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 basic SIR model

The mass-action model

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

βIS γI

Resulting equations:

S = −βIS , I = βIS − γI , R = γI

βIS γI

Solutions

5 0 5 10 150.00

5 0 5 10 150.0

tive infe

Solutions with β = 3, γ = 1

−5 0 5 10 150.00

0.35Mass ActionHomogeneousPoissonBimodalTruncated Powerlaw

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

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

θθ θ

θ = P(v did not transmit to u)

Probability a random test individual remains susceptible is

S =∑k

= ψ(θ)

whereψ(x) =

P(k)xk

Finding S

θθ θ

Probability a random degree k test individual remains susceptible is

S =∑k

= ψ(θ)

whereψ(x) =

P(k)xk

Finding S

θθ θ

S =∑k

P(k)θk

= ψ(θ)

whereψ(x) =

P(k)xk

Finding S

θθ θ

S =∑k

P(k)θk = ψ(θ)

whereψ(x) =

P(k)xk

Finding θ

θθ θ

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

θθ θ

φS =∑k

Pn(k)θk−1

=ψ′(θ)

ψ′(1)

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

ψ′(1)

Finding θ

θθ θ

φS =∑k

〈K 〉θk−1

=ψ′(θ)

ψ′(1)

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

ψ′(1)

Finding θ

θθ θ

φS =∑k

〈K 〉θk−1 =

ψ′(θ)

ψ′(1)

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

ψ′(1)

Finding θ

θθ θ

φS =∑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)

θ = 1− T + Tψ′(θ)

ψ′(1)

Calculating Dynamics

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

I = 1− S − R , R = γI

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

Calculating Dynamics

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

I = 1− S − R , R = γI

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

Revisiting the test individual

• 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)

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

Finding S(t)

θθ θ

θ(t) = P(v not yet transmitted to u)

Probability a random test individual still susceptible is

S(t) =∑k

θ(t)k

= ψ(θ(t))

whereψ(x) =

P(k)xk

Finding S(t)

θθ θ

Probability a random degree k test individual still susceptible is

S(t) =∑k

θ(t)k

= ψ(θ(t))

whereψ(x) =

P(k)xk

Finding S(t)

θθ θ

S(t) =∑k

P(k)θ(t)k

= ψ(θ(t))

whereψ(x) =

P(k)xk

Finding S(t)

θθ θ

S(t) =∑k

P(k)θ(t)k = ψ(θ(t))

whereψ(x) =

P(k)xk

How does θ evolve?

1 − θ

• θ = φS + φI + φR .

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

How does θ evolve?

1 − θ

• θ = φS + φI + φR .

How does θ evolve?

1 − θ

• θ = φS + φI + φR .

• θ = −βφI .

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

How does θ evolve?

1 − θ

• θ = φS + φI + φR .

Finding φR(t)

1 − θ

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

Finding φS(t)

θθ θ

φS(t) =∑k

θ(t)k−1

=ψ′(θ)

ψ′(1)

Finding φS(t)

θθ θ

φS(t) =∑k

Pn(k)θ(t)k−1

=ψ′(θ)

ψ′(1)

Finding φS(t)

θθ θ

φS(t) =∑k

〈K 〉θ(t)k−1

=ψ′(θ)

ψ′(1)

Finding φS(t)

θθ θ

φS(t) =∑k

〈K 〉θ(t)k−1 =

ψ′(θ)

ψ′(1)

ψ′(θ)ψ′(1) v

γβ(1 − θ)

1 − θ

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)

ψ′(θ)ψ′(1) v

γβ(1 − θ)

1 − θ

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)

ψ′(θ)ψ′(1) v

γβ(1 − θ)

1 − θ

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

−5 0 5 10 150.00

0.35Mass ActionHomogeneousPoissonBimodalTruncated Powerlaw

Dynamic Populations

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

1 − θ

ηφRηφS ηφI

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.

1 − θ

ηφS ηφI

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

1 − θ

ηφS ηφI

The replacement partner may have any status.

1 − θ

ηφS ηφI

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

1 − θ

ηθπS ηθπIηθπRηφS ηφI

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 .

1 − θ

βφI φ

′′(θ)

ψ′(θ

The simple expressions for φR and πS break.

Finding πS

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

πS(t) =∑k

Pn(k)θ(t)k

=θψ′(θ)

ψ′(1)

Finding πS

θθ θ

πS(t) =∑k

〈K 〉θ(t)k

=θψ′(θ)

ψ′(1)

Finding πS

θθ θ

πS(t) =∑k

〈K 〉θ(t)k =

θψ′(θ)

ψ′(1)

This leads to

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

ψ′(1)

Completing the equations

1 − θ

βφI φ

′′(θ)

ψ′(θ

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)

−5 0 5 10 15 200.00

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.

1 − θ

βφI φ

′′(θ)

ψ′(θ

φD η2(θ − φD)η1φD

ηθπS ηθπI

ηθπR

η2ξS

η1πS

η2ξI

η1πI

η2ξR

η1πR

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

1 − θ

βφI φ

′′(θ)

ψ′(θ

φD η2(θ − φD)η1φD

η1φDπSπ η

1φD π

η2φR

η2φS

η 2φ I

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

πφD − η2φS

φI = βφIφSψ′′(θ)

ψ′(θ)+ η1

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

−5 0 5 10 150.00

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.

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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