA replicates itself during radiation and the number of healthy replicas N is Poisson distributed with parameter λ. The healthy replicas on their turn replicate themselves, with the same offspring distribution N. This process is again a Galton-Watson branching process and EN] = R 0 = λ, so the Poisson distribution has a parameter equal to the expected number of non-mutated replica DNA strands. The probability p ext that the healthy DNA strands die out in a certain point of time is the minimal non-negative solution to x = φ(x) = k=0 x k λ k e λ k! = e λ (xλ) k k! k=0 = e λ(x 1) (2.3) The exact solution for this equation cannot be obtained analytically, but we can obtain it numerically. For instance if λ = 2 we get p ext = , which tells us that when we expect every healthy DNA strand to generate 2 healthy replicas, the probability that the healthy DNA strands die out is about 20.3%. 9

10 3. Dicrete time SIR epidemic models As we saw in the Chapter 2, the Galton-Watson branching process is an effective model for describing the extinction rate of an infectious disease that starts at a patient zero. However, the Galton Watson branching process is hardly an effective model for describing epidemics in real life. In real life epidemics, a healthy person can extract a infectious disease from multiple infected individuals he meets. In the Galton-Watson process a healthy person can only extract a disease from his direct contact in the former generation. Furthermore, in real epidemiology, infectious diseases often do not get discovered until a certain proportion of the population is sick. One cannot easily trace the generations in which the disease has spread much less find patient zero. Hence we need to find another model that describes the spread of infectious diseases without assuming that the process starts at a patient zero and allowing the disease to reach healthy people through multiple channels. One such model is called the Reed-Frost model The Reed-Frost model The Reed-Frost model is a discrete time SIR epidemic process. A SIR epidemic process involves three types of individuals: susceptible (S), infectious (I) and resistant (R). In a population of n individuals, during each unit of time, an infected person infects each susceptible with probability p independent of other infected people. The infected person is then assigned as resistant. During the following unit of time, the process continues for the newly infectious group of people. Once resistant, a person cannot become susceptible again, since being resistant equals being immune. The process kan be represented as S I R A more detailed description is as follows. Let S(t), I(t) and R(t) be the number of susceptible, infectious and resistant people respectively at time t. From the assumptions it is clear that S(t) + I(t) + R(t) = n. Then Z(t) = (S(t), I(t), R(t)) {1, 2,..., n} 3 is a random variable that represents the state of each of the n individuals at time t. Let z, z {1, 2,..., n} 3 be two states. If the number of susceptibles S(z) in state z equals S(z) = I( z)+s( z), then it is possible to reach state z from state z in one step (notation: z z). If {Z(t)} t N is a random walk on {1, 2,..., n} 3, and z z is possible, then the 10

11 transition probabilities equal ( ) S( z) ( ) S(z) S( z) S(z) P Z(t + 1) = z : Z(t) = z] = )((1 p) I(z) 1 (1 p) I(z) (3.1) S( z) ( ) ) S( z) ) I( z) S(z) = ((1 p) (1 I(z) (1 p) I(z). S( z) }{{}}{{}}{{} (i) (ii) (iii) Explanation. (i) The number of ways to obtain S( z) susceptibles out of S(z) susceptibles. (ii) The probability that S( z) people do not get infected. Obviously, for each susceptible, the probability of not getting infected is (1 p) I(z), so for all S( z) susceptibles this probability equals term (ii). (iii) The probability that I( z) people get infected. The probability for each susceptible of getting infected is 1 (1 p) I(z), so for all I( z) this probability equals (iii). Note that one has to encounter only one infected person to possibly get sick, but we do not exclude the case where one can meet multiple infected persons. Therefore, this probability does not equal p I(z)I( z). The reader can check that these probabilities add to 1 on the set { z : z z is possible}. Remark. Suppose that {Z(t)} t N is a random walk on {1, 2,..., n} 3. Then the conditional transition probability P Z(t + 1) = z : Z(t) = z,..., Z(0) = z 0 ] depends only on the present state Z(t), i.e. P Z(t + 1) = z : Z(t) = z,..., Z(0) = z 0 ] = P Z(t + 1) = z : Z(t) = z]. This means that the future process depends only on the present state and not on the history of the process. We call this memoryless property the Markov property, making Z(t) a Markov Chain on {1, 2,..., n} 3. As can be seen from equation 3.1, the conditional random variable S(t + 1) Z(t) has a binomial distribution of parameter (S(t), (1 p) I(t) ) 1. Using the law of large numbers, it can be shown that this distribution approximates a Poisson distribution in a population where n and I(t) is a finite number, i.e. in a population with a large number of healthy individuals and a small fraction of infected people. Theorem 3.1. The Law of Large Numbers. Let t 0 be fixed and let λ = S(t)(1 p) I(t). Then for finite I(t) and n, S(t + 1) Z(t) is Poisson(λ) distributed. Proof. We write ˆp = (1 p) I(t), so that S(t + 1) Z(t) Bin(S(t), ˆp) and the generating 1 while I(t + 1) Z(t) has a binomial distribution of parameter (S(t), 1 (1 p) I(t) ) 11

12 function equals φ(s) = Es S(t+1) Z(t)] n ( ) S(t) = s k ˆp k (1 ˆp) S(t) k k k=0 = (1 + (s 1)ˆp) S(t) ( = 1 + (s 1) λ S(t) e (s 1)λ, ) S(t) as n, which is exactly the generating function of a Poisson(λ) distribution as we saw in Example 2.5. Now that we know that S(t + 1) Z(t) is Poisson distributed for large enough n, provided that S(t) increases accordingly, one might think this would make calculations involving S(t + 1) easier and consequently lead to insights into the Reed-Frost epidemic model. However, since each transition probability is dependent on the former state, the previous model fails to provide insight into the long-term infection rate. We therefore introduce a new model based on random graphs which we will prove is equivalent to the Reed-Frost model The Erdős-Rényi model. The Erdős-Rényi (E-R) random graph G(n, p) is a graph on n nodes where each two nodes are connected with probability p, independent of how other edges are connected. Our model then simulates an epidemic in the following way. A finite number of nodes, which we will refer to as I(0), is infected at random at t = 0. During the next unit of time, each infected node infects his direct neighbours after which he is assigned as resistant. This process continues until all nodes in the connected component of each infected individual are infected. Now, if S(t) and I(t) represent the number of susceptibles and infectious at time t respectively, it can be argued that the conditional transition probability P S(t + 1) = z : S(t), I(t)] in our new model equals that of the Reed-Rényi model since it is exactly (i) (ii) (iii) as in 3.1, see Example 3.2. We conclude that the two models are equivalent given that both models start with the same amount of infected people I(0). Example We now consider two states z, z, of a graph in which the nodes are assigned as susceptible, infectious or resistant, see Figure 3.1. Suppose it is possible to reach state z from state z in one timestep, i.e. all susceptible nodes in z are assigned as either susceptible or infected in state z. The probability that all nodes were connected 12

13 accordingly at t = 0 in our model equals (i) (ii) (iii) as in 3.1, since (i) There are ( S(z) S( z)) ways of connecting the infectious nodes in state z to the infectious nodes in state z. (ii) The probability that the susceptible nodes in z were not connected to the infectious ones in z equals (1 p) I(z)S( z). (iii) The probability that the infectious nodes in z were connected to the infectious ones in z equals (1 (1 p) I(z) ) I( z). z (a) z z Figure 3.1.: Two states z, z on a graph of 6 nodes. The green nodes are susceptible, the red ones are infectious and the yellow ones are resistant. At least one of the solid drawn connections have to be made at t = 0 to reach state z from state z in one timestep. Through our new model we can deduce some valuable results. We can obtain an upper bound for the number of people that get infected in terms of R 0, the expected number of people that every infected person spreads his disease on to. It is clear that R 0 equals the expected number of neighbours pn of an infected person. The theorem below states that in the subcritical regime, when R 0 < 1, all connected components are sufficiently small so that there is at most a small outbreak. In the supercritical regime a so-called giant component appears, which leads to a large outbreak. We denote by C 1 and C 2 the largest and second-largest connected component respectively in G(n, p). Let C i the number of nodes in C i. Theorem 3.3. In an E-R graph G(n, p) with R 0 = pn, the following regimes occur. (i) Subcritical regime: If R 0 < 1 then there exists a constant c depending on R 0, so that lim P C 1 c log(n)] = 1. n (ii) Critical regime: If R 0 = 1 then for all a > 3/2 and all δ > 0 ] lim P C1 δ = 0. n n α Furthermore, if p ext R 0 is the extinction probability of a Galton-Watson branching process with N P oisson(r 0 ), i.e. p ext is the unique solution to 2.3, then the following holds. 13

14 (ii) Supercritical regime: If R 0 > 1 then for some constant c depending on λ and all δ > 0 ] lim P C 1 (1 pext R n n 0 ) δ and C 2 c log(n) = 1. The proof for this theorem exceeds this chapter but we refer the interested reader to 3]. We used Netlogo (version 5.10) to simulate an E-R graph for various values of R 0. For 100 nodes, we connnected each pair with probability p = R 0 /100. The results agreed with Theorem 3.3. Indeed, the largest connected components in the subcritical and critical regimes were relatively small while in the supercritical case, one can easily see the giant component emerge. See Figure 3.2. (a) R 0 = 0.5, C 1 = 6 (b) R 0 = 1, C 1 = 13 (c) R 0 = 2, C 1 = 82 Figure 3.2.: Simulation of a E-R random graph on 100 nodes. The red nodes are part of the largest component C 1. Only the connections of C 1 are shown. The previous theorem leads to some interesting results regarding the number of people that get infected when we simulate the spread of an epidemic. We will discuss this for each case: (i) For R 0 < 1 the largest component is at most of logarithmic size with respect to the total population almost surely when n is large, that is C 1 c log(n) a.s. for some c > 0. This implies that every component is at most of logarithmic size. When a finite number I(0) of the nodes are infected, a maximum of I(0) c log(n) people eventually become infected. Hence, the outbreak is restricted to a logarithmic sized proportion of the population. Seeing as I(0) c log(n) n 0 as n this is a negligible proportion of the total population. Therefore this will result in a small outbreak. (ii) For R 0 = 1 the size of the largest component is at most of order n 2/3 with high probability for large n, i.e. C 1 cn 2/3 with high probability for some c > 0. Therefore, 14

15 with high probability a maximum of I(0) c n 2/3 people are infected for some c > 0 and there is only a small outbreak of the disease, as I(0) c n 2/3 n 0 as n. (iii) For R 0 > 1, the size of the largest component in proportion to n converges in probability to 1 p ext R 0 as n. That is, in large populations a proportion of 1 p ext R 0 of the population is interconnected i.e. C 1 n (1 pext R 0 ) as n. We call C 1 the giant compontent. All other components including the second largest component are at most of logarithmic size almost surely. If a person in these logaritmic sized components is infected, the disease will be limited to a negligible proportion of the total population as n as we saw in (i). If a person in the giant component is infected this will almost surely lead to a large outbreak of the disease where at least a proportion of 1 p ext R 0 of the population is infected Simulation. We ran a simulation to find out more about the amount of sick people in a population of n people and to verify the previously stated theoretic bounds. In this section, we are interested in the mean number of infected people for R 0 = 0.5, 1 and 2 as a function of n. First, we used Matlab (version R2014b) to simulate the spread of an epidemic based on the Reed-Frost epidemic model. For every R 0, we chose a certain n and p = R 0 /n accordingly. Also, we chose I(0) = 10. For our simulation, we allowed the infected people to infect the susceptibles in each iteration, after which they were labeled as resistant. Every new amount of susceptibles was selected out of all susceptibles by using the rand function and transition probabilities stated in 3.1. See Appendix C.1 for the script. Although this procedure ran flawlessly for small n, the program failed to accurately produce all transition probabilities for n > 68. One might explain this by looking back at the transition probabilities in 3.1 and noticing that for S(z) > 68 and some values of S( z) ( ) S(z) > S( z) Any value stored as a double requires 64 bits. Hence values larger than can not be accurately stored as a double. It is therefore favorable to simulate the E-R graph, which is equivalent to the Reed-Frost model but does not require the same calculations. Again, we used Matlab to generate an E-R graph. By using the binornd function, we generated a symmetric n n adjacency matrix in which every two nodes are connected 15

16 with probability p. Next, 10 nodes were assigned as infected and during the next iteration we allowed them to infect their direct neighbours. During the iterations, the infected nodes infected their direct neighbours until there were no infected or healthy nodes left. See Appendix C.2 for the script. For every value of R 0, we repeated the previous script for increasing values of n with p = R 0 /n until the script exceeded a cpu of 300s. For every n, which we chose as a power of two starting from n = 2 4 = 16, we ran the script 100 times and calculated the mean number of infected people. See Appendix C.3 for this script. Results In the subcritical regime, the mean number of infected people was indeed bounded by c log(n) for c = 4. When fitted to an logarithmic function, the data was best fitted to Y = 1.64 log(n) Though as can be seen from Figure 3.3, the data disagrees with this fitting for some n. This could be explained by variance or maybe the data is best fit to another, non-logarithmic function. In the critical regime, the mean number of infected people was clearly bounded by and well-fitted to a function of the form c n 2/3, see Figure 3.3. The data appeared to fit the function Y = 1.10 n 2/ very well, confirming the theoretic upper bounds from Theorem 3.3 for c = Figure 3.3.: The mean number of infected nodes in an E-R graph for I(0) = 10. The blue markers represent the mean number of infected people, and the dotted lines represents the upper bounds stated in Theorem 3.3 for various values of c. Above row: R 0 = 0.5, total cpu = 379s. Lower row: R 0 = 1, total cpu = 399s. In the supercritical regime, the mean proportion infected people appeared to be converging towards 1 p ext for increasing values of n, suggesting the appearence of a giant component. See Figure 3.4. The difference δ between the data and 1 p ext is positive for every n, which tells us that in proportion to n even more people are infected than 1 p ext. The data does appear to be converging to 1 p ext, as we can see by the rapidly decreasing value of δ. 16

17 Figure 3.4.: The mean proportion of infected nodes in an E-R graph for R 0 = 2 and I(0) = 10. The blue markers represent the mean proportion of infected people, and the dotted line is the theoretic proportion of people in the giant component. The difference δ between the data and 1 p ext, is shown on the right. Total cpu = 379s. The size of I(0). The probability of a large outbreak of an infectious disease in supercritical regime is dependent on the size of I(0). If we take I(0) large enough, there is a substantial probability of infecting a node in C 1 and causing a large outbreak. In contrast, if we take I(0) small, there might not be a large outbreak. The probability of a large outbreak is dependent on I(0) as follows. Suppose we assign each of the initial infections randomly to a node. The probability of assigning an infection to a node that is not in the giant component is equal to p ext R 0, since the giant component is of size 1 p ext R 0 with respect to n. Therefore with probability (p ext R 0 ) I(0), none of the initial infections are assigned to nodes in the giant component. The probability that at least one of the infected nodes I(0) is in the giant component is 1 (p ext R 0 ) I(0). Consequently, with probability 1 (p ext R 0 ) I(0), there is a large outbreak in which at least a proportion of 1 p ext R 0 of the total population is infected. Example 3.4. Suppose R 0 = 2 i.e. the probability of infection is p = 2 n. Then pext = (see Example 2.5). Suppose I(0) = 1. For a population of size n, a proportion of 1 p ext = of the total population is infected with probability 1 p ext = as n. For I(0) = 2 at least the same proportion of the population is infected with probabiliy 1 (p ext ) 2 = Earlier, we assumed that I(0) was finite. Yet, there is a way of writing I(0) as a function of n such that the previous results still hold. For instance if we take I(0) = n α (log(n)) 1 for α < 1 in the subcritical case, a maximum of I(0) log(n) = n α people are infected. Since n α lim n n = lim n nα 1 0 this is still a negligible proportion of the total population. So this results in just a small outbreak. The same goes for I(0) = n α 2/3 with α < 1 in the critical regime, since again a maximum of I(0) n 2/3 = n α people are infected and this proportion again equals n α 1 0 as n. 17

18 The SIS epidemic process. We now consider the spread of a disease in which each infected person recovers and becomes susceptible again as opposed to becoming resistant as we saw in the SIR model. S I S An example of a SIS epidemic is the well-known common flu. We will now deduce some results based on previous observations. For I( z) = k the reader can check that the transition probabilities equal ( ) S(z) P Z(t + 1) = z : Z(t) = z] = (1 p) I(z)(S(z) k) {1 (1 p) I(z) } k. (3.2) k which implies that the conditional random variable Z(t + 1) Z(t) Bin(S(z), 1 (1 p) I(z) ). We now want to deduce some results that are similar to Theorem 3.3 that will predict whether there will be a large or small outbreak. Therefore we first have to find a random graph that will be equivalent to the SIS epidemic process. An obvious choice is a similar to the E-R graph, but with a slight alteration: each infected node infects his direct neighbours after which he becomes susceptible again. This model indeed has the same transition probabilities as 3.2, but unfortunately does not depict the same process. In the SIS epidemic model, every time a person becomes susceptible after infection, he becomes infected again in one timestep with probability (1 p) I(z). The same probability in our altered E-R graph is either 1 if the person has neighbours (seeing as they are surely infected), or 0 if the person has no neighbours. To properly adjust our altered E-R graph, we have to randomly reassign neighbours to every person that gets susceptible again after infection with probability p. Gaining further insight into this model will be complicated and unnessasary as there are easier ways of examining the SIS model. In Chapter 4 we will show that if the number of infected individuals in our SIS model is represented by a Markov chain X n (t) on N, we can construct a differential equation for which 1 X n n(t) approximates its solution for large values of n. 18

19 4. Continuous time epidemic models We now look into some continuous time epidemic models. In continuous time epidemic models, just like in real life, people can get infected at any point in time and stay infected for any amount of time. We will refer to the rate at which an infectious individual infects each susceptible as β, and the rate at which an infected person becomes immune as γ. Our main goal for this Chapter is to examine how β and γ contribute to the occurence of an epidemic and how many people are infected. In order to do so, we will have to look into large populations seeing as continuous time epidemic models are more advanced in small populations. The first continuous time model we discuss is a deterministic model which, as opposed to all previous models, contains no stochastic elements. Deterministic models are often described by differential equations and the output is fully determined by parameter values and initial conditions. Second, we will discuss a stochastic model where the number of infected people is represented by a Markov jump process. By applying Kurtz s theorem we argue that in large populations, the number of infected people approximates the solution to a differential equation. We will finalize this Chapter with a proof for Kurtz s theorem as well as a simulation of the above results The deterministic model. The SIR model We denote by s, i, r the proportion of susceptible, infectious and resistant people relative to n, i.e. s = S/n, i = I/n and r = R/n, and s(t)+i(t)+r(t) = 1 for all t 0. Clearly, this process can be represented as s β i γ r. We stated that an infectious individual infects a susceptible at rate β, where there are s susceptible and i infected people relative to n. Consequently, the rate at which all infected people infect all susceptibles equals βis. Also, since γ is the rate at which each infected person becomes immune, γi is the rate at which all infected people get immune. Therefore we obtain the following system of differental equations ds(t) = βi(t)s(t) dt (4.1) di(t) = βi(t)s(t) γi(t) dt (4.2) dr(t) = γi(t). dt 19

20 By setting ds(t) = di(t) = 0, we conclude that an equilibrium arises only when i(t) = 0, dt dt i.e. when the disease has died out. We consider two cases which eventually lead to an equilibrium, assuming that i(0) > 0. First note that the mean duration of an infection is γ 1 units of time. Since an infected individual infects each susceptible at rate β, the expected proportion of susceptibles he infects during his infectious period equals R 0 = β/γ. Seeing as R 0 is the reproductive number for this model, it is not surprising that the two cases we consider are related to the (sub)critical and the supercritical regimes. (i) s(0) R0 1. Note that R 0 1 always satisfies this condition. Since s(t) is decreasing for t > 0, one can deduce that { di = 0 if s(0) = R0 1 or i(t) = 0 (t) i(t)(βs(0) γ) dt < 0 if s(0) < R0 1. Seeing as s(0) = R0 1 is not an equilibrium, we conclude that the number of infected people will strictly decrease until i(t) = 0 and the disease has died out. No epidemic occurs. (ii) s(0) > R0 1. Note that this implies R 0 > 1. Clearly, di (t) = i(t)(βs(t) γ) dt > 0 if s(t) > R0 1 = 0 if s(t) = R0 1 or i(t) = 0 < 0 if s(t) < R0 1 Seeing as s(0) > R0 1, we deduce that the number of infected people will increase until s(t) = R0 1 for some t. Hence, an epidemic occurs. Since s(t) = R0 1 is not an equilibrium, the number of infective people then decreases until i(t) = 0 and the disease has died out. Obviously, no more than R0 1 of the susceptibles survive the epidemic. See Figure 4.1 how this event takes place in the subcritical and the supercritical regime. Figure 4.1.: Phase portrait for the equations 4.1 and 4.2 in the case where R 0 = 0.5 (left) and R 0 = 2 (right). 20

21 The SIS-model For our SIS-model we get a similar result, with the alteration that infected people become susceptible, as apposed to resistant, at rate γi. ds(t) = βi(t)s(t) + γi(t) dt di(t) = βi(t)s(t) γi(t) dt It is clear that an equilibrium occurs in either of the following conditions (i) i(t) = 0 and s(t) = 1 or (ii) i(t) = 1 R0 1 and s(t) = R0 1. Clearly di(t) > 0 if s(t) > R0 1 = i(t)(βs(t) γ) = 0 if s(t) = R0 1 or i(t) = 0 dt < 0 if s(t) < R0 1 implies that in the case where s(0) < R0 1 the number of infected people decreases until either s(t) = R0 1 and i(t) = 1 R0 1 if R 0 > 1 or in the case where R 0 < 1, the disease has died out. If R 0 1, this will imply that the disease has died out. In the case where s(0) > R0 1, and therefore R 0 > 1, the number of infected people increases until i(t) = 1 R0 1. We conclude that for all values of s(0), i(t), an equilibrium occurs between the proportion of susceptible and infected people where s(t) = R0 1 and i(t) = 1 R Kurtz s Theorem. Next, we will discuss a continuous time SIS model based on a continuous time stochastic process. We will first introduce the concept of a continuous time Markov chain. Let (X(t)) t 0 be a right-continuous process with values in a set N, which is called the state space. Let Λ = (λ x,y ) x,y N be a Q-matrix with jump martix Π. Definition 4.1. We call (X(t)) t 0 a Markov chain on a N with jump rates (λ x,y ) x,y N Λ, if for all t, h 0 conditional on X(t) = x, X(t + h) is independent of (X(s)) s t and as h 0 uniformly in t for all y N P X(t + h) = y X(t) = x] = δ x,y + λ x,y h + o(h), where δ x,y is the Kronecker delta, and o(h) is the little-o of h Remark. Recall that f(h) o(h) if and only if f(h) lim h 0 h = 0. 21

22 This implies that for any Markov chain (X(t)) t 0 on N and for any states x, y N the process jumps from x to y with certain probability upon expiration of a exponential distributed timer, whose parameter depends solely on the pair (x, y). In the remaining part of this chapter, we consider a family of Markov chains X n for n N on state space N. We will assume that for all n, X n has only a finite number k of jump directions, i.e. (e i ),...,k Z. For every n, wel define the functions λ i : R R + such that the transition rate of jumping to x + e i from x equals nλ i (x/n) for any state x N of X n. In this case, we call X n a birth and death process and the following holds for all x N P X n (t + h) = x + e i X n (t) = x] = nhλ i (x/n) + o(h) k P X n (t + h) = x X n (t) = x] = 1 nh λ i (x/n) + o(h). Example 4.2. Suppose X n (t) is a Markov chain on N that represents the number of infectives at time t in a population of n individuals. Then the jump directions are restricted to belong to { 1, 0, 1}. As we saw in the former paragraph, for jumping directions e + = +1 and e = 1 the corresponding functions λ ± equal λ (I/n) = γi/n and λ + (I/n) = βi/n(1 I/n). Therefore, for every state I N, the rate of jumping to I 1 equals nλ (I/n) = γi and the rate of jumping to I+1 equals nλ + (I/n) = β I(n I). n We obtain the following equations: P X n (t + h) = I + 1 X n (t) = I] = β I(n I)h + o(h) n (4.3) P X n (t + h) = I 1 X n (t) = I] = γih + o(h) (4.4) P X n (t + h) = I X n (t) = I] = 1 ( β I(n I) + γi n ) h + o(h) (4.5) We will now show that when n goes to infinity, the trajectories t 1 n X n(t) converge in probability to the solution x(t) of a ordinary differential equation. In particular this convergence is uniform and a.s. on 0, T ] for any T > 0. Theorem 4.3. Kurtz s Theorem Let ē := max{ e i : i = 1,..., k} for some arbitrary norm. Assume that λ = max λ i (x),...,k sup x R is finite and that the function F : R R defined by: F (x) = k e i λ i (x) 22

23 is Lipschitz-continuous, i.e. there exists a constant M such that F (x) F (y) M x y for all x, y R. 1 Assume that lim n X n n(0) = x(0) almost surely, and let x : R + R be the solution to the integral equation x(t) = x(0) + t 0 F (x(s))ds. (4.6) Then for any fixed ɛ, T > 0, for sufficiently large n, ] ( ( )) P sup 1 n X n(t) x(t) ɛ 2k exp nt λh ɛe MT 2kT λē, (4.7) where h(t) = (1 + t)log(1 + t) t. Moreover, lim 1 n n n(t) x(t) = 0 a.s. (4.8) This theorem provides a law of large numbers for a birth and death process. Equation 4.7 tells us that the right hand side of the equation converges to zero for n, we deduce that 1 n X n(t) converges in probability to x(t) for all fixed t. Moreover, this convergence is a.s. and uniform for all t 0, T ] for any T > 0, i.e. on all compact subsets of R 0. Hence, Kurtz s Theorem provides us with a limit of 1 n X n as n, as well as a characterisation of the limiting process. Before we prove Kurtz s Theorem, we will first apply Kurtz s Theorem on our SIS-model. Solution for SIS-model. For our SIS-model this implies the following. Equations 4.3, 4.4 and 4.5 imply that F equals F (x) = λ + (x) λ (x) = βx(1 x) γx. We will now provide a solution to dx = F (x) for γ = 1. For β 1 and some y R the dt solution to dx = F (x) = βx(1 x) x, x(0) = y. dt equals (β 1)y x(t) = (β 1 βy)e (1 β)t + βy. (4.9) It is clear that if β > 1 then (β 1)y lim x(t) = t 0 + βy = β 1 β, and if β < 1, then lim t x(t) = 0. Since Kurtz s Theorem tells us that x(t) is the probability limit of 1 n X n(t) as n, i.e. the proportion of infected people in a population of infinite size, we deduce that an epidemic occurs almost surely if β > 1, i.e. if R 0 > 1. In the subcritical regime, when R 0 < 1, no epidemic occurs. See Figure 4.2 for the value of x(t) in the subcritical and supercritical regime. 23

24 Figure 4.2.: Solution to 4.6 for our SIS-model in the case where R 0 R 0 = 2 (right). = 0.5 (left) and Proof. Kurtz s Theorem. Let (N i ),...,k be a family of independent unit-rate Poisson processes. That is, for each t > 0, N i (t) is Poisson distributed with parameter t for all i = 1,... k. By proof by induction it can be shown that we can write k ( t ) X n (t) = X n (0) + e i N i nλ i (X n (s)/n)ds. The proof of this statement is beyond the scope of this thesis. We now introduce the notation Y n (t) = 1 n X n(t), then it follows that Y n (t) = Y n (0) + = Y n (0) + (i) = Y n (0) + = Y n (0) + k k k k ( e t ) i n N i nλ i (X n (s)/n)ds 0 ( t e i n N i e i n N i e i n N i 0 ( t 0 ( t 0 ) nλ i (X n (s)/n)ds + ) nλ i (X n (s)/n)ds + ) nλ i (X n (s)/n)ds + 0 k t 0 t 0 e t i n 0 ( k nλ i (X n (s)/n)ds ) e i λ i (X n (s)/n) ds F (Y n (s))ds, (4.10) where N i (t) = N i (t) t, which is the centered Poisson process. For (i) we used the linearity of the integral. Clearly we are looking for an expression of Y n (t) x(t). Combining equation 4.6 and 4.10 gives us Y n (t) x(t) Y n (0) x(0) + + t 0 k F (Y n (s)) F (x(s)) ds e i n ( N i n t 0 λ i (X n (s))ds). 24

25 Now we can use the fact that lim n Y n (0) = x(0) a.s. and F is M Lipschitz to obtain the following equation for large n Y n (t) x(t) ɛ + M t 0 Y n (s) x(s) ds + k 1 n ɛ n,i(t), (4.11) ( where ɛ n,i (t) = e i N i n t λ 0 i(x n (s))ds). We are going to prove that the right side of this equation is small for large values of n. First, we will show that k 1 ɛ n n,i(t) is indeed small for large n. To obtain this result we need a Lemma 4.5 on page 27 to obtain the an upper bound for k 1 ɛ n n,i(t). Seeing as ɛ n,i (t) are non-negative for all t and i = 1,..., k we know that { sup Hence we deduce that P sup k k } { 1 n ɛ n,i(t) ɛ i : sup ] { 1 n ɛ n,i(t) ɛ P i : sup k { = P k P sup 1 n ɛ n,i(t) ɛ }. k 1 n ɛ n,i(t) ɛ }] k 1 n ɛ n,i(t) ɛ }] k sup ɛ n,i (t) ɛn k ]. (4.12) Note that λ 1 (t) λ 2 (t) for all t 0 implies that P N i (λ 1 (t)) ɛ] P N i (λ 2 (t)) ɛ] and therefore P N i (λ 1 (t)) t ɛ] P N i (λ 2 (t)) t ɛ]. Since we know that t λ 0 i(x(s))ds λt for all t 0 we deduce the following k P sup ɛ n,i (t) ɛn ] k k ( t ) P sup ē N i n λ i (X n (s))ds ɛn ] 0 k k ( ) P sup ē N i n λt ɛn ] k k P sup N i (t) ɛn ], (4.13) 0 t nt λ kē where ē := max{ e i : i = 1,..., k}. Since N i (t) = N(t) t, and N(t) is a unit-rate Poisson process we can apply Lemma 4.5 in estimate (ii) below to obtain an upper bound. Let Z n (t) = Y n (t) x(t). Combining equation 4.11 with 4.12 and 4.13 gives us 25

26 the following equation for large n { t } ] k ] P sup Z n (t) M Z n (s)ds 2ɛ = P sup 0 n ɛ n,i(t) ɛ k P sup N i (t) ɛn ] 0 t nt λ kē ( ( )) (ii) 2k exp (nt λ)h ɛn kē(nt ( ( )) λ) = 2k exp nt λh ɛ kēt λ, (4.14) where h(t) = (1 + t) log(1 + t) t. To finalize our proof, we want to show that Z n (t) is small for all t 0, T ] and large n. In particular, we want to show that ] P sup Z n (t) ɛ is small for large n. To obtain this result, we apply Grönwall s lemma on Z n (t). To see that Z n (t) is bounded on 0, T ], note that x(t) is a continuous function and therefore bounded on closed intervals. As Y n (t) is a Poisson process, it is clearly bounded on 0, T ]. Therefore Z n (t) = Y n (t) x(t) is bounded on 0, T ]. By Grönwall s lemma 1 implies Clearly this implies that { sup Hence, we deduce that { sup Z n (t) 2ɛ + M { Z n (t) M { Z n (t) M t 0 Z n (s)ds for all t 0, T ] Z n (t) 2ɛe Mt for all t 0, T ]. t 0 t 0 } } { Z n (s)ds 2ɛ sup Z n (t) 2ɛe }. MT } } { Z n (s)ds 2ɛ sup Z n (t) 2ɛe }. MT We finalize the proof for equation 4.7 by ] { t P sup Z n (t) 2ɛe MT P sup Z n (t) M 0 ( ( k exp nt λh ɛ kēt λ 1 see Appedix B.1 } ] Z n (s)ds 2ɛ )). 26

27 Or alternatively, if we substitute ɛ by ɛ 2 e MT ] ( ( )) P sup Z n (t) ɛ 2k exp nt λh ɛe MT 2kēT λ, which proves equation 4.7. To deduce that 4.15 holds, note that for any ɛ > 0 ] P sup Z n (t) ɛ 2k e nc(ɛ) 1 = 2k 1 e < c(ɛ) n N n N where ( ) c(ɛ) = T λh ɛe MT 2kēT λ > 0. We apply the Borel-Cantelli Lemma 2 to deduce that for any ɛ > 0 ] P lim sup n sup Z n (t) ɛ = 0, which tells us that sup Z n (t) converges to 0 almost surely. Hence lim sup 1 n n X n(t) x(t) = 0 a.s. (4.15) must be true Proof of Lemma 4.5 The next proof requires some knowledge of martingales. Definition 4.4. A real-valued process {X(t)} t 0 which satisfies EX(t) {X(u)} u s ] = X(s), for all 0 s t is called a martingale. Recall that we still need to prove the following statement. Lemma 4.5. Let N be a unit-rate Poisson process. Then for any ɛ > 0 and T > 0, ] P sup N(t) t ɛ 2e T h(ɛ/t ), where h(t) = (1 + t) log(1 + t) t. 2 see Appedix B.2 27

28 Proof. Let θ > 0 be fixed. Seeing as the exponential is an increasing function we have (a) ] { (i) }} ]{{ }} ]{ P sup N(t) t ɛ P sup (N(t) t) ɛ + P sup (t N(t)) ɛ ] = P sup e θ(n(t) t) e θɛ + P sup e θ(t N(t)) e ], θɛ where we used Boole s inequality for (i). Next, we are going to find an upper bound for (a) and (b). Example 4.6. {N(t) t} t 0 and {t N(t)} t 0 are martingales. The proof of this statement is beyond the scope of this chapter and is left to the reader. Example 4.7. {e θ(n(t) t) } t 0 and {e θ(t N(t)) } t 0 are non-negative submartingales by Example 4.6 and Jensen s inequality 3. The proof of this statement again exceeds this chapter and is left to the reader. By applying Doob s inequality 4 we obtain ] P sup e θ(n(t) t) e θɛ e θɛ Ee θ(n(t ) T ) ], and ] P sup e θ(t N(t)) e θɛ e θɛ Ee θ(n(t ) T ) ]. Seeing as N(T ) Poisson(T ), we know that by Example 2.5 that Ee θn(t ) ] = φ(e θ ) = exp(t (e θ 1)). Hence we deduce ] P sup e θ(n(t) t) e θɛ e θ(ɛ+t ) Ee θn(t ) ] = e θ(ɛ+t ) exp(t (e θ 1)) = e θ(ɛ+t ) exp(t (e θ 1)) (b) = exp( θ(ɛ + T ) + T (e θ 1)) We now want to minimize the exponent θ(ɛ + T ) + T (e θ 1) over θ. We will obtain this minimum by setting 0 = d dθ exp( θ(ɛ + T ) + T (eθ 1)) = ( (ɛ + T ) + T e θ ) exp( θ(ɛ + T ) + T (e θ 1)) = (ɛ + T ) + T e θ = 0 3 see Appendix B.3 4 see Appedix B.4 28

29 Thus, θ = log(1 + ɛ/t ). Combining the above yields the following upper bound for (a) ] P sup (N(t) t) ɛ exp( log(1 + ɛ/t )(ɛ + T ) + ɛ) = exp( T (log(1 + ɛ/t )(1 + ɛ/t ) ɛ/t )) = exp( T h(ɛ/t )), where h(t) = (1 + t) log(1 + t) t. We find an upper bound for (b) in a similar fashion ] P sup e θ(t N(t)) e θɛ exp(θ(t ɛ) + T (e θ 1)) Since the right side of the inequality attains its minimal at θ = log(1 ɛ/t ). ] P sup (t N(t)) ɛ exp( log(1 ɛ/t )(T ɛ) + T (e log(1 ɛ/t ) 1)) exp( T (log(1 ɛ/t )(1 ɛ/t ) ɛ/t )) = exp( T h( ɛ/t )) Clearly h( t) h(t) for all t 0, 1]. Therefore, we conclude that ] P sup N(t) t ɛ e T h(ɛ/t ) + e T h( ɛ/t ) T h(ɛ/t ) 2e holds Simulation. We ran a simulation in Matlab to comfirm that the Markov chain X n (t) described by equations 4.3, 4.4 and 4.5 indeed converges in probability to x(t) in 4.9 for large n. Again we ran the program for R 0 = 0.5 and R 0 = 2 for γ = 1, i.e. β = R 0 /γ. The jump matrix and mean sojourn times of X n (t) were constructed the according to equations 4.3, 4.4 and 4.5. We chose the number of infected people at t = 0 as ten percent of the population, i.e. I(0) = N/10, and constructed the initial distribution of X n (t) accordingly. Jump times were generated according to the sojourn times using the rand function. Jump directions were generated using the rand function to chose a direction according to the jumping prababilities. See Appendix C.4 and C.5 for the scripts. For every value of R 0, we repeated the previous script for increasing values of n, which we chose as 2 10, 2 11 and

30 Results In the subcritical as well as the supercritical regime, the value of X n (t) was indeed close to x(t) for all 3 simulations. Naturally, an epidemic occured for all simulations the supercritical regime where R 0 = 2, while the disease decreased to a small fraction in all simulations of the subcritical regime where R 0 = 0.5. See Figures 4.3 and 4.4 The difference between x(t) and X n (t) did not seem to decrease for larger n. An explanation for this could be that our largest value of n, 2 12, may not be sufficiently large to show convergence. Figure 4.3.: Number of infected people in a Markov jumping process for R 0 = 0.5. The blue line equals x(t), the others equal 1 n X n(t) for n = 2 10 (yellow), n = 2 11 (orange) and n = 2 12 (red). Figure 4.4.: Number of infected people in a Markov jumping process for R 0 = 2. The blue line equals x(t), the others equal 1 n X n(t) for n = 2 10 (yellow), n = 2 11 (orange) and n = 2 12 (red). 30