part 5 percolation approaches to disease spreadtjhladish.github.io/sismid/sismid_lecture5.pdf ·...

Part 5 — Percolation approaches to diseasespread

Joel C. Miller & Tom Hladish

18–20 July 2018

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SIR and percolation

SIS disease

References

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Percolation

We are going to explore a relationship between SIR disease andpercolation.This will lead to methods to

I predict epidemic probability from a single infection.

I predict final size of an epidemic.

I predict the dynamics of an epidemic.

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Recall SIR behavior

0 1 2 3 4-6

log10 number infected m

N = 100

N = 400

N = 1600

N = 6400

N = 25600

0 20 40 60 80 100-4

number infected m

N = 100

N = 400

N = 1600

N = 6400

N = 25600

0.0 0.2 0.4 0.6 0.8 1.00

proportion infected m/N

N = 100

N = 400

N = 1600

N = 6400

N = 25600

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Modified model

We have a network

I An edge represents a potential transmission path (unweighted,bidirectional).

I An infected node remains infected for a single time step.

I An infected node transmits to a neighbor with probability p.

I Warning — no longer assuming continuous time

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Modified model

We have a network

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Modified model

We have a network

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Modified model

We have a network

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Modeling Disease Spread in a network

(1− p)2

p(1−

1−p)(1−

p2(1− p)

(1−p

p(1− p)

p(1− p)2

(1−p

p2(1− p)

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(1− p)2

p(1−

1−p)(1−

p2(1− p)

(1−p

p(1− p)

p(1− p)2

(1−p

p2(1− p)

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(1− p)2

p(1−

1−p)(1−

p2(1− p)

(1−p

p(1− p)

p(1− p)2

(1−p

p2(1− p)

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(1− p)2

p(1−

1−p)(1−

p2(1− p)

(1−p

p(1− p)1

p(1− p)2

(1−p

p2(1− p)

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(1− p)2

p(1−

1−p)(1−

p2(1− p)

(1−p

p(1− p)1

p(1− p)2

(1−p

p2(1− p)

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(1− p)2

p(1−

1−p)(1−

p2(1− p)

(1−p

p(1− p)1

p(1− p)2

(1−p

p2(1− p)

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(1− p)2

p(1−

1−p)(1−

p2(1− p)

(1−p

p(1− p)1

p(1− p)2

(1−p

p2(1− p)

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Alternative perspective

(1− p)2

p(1−

1−p)(1−

p2(1− p)

(1−p

p(1− p)

p(1− p)2

(1−p

p2(1− p)

At each step, if thereis an edge to cross, itis crossed with probabil-ity p. No edge is evercrossed twice.

I It is equivalent to decide in advance whether the edges will becrossed if encountered.

(1− p)3

p(1− p)2

p2(1− p)

p(1− p)2

p2(1− p)

p(1− p)2

p2(1− p)

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(1− p)2

p(1−

1−p)(1−

p2(1− p)

(1−p

p(1− p)

p(1− p)2

(1−p

p2(1− p)

(1− p)3

p(1− p)2

p2(1− p)

p(1− p)2

p2(1− p)

p(1− p)2

p2(1− p)

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(1− p)2

p(1−

1−p)(1−

p2(1− p)

(1−p

p(1− p)

p(1− p)2

(1−p

p2(1− p)

(1− p)3

p(1− p)2

p2(1− p)

p(1− p)2

p2(1− p)

p(1− p)2

p2(1− p)

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(1− p)2

p(1−

1−p)(1−

p2(1− p)

(1−p

p(1− p)

p(1− p)2

(1−p

p2(1− p)

(1− p)3

p(1− p)2

p2(1− p)

p(1− p)2

p2(1− p)

p(1− p)2

p2(1− p)

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Percolation in different size networksComparison of largest (red) and second largest (blue) componentsin different size networks below and above percolation threshold.

I Below threshold largest and second largest in a network areabout the same size as each other and similar size in bothnetworks

I Above threshold largest is proportional to network size.8 / 49

More detailed comparison of network size

0.0 0.2 0.4 0.6 0.8 1.00.0

N = 100

N = 1000

N = 10000

N = 100000

I Above the threshold, an epidemic occurs if the initial node isin the giant component.

I The entire component containing the index is infected.

I For a large network with given p, the giant component’s sizeis remarkably consistent. So the probability of an epidemicequals the proportion infected.

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0 20 40 60 80 1000.00

Number infected

0 5 10 15 200.00

0 50 100 150 200 250 300 350 4000.000

Number infected

0 5 10 15 200.00

0 200 400 600 800 1000 1200 1400 16000.0000

0.0005

0.0010

0.0015

0.0020

0.0025

0.0030

0.0035

Number infected

N=1600

0 5 10 15 200.00

0 1000 2000 3000 4000 5000 60000.0000

0.0002

0.0004

0.0006

0.0008

Number infected

N=6400

0 5 10 15 200.00

N P A100 0.237 0.423

400 0.340 0.387

1600 0.339 0.350

6400 0.365 0.366

25600 0.368 0.368

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0 20 40 60 80 1000.00

Number infected

0 5 10 15 200.00

0 50 100 150 200 250 300 350 4000.000

Number infected

0 5 10 15 200.00

0 200 400 600 800 1000 1200 1400 16000.0000

0.0005

0.0010

0.0015

0.0020

0.0025

0.0030

0.0035

Number infected

N=1600

0 5 10 15 200.00

0 1000 2000 3000 4000 5000 60000.0000

0.0002

0.0004

0.0006

0.0008

Number infected

N=6400

0 5 10 15 200.00

N P A100 0.237 0.423

400 0.340 0.387

1600 0.339 0.350

6400 0.365 0.366

25600 0.368 0.368

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Now return back to transmitting with rate β and recovering withrate γ.

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Transmission probability

I We’ll want to know the probability an infected node vtransmits to its neighbor u (assuming v is infected).

I Transmission is at rate β, and recovery is at rate γ. Theprobability of transmitting before recovering is β/(β + γ).

I Note: v transmitting to u and to w are correlated events(both depend on duration of v ’s infection), but transmissionsfrom different nodes to a single node are independent.

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Directed percolation analogy

Given a network G , I want to simulate the spread of an SIR diseasewith given β and γ

I I use Tom as a random number generator.

I When a node u becomes infected, I ask Tom: “how long willits infection last?”

I Then for each neighbor v I ask “will u transmit to v? When?”

I Tom decides he doesn’t like the rush to generate a randomnumber on the fly. So he does it in advance.

I For each node Tom assigns the duration its infection will last ifinfected.

I Once the duration is chosen, Tom decides which neighbors itwill transmit to and how long it will take.

I Then he reports those to me when I ask.

I Is it possible for me to know whether he is calculating inadvance or not?

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I I use Tom as a random number generator.I When a node u becomes infected, I ask Tom: “how long will

its infection last?”I Then for each neighbor v I ask “will u transmit to v? When?”

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a0.084

b0.006

c4.237

d0.653

0.0450.38

Every number that Tom gives me is a random number that isgenerated independently of every other number. It doesn’t matterwhen he generates it.

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a0.084

b0.006

c4.237

d0.653

0.0450.38

Every number that Tom gives me is a random number that isgenerated independently of every other number. It doesn’t matterwhen he generates it.

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Typical structure

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Directed Percolation Equivalence

The following processes produce indistinguishable output:I Standard epidemic simulation:

I Take a network G .I Choose an initial infected individual.I Allow edges to transmit until the random time the individual

recovers.

I Percolation-based simulation:

I Take a network G .I Generate a new directed network H:

I For each individual u, assign a duration d of infection.I For each edge from u, determine delay t until transmitting.I If t < d , place directed edge into network with associated

I Choose an initial infected individual.I Trace the disease spread following edges in H, transmitting

after the given time.

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recovers.

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recovers.

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recovers.

I Percolation-based simulation:I Take a network G .

I Generate a new directed network H:

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recovers.

I Percolation-based simulation:I Take a network G .I Generate a new directed network H:

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recovers.

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recovers.

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Comments on directed percolation

I Directed percolation can be used more generally when thereare other sources of heterogeneity in infectiousness and/orsusceptibility.

I The eventually infected nodes are exactly those nodes in theout-component of the index case.

I The probability a random node is infected follows from thesize of its in-component.

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Typical structure

I We can understand the dynamics with a “bowtie” diagram.

I Above a threshold there is a Giant Strongly ConnectedComponent HSCC

I It has an in-component HIN and an out-component HOUT .I If the index case is in HIN or HSCC then all of HSCC and

HOUT are eventually infected.I So Epidemic Probability P = E(|HIN ∪ HSCC |)/N and Attack

rate A = E(|HSCC ∪ HOUT |)/N.

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Typical structure

I We can understand the dynamics with a “bowtie” diagram.I Above a threshold there is a Giant Strongly Connected

Component HSCC

rate A = E(|HSCC ∪ HOUT |)/N.

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Typical structure

Component HSCC

I It has an in-component HIN and an out-component HOUT .

I If the index case is in HIN or HSCC then all of HSCC andHOUT are eventually infected.

I So Epidemic Probability P = E(|HIN ∪ HSCC |)/N and Attackrate A = E(|HSCC ∪ HOUT |)/N.

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Typical structure

Component HSCC

HOUT are eventually infected.

I So Epidemic Probability P = E(|HIN ∪ HSCC |)/N and Attackrate A = E(|HSCC ∪ HOUT |)/N.

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Typical structure

Component HSCC

rate A = E(|HSCC ∪ HOUT |)/N.18 / 49

Some consequences

The dynamic process of the epidemic is now encoded in a staticnetwork H. Studying H gives us some insight into what ishappening.

I There is a symmetry between epidemic probability andepidemic final size.

I Because edges out of a node are correlated and edges in to anode are not, P 6= A.

I The probability of an epidemic is the proportion of nodes fromwhich there is a long chain of transmissions in H.

I The final size of an epidemic with a very small initialproportion infected is the proportion of nodes which are thetarget of a long chain of transmissions in H.

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Some consequences

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Some consequences

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Some consequences

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Some consequences

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SIR epidemics in Configuration Model networks

I Consider a Configuration Model network in which we infect a(probably small) fraction of the population ρ.

I Allow the SIR disease to spread.

I We assume ρN is large enough that stochastic die-out doesnot play a major role.

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Recall our key questions

For SIR:

I P, the probability of an epidemic.

I A, the “attack rate”: the fraction infected if an epidemichappens (better named the attack ratio).

I R0, the average number of infections caused by those infectedearly in the epidemic.

I I (t), the time course of the epidemic.

For SIS:

I PI I (∞), the equilibrium level of infection

I I (t)

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R0 calculationFor SIR disease:

I The probability a newly infected individual has degree k isPn(k).

I The expected number of infections it causes given k is(k − 1) β

β+γ [it cannot reinfect the source of its infection].I So

R0 = E(number infections caused|infected early)

P(k |infected early)E(number infections|k)

Pn(k)(k − 1)β

β + γ

kP(k)(k − 1)

〈K 〉

β + γ

⟨K 2 − K

⟩〈K 〉

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Pn(k)(k − 1)β

β + γ

kP(k)(k − 1)

〈K 〉

β + γ

⟨K 2 − K

⟩〈K 〉

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Pn(k)(k − 1)β

β + γ

kP(k)(k − 1)

〈K 〉

β + γ

⟨K 2 − K

⟩〈K 〉

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Pn(k)(k − 1)β

β + γ

kP(k)(k − 1)

〈K 〉

β + γ

⟨K 2 − K

⟩〈K 〉

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Pn(k)(k − 1)β

β + γ

kP(k)(k − 1)

〈K 〉

β + γ

⟨K 2 − K

⟩〈K 〉

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Pn(k)(k − 1)β

β + γ

kP(k)(k − 1)

〈K 〉

β + γ

⟨K 2 − K

⟩〈K 〉

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Pn(k)(k − 1)β

β + γ

kP(k)(k − 1)

〈K 〉

β + γ

⟨K 2 − K

⟩〈K 〉

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For SIR:

For SIS:

I I (t)

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Changing the final size question

Instead of asking what proportion end up susceptible or recoveredask:

What is the probability a random node does not have atransmission path to it from one of the index nodes?

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Changing the final size question

Instead of asking what proportion end up susceptible or recoveredask:

What is the probability a random node does not have atransmission path to it from one of the index nodes?

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ΘΘ Θ

Θ = P(v did not transmit to u)

Probability a random degree k test individual is susceptible at theend is

N=∑k

(1− ρ)Θk

= (1− ρ)ψ(Θ)

whereψ(x) =

P(k)xk

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ΘΘ Θ

N=∑k

(1− ρ)Θk

= (1− ρ)ψ(Θ)

whereψ(x) =

P(k)xk

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ΘΘ Θ

N=∑k

P(k)(1− ρ)Θk

= (1− ρ)ψ(Θ)

whereψ(x) =

P(k)xk

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ΘΘ Θ

N=∑k

P(k)(1− ρ)Θk = (1− ρ)ψ(Θ)

whereψ(x) =

P(k)xk

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Finding Θ

ΘΘ Θ

Probability a random degree k partner never infected is

φS =∑k

(1− ρ)Θk−1

= (1− ρ)ψ′(Θ)

〈K 〉

Given β and γ, partner does not transmit to u with probability

Θ = φS +

(1− β

β + γ

)(1−φS)

= 1− β

β + γ+

β + γ

(1− ρ)ψ′(Θ)

〈K 〉

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Finding Θ

ΘΘ Θ

φS =∑k

Pn(k)(1− ρ)Θk−1

= (1− ρ)ψ′(Θ)

〈K 〉

Θ = φS +

(1− β

β + γ

)(1−φS)

= 1− β

β + γ+

β + γ

(1− ρ)ψ′(Θ)

〈K 〉

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Finding Θ

ΘΘ Θ

φS =∑k

〈K 〉(1− ρ)Θk−1

= (1− ρ)ψ′(Θ)

〈K 〉

Θ = φS +

(1− β

β + γ

)(1−φS)

= 1− β

β + γ+

β + γ

(1− ρ)ψ′(Θ)

〈K 〉

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Finding Θ

ΘΘ Θ

φS =∑k

〈K 〉(1− ρ)Θk−1 = (1− ρ)

ψ′(Θ)

〈K 〉

Θ = φS +

(1− β

β + γ

)(1−φS)

= 1− β

β + γ+

β + γ

(1− ρ)ψ′(Θ)

〈K 〉

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Finding Θ

ΘΘ Θ

φS =∑k

〈K 〉(1− ρ)Θk−1 = (1− ρ)

ψ′(Θ)

〈K 〉

Θ = φS +

(1− β

β + γ

)(1−φS)

= 1− β

β + γ+

β + γ

(1− ρ)ψ′(Θ)

〈K 〉

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Finding Θ

ΘΘ Θ

φS =∑k

〈K 〉(1− ρ)Θk−1 = (1− ρ)

ψ′(Θ)

〈K 〉

Θ = φS +

(1− β

β + γ

)(1−φS) = 1− β

β + γ+

β + γ

(1− ρ)ψ′(Θ)

〈K 〉

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Final Size

SoA = 1− (1− ρ)ψ(Θ)

Θ =γ

β + γ+

β + γ(1− ρ)

ψ′(Θ)

〈K 〉

A more rigorous definition would be that Θ is the probability that the given edge isn’t the final edge of a directedpath from an index node to u in the percolated network H.

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Final Size

SoA = 1− (1− ρ)ψ(Θ)

Θ =γ

β + γ+

β + γ(1− ρ)

ψ′(Θ)

〈K 〉A more rigorous definition would be that Θ is the probability that the given edge isn’t the final edge of a directedpath from an index node to u in the percolated network H.

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For SIR:

For SIS:

I I (t)

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Finding S(t) for SIR disease

θθ θ

θ(t) = P(v not yet transmitted to u)

Probability a random test individual still susceptible is

N=∑k

(1− ρ)θ(t)k

= (1− ρ)ψ(θ(t))

whereψ(x) =

P(k)xk

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θθ θ

Probability a random degree k test individual still susceptible is

N=∑k

(1− ρ)θ(t)k

= (1− ρ)ψ(θ(t))

whereψ(x) =

P(k)xk

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θθ θ

N=∑k

P(k)(1− ρ)θ(t)k

= (1− ρ)ψ(θ(t))

whereψ(x) =

P(k)xk

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θθ θ

N=∑k

P(k)(1− ρ)θ(t)k = (1− ρ)ψ(θ(t))

whereψ(x) =

P(k)xk

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How does θ evolve?

1 − θ

I θ = φS + φI + φR .

I θ = −βφI .I Our goal is to find φI in terms of θ.

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How does θ evolve?

1 − θ

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How does θ evolve?

1 − θ

I θ = −βφI .

I Our goal is to find φI in terms of θ.

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How does θ evolve?

1 − θ

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Finding φR(t)

1 − θ

Because derivatives are proportional, φR = γβ (1− θ)

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Finding φS(t)

θθ θ

Probability a random degree k partner still susceptible is

φS(t) =∑k

(1− ρ)θ(t)k−1

= (1− ρ)ψ′(θ)

〈K 〉

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Finding φS(t)

θθ θ

φS(t) =∑k

Pn(k)(1− ρ)θ(t)k−1

= (1− ρ)ψ′(θ)

〈K 〉

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Finding φS(t)

θθ θ

φS(t) =∑k

〈K 〉(1− ρ)θ(t)k−1

= (1− ρ)ψ′(θ)

〈K 〉

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Finding φS(t)

θθ θ

φS(t) =∑k

〈K 〉(1− ρ)θ(t)k−1 = (1− ρ)

ψ′(θ)

〈K 〉

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(1 − ρ)ψ′(θ)〈K〉 v

γβ(1 − θ)

1 − θ

We have

φI = θ − φS − φR

= θ − (1− ρ)ψ′(θ)

〈K 〉− γ

β(1− θ)

θ = −βφI

= −βθ + β(1− ρ)ψ′(θ)

〈K 〉+ γ(1− θ)

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(1 − ρ)ψ′(θ)〈K〉 v

γβ(1 − θ)

1 − θ

We have

φI = θ − φS − φR = θ − (1− ρ)ψ′(θ)

〈K 〉− γ

β(1− θ)

θ = −βφI

= −βθ + β(1− ρ)ψ′(θ)

〈K 〉+ γ(1− θ)

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(1 − ρ)ψ′(θ)〈K〉 v

γβ(1 − θ)

1 − θ

We have

φI = θ − φS − φR = θ − (1− ρ)ψ′(θ)

〈K 〉− γ

β(1− θ)

θ = −βφI = −βθ + β(1− ρ)ψ′(θ)

〈K 〉+ γ(1− θ)

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Final System

We finally have

θ = −βθ + β(1− ρ)ψ′(θ)

〈K 〉+ γ(1− θ)

R = γI S = (1− ρ)Nψ(θ) I = N − S − R

Compare with

θ = −βθ + βθ2 (1− ρ)ψ′(θ)

〈K〉− θγ ln θ

R = γI , S = (1− ρ)Nψ(θ), I = N − S − R

More details in [1, 2, 3]

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0 5 10 15 20

ion infe

Truncated Power Law

Bimodal

Poisson

Homogeneous

35 / 49

A good exercise

Repeat this derivation for a model in which infections last for onetime step and transmission occurs with probability p.

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Epidemic probability

I To calculate epidemic probability, we consider a singleintroduced node, randomly chosen in the population.

I ρ = 0.

I ψ(x) =∑

k P(k)xk is the probability generating function forthe degree distribution.

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I ρ = 0.

I ψ(x) =∑

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I ρ = 0.

I ψ(x) =∑

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Calculating epidemic probability

Ω(D)Ω(D)Ω(D)

Ω(D) = P(u does not transmit to a neighbor|D) + P(u transmits, but neighbor doesn’t lead to an epidemic)

Probability a random degree k index case whose infection durationis D does not start an epidemic is

1−P =

∫ ∞0

γe−γD∑k

Ω(D)k

∫ ∞0

γe−γDψ(Ω(D))dD

whereψ(x) =

P(k)xk

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Ω(D)Ω(D)Ω(D)

1−P =

∫ ∞0

γe−γD∑k

Ω(D)k

∫ ∞0

whereψ(x) =

P(k)xk

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Ω(D)Ω(D)Ω(D)

1−P =

∫ ∞0

γe−γD

P(k)Ω(D)k

∫ ∞0

whereψ(x) =

P(k)xk

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Ω(D)Ω(D)Ω(D)

1−P =

∫ ∞0

γe−γD∑k

P(k)Ω(D)k dD

∫ ∞0

whereψ(x) =

P(k)xk

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Ω(D)Ω(D)Ω(D)

1−P =

∫ ∞0

γe−γD∑k

P(k)Ω(D)k dD =

∫ ∞0

whereψ(x) =

P(k)xk

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Finding Ω

Ω(D)Ω(D)Ω(D)

Ω(D)Ω(D)

Probability a random partner of the index case having degree kwhose infection duration is D does not start an epidemic is

Ω(D) =

[1− p(D)] +

∫ ∞0

γe−γD∑k

p(D)Ω(D)k−1

p(D) = 1− e−βD is the probability of transmitting given infection duration of D

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Finding Ω

Ω(D)Ω(D)Ω(D)

Ω(D)Ω(D)

Ω(D) =

[1− p(D)] +

∫ ∞0

γe−γD∑k

p(D)Ω(D)k−1

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Finding Ω

Ω(D)Ω(D)Ω(D)

Ω(D)Ω(D)

Ω(D) =

[1− p(D)] +

∫ ∞0

γe−γD

Pn(k)p(D)Ω(D)k−1

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Finding Ω

Ω(D)Ω(D)Ω(D)

Ω(D)Ω(D)

Ω(D) = [1− p(D)] +

∫ ∞0

γe−γD∑k

Pn(k)p(D)Ω(D)k−1 dD

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Finding Ω

Ω(D)Ω(D)Ω(D)

Ω(D)Ω(D)

Ω(D) = [1− p(D)] + p(D)

∫ ∞0

γe−γD∑k

Pn(k)Ω(D)k−1 dD

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Finding Ω

Ω(D)Ω(D)Ω(D)

Ω(D)Ω(D)

Ω(D) = [1− p(D)] + p(D)

∫ ∞0

γe−γD∑k

〈K 〉Ω(D)k−1 dD

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Finding Ω

Ω(D)Ω(D)Ω(D)

Ω(D)Ω(D)

Ω(D) = [1− p(D)] + p(D)

∫ ∞0

γe−γD∑

k kP(k)Ω(D)k−1

〈K 〉dD

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Finding Ω

Ω(D)Ω(D)Ω(D)

Ω(D)Ω(D)

Ω(D) = [1− p(D)] + p(D)

∫ ∞0

γe−γDψ′(Ω(D))

〈K 〉dD

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We arrive at

1− P =

∫ ∞0

Ω(D) = e−βD +(

1− e−βD)∫ ∞

0γe−γD

ψ′(Ω(D))

〈K 〉dD

In general we can only solve this numerically, but it isstraightforward. We start with a guess that Ω(D) = 1, plug it inand iterate.In fact the nth iteration will give the probability that the diseasespreads at least n generations.

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SIR and percolation

SIS disease

References

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SIS disease

It is very difficult to write down an analytic model of SIS disease innetworks that accounts for partnership duration.

I The difficulty results from the fact that a node can infect itsneighbors, thus changing the exposure it receives after itrecovers.

I This effect is real: a person who has recovered from, say,MRSA but passed it on to his/her family is at higher risk ofreaquiring MRSA. Ignoring this weakens our ability to drawconclusions.

I This has policy implications: how much will it reduce MRSAtransmission if we clear the disease from a hospital or a prison?

I So for SIS disease simulation is likely to play a major role.

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SIS disease

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SIS disease

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SIS disease

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SIS disease

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Percolation-like results and SIS

I It is possible to use percolation-like results to for rigorousconclusions about SIS disease.

I As a general rule, these rigorous results do not generalize if wedo not assume constant infection and transmission rates.

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Percolation-like results and SIS

I It is possible to use percolation-like results to for rigorousconclusions about SIS disease.

I As a general rule, these rigorous results do not generalize if wedo not assume constant infection and transmission rates.

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A percolation-like approach

BobbieAlex Charlie

I Find transmission events as Poisson process

I Find recovery events as Poisson process

I Trace out from initial infection

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BobbieAlex Charlie

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BobbieAlex Charlie

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BobbieAlex Charlie

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Now invert the picture

BobbieAlex Charlie

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BobbieAlex Charlie

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BobbieAlex Charlie

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Some conclusions

I An infection of u at time 0 leads to an infection of v at time tiff there is a path that doesn’t go through a recovery event.

I That reversed path also works. So any node infected at time twould cause infection of the initial node at time t in thereversed process.

I So the expected number of nodes infected at time t startingfrom infection of u at time 0 is equal to the probability u isinfected at time t if we infect a random individual at time 0.

I The equilibrium size of an SIS epidemic with Poissoniantransmission and recovery equals the probability that anepidemic occurs.

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Some conclusions

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Some conclusions

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Some conclusions

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SIS size vs Probability

1000 simulations starting with a single randomly chosen node in aConfiguration model network with P(1) = P(5) = 0.5.

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0 5 10 15 20 25 30

β = 0.5, γ = 1273 did not die out

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0 5 10 15 20 25 30

β = 0.75, γ = 1448 did not die out

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0 5 10 15 20 25 30

β = 1, γ = 1537 did not die out

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SIR and percolation

SIS disease

References

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References I

[1] Joel C. Miller, Anja C. Slim, and Erik M. Volz.Edge-based compartmental modelling for infectious disease spread.Journal of the Royal Society Interface, 9(70):890–906, 2012.

[2] Joel C. Miller.Epidemics on networks with large initial conditions or changing structure.PLoS ONE, 9(7):e101421, 2014.

[3] Istvan Z Kiss, Joel C Miller, and Peter L Simon.Mathematics of epidemics on networks: from exact to approximate models.IAM. Springer, 2017.

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part 5 percolation approaches to disease spreadtjhladish.github.io/sismid/sismid_lecture5.pdf ·...

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