Joel C. Miller - Harvard Universityscholar.harvard.edu/files/joelmiller/files/2012_09_07...Intro to...

Intro to dynamic modeling of epidemics Static Networks Dynamic Populations

Edge-based compartmental modeling

Joel C. Miller& Erik Volz

RAPIDD; Harvard School of Public Health; Penn State University

7 Sept 2012


Intro to dynamic modeling of epidemics

Static NetworksNetwork AssumptionsFinal SizeDynamics

Dynamic PopulationsUnchanging DegreesDormant contacts (binding sites)


The basic SIR model

• Individuals begin susceptible,

• become infected from contacting infected individuals,

• and eventually recover with immunity.


The mass-action model

Let S be the proportion susceptible, I be the proportion infected,and R be the proportion recovered.

βIS γI

S I R

Resulting equations:

S = −βIS , I = βIS − γI , R = γI


Solutions

S = −βIS , I = βIS − γI , R = γI

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Solutions with β = 3, γ = 1


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Inserting partnership structure into the SIR model

• Different people have different numbers of partners.

• Partners are chosen randomly (pop is unclustered).

P(3) = P(1) = 0.5

• Assume static partnerships.

• Assume few initial infections.• Assume disease process occurs at simple rates:

• Transmit at rate β per partnership.• Recover at rate γ.

• Assume closed populations.





P(3) = P(1) = 0.5


• Assume few initial infections.

• Assume disease process occurs at simple rates:• Transmit at rate β per partnership.• Recover at rate γ.






P(3) = P(1) = 0.5


• Assume few initial infections.• Assume disease process occurs at simple rates:

• Transmit at rate β per partnership.• Recover at rate γ.



The final size of an epidemic in a network

• Consider a randomly chosen test individual u in thepopulation.

• Disallow infection from u to its partners (allows independenceassumption for partners).

• The probability u is Susceptible or Recovered at the end ofthe epidemic is affected by the status of its partners.

• The fraction of the population that is susceptible S equals theprobability u is susceptible.

S = P(u is susceptible)

• Let v be a random partner of u.

• Defineθ = P(v did not transmit to u)


Finding S

u

θθ θ

θθ

θ = P(v did not transmit to u)

Probability a random test individual remains susceptible is

S =∑k

P(k)

θk

= ψ(θ)

whereψ(x) =

∑k

P(k)xk


Finding S

u

θθ θ

θθ


Probability a random degree k test individual remains susceptible is

S =∑k

P(k)

θk

= ψ(θ)

whereψ(x) =

∑k

P(k)xk


Finding S

u

θθ θ

θθ



S =∑k

P(k)θk

= ψ(θ)

whereψ(x) =

∑k

P(k)xk


Finding S

u

θθ θ

θθ



S =∑k

P(k)θk = ψ(θ)

whereψ(x) =

∑k

P(k)xk


Finding θ

v

u

θθ θ

θθ

θθ

Probability a random degree k partner still susceptible is

φS =∑k

θk−1

=ψ′(θ)

ψ′(1)

If T = β/(β+γ), then probability partner does not transmit to u is

θ = φS + (1− T )(1− φS) = 1− T + Tψ′(θ)

ψ′(1)


Finding θ

v

u

θθ θ

θθ

θθ


φS =∑k

Pn(k)θk−1

=ψ′(θ)

ψ′(1)


θ = φS + (1− T )(1− φS) = 1− T + Tψ′(θ)

ψ′(1)


Finding θ

v

u

θθ θ

θθ

θθ


φS =∑k

kP(k)

〈K 〉θk−1

=ψ′(θ)

ψ′(1)


θ = φS + (1− T )(1− φS) = 1− T + Tψ′(θ)

ψ′(1)


Finding θ

v

u

θθ θ

θθ

θθ


φS =∑k

kP(k)

〈K 〉θk−1 =

ψ′(θ)

ψ′(1)


θ = φS + (1− T )(1− φS) = 1− T + Tψ′(θ)

ψ′(1)


Final Size

SoR = 1− ψ(θ)

where

θ = 1− T + Tψ′(θ)

ψ′(1)


Calculating Dynamics

The network structure alters the infection process (but not therecoveries)

? γI

S I R

I = 1− S − R , R = γI

We will switch to a partnership-based perspective to find S(t).


Revisiting the test individual

• Consider a randomly chosen test individual u in thepopulation.

• Disallow infection from u to its partners (allows independenceassumption for partners).

• The probability u is Susceptible, Infected, or Recovered attime t is affected by the status of its partners.

• The fraction of the population that is susceptible S(t) equalsthe probability u is susceptible.

S(t) = P(u is susceptible)

• Let v be a random partner of u.

• Defineθ(t) = P(v not yet transmitted to u)


Finding S(t)

u

θθ θ

θθ

θ(t) = P(v not yet transmitted to u)

Probability a random test individual still susceptible is

S(t) =∑k

P(k)

θ(t)k

= ψ(θ(t))

whereψ(x) =

∑k

P(k)xk


Finding S(t)

u

θθ θ

θθ


Probability a random degree k test individual still susceptible is

S(t) =∑k

P(k)

θ(t)k

= ψ(θ(t))

whereψ(x) =

∑k

P(k)xk


Finding S(t)

u

θθ θ

θθ



S(t) =∑k

P(k)θ(t)k

= ψ(θ(t))

whereψ(x) =

∑k

P(k)xk


Finding S(t)

u

θθ θ

θθ



S(t) =∑k

P(k)θ(t)k = ψ(θ(t))

whereψ(x) =

∑k

P(k)xk


How does θ evolve?

v

u

θ

θ

v

u

φS v

u

φI v

u

φR

v

u

1 − θ

• θ = φS + φI + φR .

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


How does θ evolve?

θ

v

u

φS v

u

φI v

u

φR

{ v

u

1 − θ

• θ = φS + φI + φR .



How does θ evolve?

θ

v

u

φS v

u

φI v

u

φR

{ v

u

1 − θ

βφI

γφI

• θ = φS + φI + φR .

• θ = −βφI .

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


How does θ evolve?

θ

v

u

φS v

u

φI v

u

φR

{ v

u

1 − θ

βφI

γφI

• θ = φS + φI + φR .



Finding φR(t)

θ

v

u

φS v

u

φI v

u

φR

{ v

u

1 − θ

βφI

γφI

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


Finding φS(t)

v

u

θθ θ

θθ

θθ


φS(t) =∑k

θ(t)k−1

=ψ′(θ)

ψ′(1)


Finding φS(t)

v

u

θθ θ

θθ

θθ


φS(t) =∑k

Pn(k)θ(t)k−1

=ψ′(θ)

ψ′(1)


Finding φS(t)

v

u

θθ θ

θθ

θθ


φS(t) =∑k

kP(k)

〈K 〉θ(t)k−1

=ψ′(θ)

ψ′(1)


Finding φS(t)

v

u

θθ θ

θθ

θθ


φS(t) =∑k

kP(k)

〈K 〉θ(t)k−1 =

ψ′(θ)

ψ′(1)


θ

v

u

ψ′(θ)ψ′(1) v

u

φI v

u

γβ(1 − θ)

{ v

u

1 − θ

βφI

γφI

Since φI = θ − φS − φR

= θ − ψ′(θ)ψ′(1) −

γβ (1− θ), we have

θ = −βφI = −βθ + βψ′(θ)

ψ′(1)+ γ(1− θ)


θ

v

u

ψ′(θ)ψ′(1) v

u

φI v

u

γβ(1 − θ)

{ v

u

1 − θ

βφI

γφI

Since φI = θ − φS − φR = θ − ψ′(θ)ψ′(1) −

γβ (1− θ)

, we have

θ = −βφI = −βθ + βψ′(θ)

ψ′(1)+ γ(1− θ)


θ

v

u

ψ′(θ)ψ′(1) v

u

φI v

u

γβ(1 − θ)

{ v

u

1 − θ

βφI

γφI

Since φI = θ − φS − φR = θ − ψ′(θ)ψ′(1) −

γβ (1− θ), we have

θ = −βφI = −βθ + βψ′(θ)

ψ′(1)+ γ(1− θ)


Final System

We finally have

θ = −βθ + βψ′(θ)

ψ′(1)+ γ(1− θ)

R = γI S = ψ(θ) I = 1− S − R


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Dynamic Populations

We now modify the population to have dynamic partnerships.Existing partnerships break at rate η and are immediately replacedby new connections.

7→


θ

v

u

φSv

u

φIv

u

φR

v

u

1 − θ

βφI

γφI

ηφRηφS ηφI

πS

πI

πR

γπI

We begin with the previous diagram, but φS , φI , and φR are theprobabilities a stub has not received infection from a partner andcurrently connects to a susceptible, infected, or recovered partner.


θ

v

u

φSv

u

φIv

u

φR

v

u

1 − θ

βφI

γφI

ηφR

ηφS ηφI

πS

πI

πR

γπI

Consider an edge with a recovered partner breaking. Rate = ηφR .


θ

v

u

φSv

u

φIv

u

φR

v

u

1 − θ

βφI

γφI

ηφR

ηφS ηφI

πS

πI

πR

γπI

The replacement partner may have any status.


θ

v

u

φSv

u

φIv

u

φR

v

u

1 − θ

βφI

γφI

ηφR

ηθ

ηφS ηφI

πS

πI

πR

γπI

The same is true for any edge that breaks. The total rate stubsthat have not received infections disconnect is ηθ.


θ

v

u

φSv

u

φIv

u

φR

v

u

1 − θ

βφI

γφI

ηφR

ηθ

ηθπS ηθπIηθπRηφS ηφI

πS

πI

πR

γπI

The probability the new partner is susceptible, infected, orrecovered is equal to the proportion of stubs belonging toindividuals of that type: πS , πI , and πR .


θ

v

u

φSv

u

φIv

u

φR

v

u

1 − θ

βφI

βφI φ

Sψ

′′(θ)

ψ′(θ

)

γφI

ηφR

ηθ


πS

πI

πR

γπI

The simple expressions for φR and πS break.


Finding πS

θθ θ

θθ

θθ

θ

Probability a random stub belongs to a degree k susceptibleindividual is

πS(t) =∑k

Pn(k)θ(t)k

=θψ′(θ)

ψ′(1)


Finding πS

θθ θ

θθ

θθ

θ


πS(t) =∑k

kP(k)

〈K 〉θ(t)k

=θψ′(θ)

ψ′(1)


Finding πS

θθ θ

θθ

θθ

θ


πS(t) =∑k

kP(k)

〈K 〉θ(t)k =

θψ′(θ)

ψ′(1)


πS

πI

πR

γπI

This leads to

πR = γπI , πI = 1− πS − πR , πS =θψ′(θ)

ψ′(1)


Completing the equations

θ

v

u

φSv

u

φIv

u

φR

v

u

1 − θ

βφI

βφI φ

Sψ

′′(θ)

ψ′(θ

)

γφI

ηφR

ηθ


φS = ηθπS − ηφS − βφIφSψ′′(θ)/ψ′(θ)

φI = βφIφSψ′′(θ)/ψ′(θ) + ηθπI − (η + γ + β)φI

θ = −βφI


Final Equations

φS = ηθπS − ηφS − βφIφSψ′′(θ)

ψ′(θ)

φI = βφIφSψ′′(θ)

ψ′(θ)+ ηθπI − (η + γ + β)φI

θ = −βφI

πR = γπI , πI = 1− πS − πR , πS =θψ′(θ)

ψ′(1)

R = γI , I = 1− S − R , S = ψ(θ)


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Comparison of theory (dashed) with average of 102 simulations ina population of 104 individuals (solid).Negative binomial degree distribution: P(k) =

(k+r−1k

)(1− p)rpk

for r = 4, p = 1/3. ψ(x) = [2/(3− x)]4.β = 5/4, γ = 1, η = 1/2.


Introducing “dormant” contacts

• Consider a population in which individuals have k stubs (or“binding sites”), not all of which are active at any time.

• A dormant stub becomes active at rate η1.

• An active stub becomes dormant at rate η2.


θ

v

u

φSv

u

φIv

u

φR

v

u

1 − θ

βφI

βφI φ

Sψ

′′(θ)

ψ′(θ

)

γφI

u

φD η2(θ − φD)η1φD

ηθ

ηθπS ηθπI

ηθπR

ηφR

ηφS

ηφI

πS

πI

πR

γπI

ξS

ξI

ξR

γξI

η2ξS

η1πS

η2ξI

η1πI

η2ξR

η1πR

Introduce φD : probability stub is dormant, and let π variablesrepresent dormant stubs, ξ variables represent active stubs.


θ

v

u

φSv

u

φIv

u

φR

v

u

1 − θ

βφI

βφI φ

Sψ

′′(θ)

ψ′(θ

)

γφI

u

φD η2(θ − φD)η1φD

η1φDπSπ η

1φD π

Iπ

η1φ

D πR

π

η2φR

η2φS

η 2φ I

πS

πI

πR

γπI

ξS

ξI

ξR

γξI

η2ξS

η1πS

η2ξI

η1πI

η2ξR

η1πR

Introduce φD : probability stub is dormant, and let π variablesrepresent dormant stubs, ξ variables represent active stubs.


Equations

θ = −βφI

φS = −βφIφSψ′′(θ)

ψ′(θ)+ η1

πS

πφD − η2φS

φI = βφIφSψ′′(θ)

ψ′(θ)+ η1

πI

πφD − (η2 + β + γ)φI

φD = η2(θ − φD)− η1φD

ξR = −η2ξR + η1πR + γξI , ξS = (θ − φD)ψ′(θ)

ψ′(1), ξI = ξ − ξS − ξR

πR = η2ξR − η1πR + γπI , πS = φDψ′(θ)

ψ′(1), πI = π − πS − πR

ξ =η1

η1 + η2, π =

η2η1 + η2

R = γI , S = ψ(θ) , I = 1− S − R


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Hierarchy

DormantContacts

DynamicFixed-Degree

DynamicVariable-Degree

ConfigurationModel

Mixed Pois-son

Mean FieldSocial Het-erogeneity

Mass Action


Conclusions

Fairly complex population structure can be captured by a smallnumber of equations.

PS: mass action can be written as

ξ = βI

S = S(0)e−ξ , R = R(0) +γξ

β, I = 1− S − R

where ξ(0) = 0.

Joel C. Miller - Harvard Universityscholar.harvard.edu/files/joelmiller/files/2012_09_07...Intro to...

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