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Prevalence and Sibling Recurrence Risk Based on the Multi-Risk Model
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We let Xi represent the dummy variable denoting the affectedness status of a sibling, where Xi
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= 1 if sibling “i” is affected and Xi = 0 if sibling “i” is not affected, for i = 1, 2, …, n.
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Similarly, we let Yi represent the dummy variable denoting sex of a sibling, where Yi = 1 if
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sibling “i” is male and Yi = 0 if sibling “i” is female, for i = 1, 2, …, n.
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It follows, using the notation of this article, that:
K
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a x
P(Xi =1|Yi =1) =
i i
i 1
(= Rm)
K
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P(Xi =1|Yi =0) = p  ai xi (= pRm)
i 1
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When the male proportion, P(Yi =1), and the female proportion, P(Yi =0), are denoted by qm
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and qf, respectively, then the prevalence, R = P(Xi =1), equals:
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P(Yi =1)  P(Xi =1|Yi =1) + P(Yi =0)  P(Xi =1|Yi =0)
K
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= (qm  pq f ) ai xi ( = (qm+pqf)Rm)
(3)
i 1
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Moreover, it follows, using the notation of this article, that:
K
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P(Xi =1, Xj =1|Yi =1, Yj =1) =
a x
i 1
i
2
i
( = Sm)
K
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P(Xi =1, Xj =1|Yi =1, Yj =0) = P(Xi =1, Xj =1|Yi =0, Yj =1) = p  ai xi2 ( = pSm)
i 1
K
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P(Xi =1, Xj =1|Yi =0, Yj =0) = p 2  ai xi2 ( = p2Sm)
i 1
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When the risk that the sibling of the proband will be affected, given that the proband is male,
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is termed as the sibling recurrence risk given by the male proband, this equals:
1
1
P(Xj =1|Xi =1, Yi =1)
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= {P(Xi =1, Xj =1,Yi =1, Yj =1)+ P(Xi =1, Xj =1,Yi =1, Yj =0)}/ P(Xi =1,Yi =1)
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= {P(Xi =1, Xj =1|Yi =1, Yj =1)  qm2 + P(Xi =1, Xj =1|Yi =1, Yj =0)  qm qm}/ qmRm
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= (qm+pqf)Sm/Rm
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Similarly, the sibling recurrence risk given by the female proband is:
P(Xj =1|Xi =1, Yi =0)
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= {P(Xi =1, Xj =1,Yi =0, Yj =1)+ P(Xi =1, Xj =1,Yi =0, Yj =0)}/ P(Xi =1,Yi =0)
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= {P(Xi =1, Xj =1|Yi =0, Yj =1)  qmqf + P(Xi =1, Xj =1|Yi =0, Yj =0)  qf2}/ pqfRm
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= (qm+pqf)Sm/Rm
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Moreover, the joint probability that both siblings i and j are affected equals:
P(Xj =1, Xi =1)
= P(Xi =1, Xj =1|Yi =1, Yj =1)  qm2 + 2P(Xi =1, Xj =1|Yi =0, Yj =1)  qm qm
+ P(Xi =1, Xj =1|Yi =0, Yj =0)  qf2
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= (qm2 + 2pqm qm + p2qf2) Sm
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= (qm+ pqf)2 Sm
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Then, the sibling recurrence risk, S, equals:
P(Xj =1|Xi =1) = P(Xj =1, Xi =1)/ P(Xi =1)
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= (qm+pqf)2 Sm/(qm+pqf)Rm
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= (qm+pqf)Sm/Rm
2
K
1
i 1
0
= (qm  pq f ) ai xi /  x 2 f ( x)dx
1
(4)
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Therefore, the sibling recurrence risk, the sibling recurrence risk given by the male proband
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and the sibling recurrence risk given by the female proband are all equal under the model
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used.
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Log-likelihood Including the Prevalence Information, LLsample+supp
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Suppose that the number of families in the catchment area is T and that the families in the
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catchment area are partitioned into sampled families, A, and the other families, Ac. Then
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consider the decomposition:
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T
T
i 1
i 1
 P(n i )   P(n i | mi  1) P(mi  1)  P(n i | mi  0) P(mi  0)
  P(n i | mi  1) P(mi  1)  P(n i | mi  0) P(mi  0)
iA
.
iAc
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since P(mi =1) = 0 for sampled families, and P(mi ≥ 1) = 0 for non-sampled families.
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As a first approximation on the non-sampled families, we neglect the heterogeneity of the risk
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among families and assume that the number of affected children, mi, is distributed according
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to a binomial distribution: mi ~ Bin (ni , R) . We suppose that P(mi  1)  P(mi  1) , because the
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probability of two or more siblings in a family becoming affected is small (0.0018, compared
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with the prevalence of 0.022 in the catchment area). In this case, the number of families in the
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sample (N) is approximately equal to the number of affected children (M1) that is, N  M1 ,
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because each family in the sample has one affected child. Then, P(mi  1)  ni R(1  R) ni 1 and
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1
P(mi  0)  (1  R) ni . Therefore, we obtain
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P(n i )  const   P(n i | mi  1)  R N (1  R) N1  N  const 
iA
N
 P(n
i
| mi  1)  R M1 (1  R) N1 M1 ,
i 1,m1 1
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which yields LLsample+supp(θ). Note that N1 refers to the number of all children in the
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catchment area and P(n i | mi  0)  1 under the binomial assumption.
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For the AGRE sample, suppose that the number of families in the catchment area is T and that
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the families in the catchment area are partitioned into the families with two or more affected
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children, A2, the families with one affected child, A1, and the families with no affected child,
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A0. Then consider the decomposition as in our sample:
T
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 P(n )  P(n
i
i 1
iA2
i
| mi  2) P(mi  2)  P(n i | mi  1) P(mi  1)  P(n i | mi  0) P(mi  0)
iA1
iA0
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We suppose that P(mi  2)  P(mi  2) , because the probability of three or more siblings in a
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family becoming affected is small. If we assume that the number of affected children, mi, is
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distributed according to a binomial distribution: mi ~ Bin (ni , R) , then
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P(mi  2)  ni (ni  1) R 2 (1  R) ni 2 / 2 , P(mi  1)  ni R(1  R) ni 1 , and P(mi  0)  (1  R) ni .
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Here note that P(n i |mi  1)  P(n i | mi  0)  1 under the binomial assumption. Therefore,
P(n i )  const   P(n i | mi  2)   P(mi  2)   P(mi  1)   P(mi  0)
iA2
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iA2
 const   P(n i | mi  2)  R
iA1
 
2
A2
A1
1
 (1  R)
iA2
 const   P(n i | mi  2)  R M1 (1  R) N1 M1
iA2
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which yields LLsample+supp(θ) for the AGRE sample.
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
A2
( ni  2 ) 
iA0

A1
( ni 1) 
 ni
A0
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MCMC Sampling For a Discrete Model
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Given observation X, we sample from the joint distribution of the model parameters, θ = (x,
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a, w) where {xi} and {ai} (1 ≤ i ≤ K). In brief, the approach is as follows:
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1. Update w, given x and a.
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2. Update a, given w and x.
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3. Update x, given a and w.
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These steps are iterated many times to obtain samples from a Markov Chain whose stationary
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distribution is the joint posterior distribution of all model parameters.
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1. To update w, we first propose a new value w’ from Beta(α0, β0), where Beta(α, β), α0 and
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β0 represent the beta distribution with parameters α and β, Kw  w and Kw  (1-w),
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respectively. We set α1 = Kw  w’ and β1 = Kw  (1-w’). Then, we use a Metropolis-Hasting
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step to accept or reject this new value. The new configuration is accepted with probability:
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 Beta ( w | 1 , 1 ) P( X | x, a, w' ) 

min 1,
 Beta ( w'|  0 ,  0 ) P( X | x, a, w) 
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2. To describe this update, we used the Dirichlet distribution, Dir(γ), where {γi},ｉ = 1,…,K,
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while the prior γ0 is specified as γ0 = Kγ  a. We propose a new value of a’ for a from
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Dir(γ0), and we set Kγ  a’ = γ1. The new proposed values are then accepted with
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1
2
3
probability:
 Dir ( |  1 ) P( X | x, a' , w) 

min 1,
 Dir ( '|  0 ) P( X | x, a, w) 
3. To describe this update, we introduced additional notation. Let Ui+1 denote the interval
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between xi and xi+1 (1 ≤ i ≤ K, U1 = x1 and UK+1 = 1-XK), and set μ0 = Kμ  u. We propose a
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new value of u’ from Dir(μ0), and we set Kμ  u’ = μ1. From u’, we can obtain x’. Then, the
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new proposed values are accepted with probability:
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 Dir (u | 1 ) P( X | x' , a, w) 

min 1,
 Dir (u '|  0 ) P( X | x, a, w) 
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In practice, we adjusted the values of Kw, Kγ and Kμ to 10-100, 500- 10000 and 500- 10000,
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respectively, to achieve a good mixing of the chain.
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MCMC Sampling For a Continuous Model
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Given observation X, we sample from the joint distribution of the model parameters, θ = (α,
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β, w). Briefly, the approach is as follows:
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1. Update w, given α and β.
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2. Update α, given w and β.
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3. Update β, given α and w.
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1. To update w, we first propose a new value w’ from Beta(θ0,κ0), where Beta(θ, κ), θ0 and κ0
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1
represent the beta distribution with parameters θ and κ, Kw  w and Kw  (1-w),
2
respectively. We set θ1 = Kw  w’ and κ1 = Kw  (1-w’). Then, we use a Metropolis-Hasting
3
step to accept or reject the new value. The new configuration is accepted with probability:
 Beta( w | 1 , 1 ) P( X |  ,  , w' ) 

min 1,
 Beta ( w'|  0 ,  0 ) P( X |  ,  , w) 
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2. To update α, we first propose a new value α’ from U((1-Kα)α,(1+Kα)α), where U(δ, ζ)
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represents the uniform distribution over [δ, ζ]. Then, we use a Metropolis-Hasting step to
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accept or reject the new value. The new configuration is accepted with probability:
  P( X |  ' ,  , w) 

min 1,
  ' P( X |  ,  , w) 
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3. To update β, we used the same procedure as that used for α, replacing α with β and
replacing Kα with Kβ.
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In practice, we adjusted the values of Kw , Kα and Kβ to 10-100, 0.10- 0.35 and 0.10- 0.35,
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respectively, to achieve good mixing of the chain.
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