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Immunisation Strategies for a Community of Households

Niels G Becker (with help from David Philp) National Centre for Epidemiology and Population Health Australian National University

1. Background A - A case for stochastic models and their formulation

2. Background B - Two types of infective

3. The need for different types of infective in a household setting

4. Different reproduction numbers

5. Different vaccination strategies

6. Critical immunity coverage for different vaccination strategies

7. Probability of containment

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Background A

A case for stochastic models

The deterministic SIR model has some limitations

• It and St are taken as continuous when they are really integers. (Of concern when either is small)

• They suggest that an outbreak always takes off when

R0 s0 > 1. (Not always the case.)

• They ignore the chance element in transmission. (Of particular concern when It or St are small, e.g. during the early stages, or when control is effective)

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Formulation of stochastic infectious disease models

When we allow for chance, the description of disease

transmission involves a probability distribution for the number of infectives and susceptibles at each point in time.

Equations for these distributions are easy to write down, but difficult to solve.

With the speed of modern computers stochastic formulations are often accommodated by simulation studies. That is, many realisations of the random process are generated, from which means and proportions are calculated.

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Some questions can be explored analytically, including:

1. What is the probability that an imported infection leads to a major outbreak?

2. What fraction of the community needs to be immune to ensure that an imported infection can not lead to a major outbreak?

3. What is the distribution of the eventual size of the outbreak?

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Branching processes are a central tool in such analyses

A standard Branching Process (BP) is a population model in which each individual independently generates a number of offspring. This number is randomly selected from a common probability distribution.

Some of results in the rich theory of Branching Processes are useful for studying the control of outbreaks of an infectious disease.

In disease transmission a BP is used to approximate the population dynamics of infectives, while the depletion of susceptibles remains negligible.

This applies during the early stages of an outbreak initiated by an imported infection and for the entire outbreak when the outbreak is minor.

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020406080100

0 10 20Generation

Early stages of an outbreak

Mean of offspring distribution = µ

Mean number in generation r = I0 µ r

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and after that …

0

1000

2000

3000

4000

20 25 30Generation

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Some results from Branching Processes:

1. Suppose the mean of the offspring distribution is µ.

The BP becomes extinct with probability 1 when µ < 1.

2. The mean of the offspring distribution for infectives is µ (1-v), when a fraction v of close contacts is with immune people.

The BP becomes extinct with probability 1 when µ (1-v) < 1.

Therefore, the critical immunity coverage is 1 – 1/ µ.

3. Let X be the number of offspring an infective has.

Suppose that Pr(X = r) = pr

The probability that 1 infective starts an outbreak that does

not take off is

p0 + p1 p0 + p1 p1 p0 + p2 p0 p0 + etc, etc, etc

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A simpler approach:

Let q be the probability that an outbreak started by 1 infective does not take off.

If the first infective has x offspring then each of them starts an independent BP and the probability that none of those x BPs takes off is q x.

Therefore q = p0.1 + p1.q + p2.q2 + p3.q3 + …

So q is the smallest solution ofs = φ(s),

where φ is the probability generating function of the offspring distribution.

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Illustration

Consider the Poisson offspring distribution with mean (1-v), where v is the fraction immune.

The p.g.f. is then φ(s) = exp[ (1-v)(s-1)]

With = 3, we need to fix v and solve s = exp[3(1-v)(s-1)]

The value v that achieves q is v = 1 – ln(q) /[3(q -1)]

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 v

q

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Background B

Different types of infective

Suppose there are two types of individual, perhaps children and adults, who differ in susceptibility and infectivity, and mix differently.

Now each type has a different offspring distribution.

Means are again central for determining interventions that can contain transmission, BUT we need a mean matrix.

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Offspring infective

Child Adult

Parent infective

Child 4 2

Adult 2 1

Mean matrix (with everyone susceptible):

Result from multi-type Branching Processes:

A multi-type branching process becomes extinct with probability 1 if the largest eigenvalue of the mean matrix is less than 1.For the above matrix: R = larger root of (4-x)(1-x)-2*2=0, that is R = 5.

Is this a “reproduction” number?

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If the branching process takes off, then eventually infectives will consist of 2/3 children and 1/3 adults.[Obtain this from the eigenvector corresponding to eigenvalue 5]

If, at that stage, we sample an infective at random, then the mean number of offspring is

(2/3)*6 + (1/3)*3 =5

so R is a reproduction number in that sense.

It is the eventual rate of growth of the BP.

While this ‘reproduction number’ is a useful tool for determining the requirements for containment, its interpretation does not really provide a lot of direct insights for infectious disease epidemiology.

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Suppose we immunise a fraction vc of children and a fraction va of adults. Then the mean matrix becomes

Offspring infective

AChild Adult

Parent infective

Child 4(1-vc) 2(1-va)

Adult 2(1-vc) (1-va)

This changes the reproduction number to the larger root of

[4(1-vc) –x][(1-va) –x]- 4(1-vc)(1-va) =0.

Thus Rv = 5 – va – 4vc.

Can now address questions such as:

1. What is the smallest coverage required to make Rv < 1?

` 2. What is the critical coverage when va = vc?

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Household structure

Choice of Reproduction NumberThere are numerous ways to define a reproduction number for transmission in a community of households.

Becker and Dietz (1996) define 4 reproduction numbers assuming an SIR model and that the community consists of households of size 3.

These reproduction numbers are distinguished by the way infections are attributed to infectives.

To illustrate the choice we define here 2 reproduction numbers in the simplest setting, where all households have size 2.

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A

For calculating RI0 we attribute 4 infections to A, in this example.

RI0 reproduction based on individuals infected

For this reproduction number we attribute cases actually infected

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A primary case in a household of two can infect community members AND their household partner.

A secondary case in a household of two can infect only community members.

Their potential to infect others differs, so consider them as different types of infective, P and S.

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The mean matrix is

0S

P

SP

p

)4/(2/0 pRITherefore

On average, a primary case infects individuals outside the household and p individuals within the household.

On average, a secondary case infects individuals outside the household and none within the household.

The largest eigenvalue is the larger solution of the

equation x2 – x – p = 0.

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We now attribute infections in such a way that primary and secondary cases have exactly the same potential to generate infectives.

This is convenient because there is then only one type of infective and the reproduction number is simply the mean number of infections attributed to an infective.

The Trick

Attribute to an infective A the individuals she infects in other households AND all infections that arise in those household outbreaks.

Do not attribute infections to A infectives she infects in her own household.

RH0 reproduction based on households infected

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A

When we attribute infections this way we obtain the reproduction number RH0 = µ (1+p)

Compare this with )4/(2/0 pRI

For calculating

RH0 we attribute

5 infections to

individual A ,

in this example.

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Derivation

Let X have probability distribution Pr(X =2) = p = 1-Pr(X =1)

Suppose an infective generates Y primary household cases.

The total number of offspring attributed is

W = X1+ X2 + … + XY.

E(W|Y) = E(X1+ X2 + … + XY |Y)

= Y E(X|Y)

= Y E(X)

= Y (1+p)

E[E(W|Y)] = E[Y (1+p)] = µ(1+p)

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Assume that the vaccine gives complete immunity.

Suppose we achieve a vaccination coverage of v.

Strategy H: Vaccinate both members in a fraction v of households

When allocating ‘individuals actually infected’ the mean matrix becomes

This gives

Compare this with RHv = µ (1v)(1+p) = (1v)RH0

Critical vaccination coverageWhat are these reproduction numbers when we

partially vaccinate the community?

)4/)1()1(2/)1( 22 vpvvRIv

0)1(

)1(

S

P

SP

v

pv

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Example:

µ = 3, p = 0.5

RI0 = 3.44, RH0 = 4.5

Critical vaccination coverage is

vC = 1 – 1/RH0

= 1 – 1/4.5

= 0.778

Strategy H: The critical vaccination coverage is v such that

RHv = (1v)RH0 =1. Therefore vc= 1 – 1 / RH0.

0

1

2

3

4

0 0.2 0.4 0.6 0.8 1

Vaccination coverage

Rep

rod

uct

ion

nu

mb

er.

RHv

RIv

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Strategy I:

Vaccinate a fraction v of individuals at random.

The proportion of households with 0, 1 or 2 susceptibles is v2, 2v(1-v) and (1-v)2, respectively

When allocating ‘individuals actually infected’ the mean matrix becomes

This gives RIv = (1-v) RI0

Compare this with the reproduction number for infected households, which becomes

RHv = µ (1-v) [1+p (1-v)]

0)1(

)1()1(

S

P

SP

v

vpv

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Example:

µ = 3, p = 0.5

RI0 = 3.44, RH0 = 4.5

Critical vaccination coverage is

vC = 1 – 1/RI0

= 1 – 1/3.44

= 0.7090

1

2

3

4

0 0.2 0.4 0.6 0.8 1Vaccination coverage

Rep

rod

uctio

n n

um

ber.

Strategy I: The critical vaccination coverage is v such that

RIv = (1v)RI0 =1. Therefore vc= 1 – 1 / RI0.

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Strategy O:

Vaccinate a fraction v of individuals by vaccinating

(a) 1 individual in each of a fraction 2v of households, if v ≤ 0.5;

(b) 2 individuals in a fraction 2v -1 of households, and 1 individual in the remaining households, if v > 0.5.

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(a) v ≤ 0.5

When allocating ‘individuals actually infected’ the mean matrix becomes

This gives

Also RHv = µ [1-v+p (1-2v)]

(b) v > 0.5

Now there is no within household infection and

RIv = RHv = µ (1-v)

0)1(1

21)1(

S

P

SP

vvv

pv

4/)1()21(2/)1( 22 vvpvRIv

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Example:

µ = 3, p = 0.5

RI0 = 3.44, RH0 = 4.5

Critical vaccination coverage is

vC = 1 – 1/µ

= 1 – 1/3

= 0.667

Strategy O:

0

1

2

3

4

0 0.2 0.4 0.6 0.8 1

Vaccination coverage

Rep

rod

uct

ion

nu

mb

er.

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Comparing RHv for the different strategies

0

1

2

3

4

0 0.2 0.4 0.6 0.8 1

Vaccination coverage

Rep

rod

uct

ion

nu

mb

er.

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The probability of containment

The above discussion focused on conditions such that Probability(containment) = 1.

We now show how to compute Probability(containment) in cases where it is not 1 (for a community of households)

Suppose all households are of size 2.

Consider first the case with no immunity.

One infective starts the outbreak. The probability that s/he infects the other household member is p.

The first infected household will have

1 case, with probability 1 p

2 cases, with probability p

These cases have the same offspring distribution

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The common offspring distribution

Suppose an infective generates Y primary household cases.

The total number of offspring attributed is

W = X1+ X2 + … + XY.

E[E(sW |Y)] = E[s(1-p)+s2p]Y = φY[s(1-p)+s2p]

which is exp[(s-sp+s2p1)] for a Poisson distribution.

Hence

Probability(containment) = (1-p)+ 2p

where is the smaller solution of

s = φY[s(1-p)+s2p]

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Probability(containment)

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How does the strategy affect the probability of containment?

General vaccination strategy

Number vaccinated 0 1 2

Proportion v0 v1 v2

v0 + v1 + v2 = 1 and vaccination coverage v = v2 + ½ v1

For example,

Strategy H has v0 = 1-v, v1 = 0 and v2 = v

Srategy I has v0 = (1-v)2 , v1 = 2v(1-v) and v2 = v2

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The first infected household will have

1 case, with probability

2 cases, with probability pvv

v

pvv

vvv

v

01

0

01

0

01

1

22

)1(2

22

Probability(containment) is

2

01

0

01

0

01

1

22

)1(2

22

p

vvv

pvv

vvv

v

where is the smaller solution of

01

20

01

0

01

1

22

2)1(2

2 vvpsv

vvpsv

vvsv

s Y

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Probability(containment)

For Strategy H and Strategy O

Vaccination coverage = 0.5

Difference in Probability(containment)

Strategy H

Strategy O

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References

Ball FG, Mollison D, Scalia-Tomba G (1997). Epidemics with two levels of mixing. Ann. Appl. Prob. 7 46.

Becker NG, Dietz K (1996). Reproduction numbers and critical immunity levels for epidemics in a community of households. In Athens Conference on Applied Probability and Time Series, Volume 1: Applied Probability, (Eds Heyde CC, Prohorov Yu V, Pyke R and Rachev ST) Lecture Notes in Statistics 114, 267-276.

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