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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
1
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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Early stages of an outbreak
100
80
60
40
20
0
0
10
Generation
Mean of offspring distribution = µ
6
Mean number in generation r = I0 µ
r
20
and after that …
4000
3000
2000
1000
0
20
7
25
Generation
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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
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p0 + p1 p0 + p1 p1 p0 + p2 p0 p0 + etc, etc, etc
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 of
s = 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)]
1
0.8
0.6
q
0.4
0.2
0
10
0
0.2
0.4
0.6
v
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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Mean matrix (with everyone susceptible):
Offspring infective
Parent
infective
Child
Child
4
Adult
2
Adult
2
1
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
Parent
infective
Child
AChild
4(1-vc)
Adult
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 Number
There 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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RI0  reproduction based on individuals infected
For this reproduction number we attribute cases actually
infected
A
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For calculating RI0 we attribute 4 infections to A,
in this example.
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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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.
P S
The mean matrix is
P 

S 
p

0
The largest eigenvalue is the larger solution of the equation
x2 –  x –  p = 0.
Therefore
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RI 0   / 2  (p   / 4)
RH0  reproduction based on households infected
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.
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For calculating
RH0 we attribute
5 infections to
individual A ,
in this example.
A
When we attribute infections this way we obtain the
reproduction number RH0 = µ (1+p)
Compare this with
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RI 0   / 2  (p   / 4)
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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Critical vaccination coverage
What are these reproduction numbers when we partially
vaccinate the community?
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
S
P   (1  v ) p 


S   (1  v ) 0 
R Iv   (1  v ) / 2   (1  v )p   2 (1  v )2 / 4)
Compare this with
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P
RHv = µ (1v)(1+p) = (1v)RH0
Strategy H: The critical vaccination coverage is v such that
Example:
µ = 3, p = 0.5
RI0 = 3.44, RH0 = 4.5
Critical vaccination
coverage is
vC = 1 – 1/RH0
= 1 – 1/4.5
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= 0.778
Reproduction number.
RHv = (1v)RH0 =1. Therefore vc= 1 – 1 / RH0.
4
RHv
3
RIv
2
1
0
0
0.2
0.4
0.6
0.8
Vaccination coverage
1
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
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RIv = (1-v) RI0
P
S
P   (1  v ) p (1  v ) 


S   (1  v )
0

Compare this with the reproduction number for infected
households, which becomes
RHv = µ (1-v) [1+p (1-v)]
Strategy I: The critical vaccination coverage is v such that
RIv = (1v)RI0 =1. Therefore vc= 1 – 1 / RI0.
Example:
RI0 = 3.44, RH0 = 4.5
Critical vaccination
coverage is
vC = 1 – 1/RI0
Reproduction number.
µ = 3, p = 0.5
4
3
2
1
= 1 – 1/3.44
= 0.709
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0
0
0.2
0.4
0.6
0.8
Vaccination coverage
1
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
P
S
P   (1  v ) p 1  2v
1 v
S   (1  v )
0

This gives
R Iv   (1  v ) / 2  p (1  2v )   2 (1  v )2 / 4
Also RHv = µ [1-v+p (1-2v)]
(b) v > 0.5
Now there is no within household infection and
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RIv = RHv = µ (1-v)




Strategy O:
µ = 3, p = 0.5
RI0 = 3.44, RH0 = 4.5
Critical vaccination
coverage is
vC = 1 – 1/µ
= 1 – 1/3
= 0.667
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Reproduction number.
Example:
4
3
2
1
0
0
0.2
0.4
0.6
0.8
Vaccination coverage
1
Reproduction number.
Comparing RHv for the different strategies
4
3
2
1
0
0
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0.2
0.4
0.6
0.8
Vaccination coverage
1
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
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These cases have the same offspring distribution
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+s2p1)] 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
Srategy I has
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v2 = v
v0 = (1-v)2 , v1 = 2v(1-v) and
v2 = v2
The first infected household will have
1 case, with probability
2 cases, with probability
v1

v 1  2v 0
2v 0
p
v 1  2v 0
2v 0
(1  p )
v 1  2v 0
Probability(containment) is
 v1

 2v 0
 2
2v 0
v  2v  v  2v (1  p )   v  2v p  
0
1
0
0
 1

 1

where  is the smaller solution of
 v 1s
2v 0s (1  p ) 2v 0 ps 2 

s  Y 


v 1  2v 0
v 1  2v 0 
v 1  2v 0
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Probability(containment)
Strategy H
For Strategy H and
Strategy O
Vaccination coverage = 0.5
Strategy O
Difference in
Probability(containment)
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e
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.
The End
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