Appendix A. EM algorithms for TDT1-NGS and TDT2-NGS
The incomplete-data likelihood function is written as
𝑁
15
3
𝐿(𝝁, 𝑟1 , 𝑟2 , 𝜀0 , 𝜀1 |𝑿, 𝑽) = ∏ {∑ 𝜋𝑖 (𝝁 , 𝑟1 , 𝑟2 ) ∏ 𝑓𝐵 (𝑥𝑘𝑗 |𝑣𝑘𝑗 , 𝜀0 +
𝑘=1
𝑖=1
𝑗=1
1 − 𝜀0 − 𝜀1
𝑔𝑖𝑗 )}.
2
Let 𝒛𝑘 = (𝑧1𝑘 , ⋯ , 𝑧15𝑘 ) be the missing indicator random vector representing the
membership of trio 𝑘. The value of 𝑧𝑖𝑘 = 1 if and only if trio 𝑘 has the genotype
vector 𝑮𝑖 = (𝑔𝑖1 , 𝑔𝑖2 , 𝑔𝑖3 ). Otherwise, 𝑧𝑖𝑘 = 0.
The complete-data likelihood function can be written as
𝐿𝐶 (𝝁, 𝑟1 , 𝑟2 , 𝜀0 , 𝜀1 |𝑿, 𝑽)
15
3
= ∏𝑁
𝑘=1 {∏𝑖=1 𝜋𝑖 (𝝁, 𝑟1 , 𝑟2 ) ∏𝑗=1 𝑓𝐵 (𝑥𝑘𝑗 |𝑣𝑘𝑗 , 𝜀0 +
1−𝜀0 −𝜀1
2
𝑔𝑖𝑗 )}
𝑧𝑖𝑘
The complete-data log-likelihood function is as follow:
𝑙𝐶 (𝝁, 𝑟1 , 𝑟2 , 𝜀0 , 𝜀1 |𝑿, 𝑽)
15
3
= ∑𝑁
𝑘=1 {∑𝑖=1 𝑧𝑖𝑘 [log 𝜋𝑖 (𝝁, 𝑟1 , 𝑟2 ) + ∑𝑗=1 log 𝑓𝐵 (𝑥𝑘𝑗 |𝑣𝑘𝑗 , 𝜀0 +
1−𝜀0 −𝜀1
2
𝑔𝑖𝑗 )]}
(A1)
E-step
(𝑟)
(𝑟)
Given 𝑟1 , 𝑟2 , 𝝁(𝑟) , 𝜀0 (𝑟) , 𝜀1 (𝑟) that are obtained form the 𝑟 𝑡ℎ M-step of the algorithm
and the read counts 𝒙𝑘 , the missing random vector is distributed as a multinomial
distribution with the posterior probability estimates of trio k having the genotype
category 𝑮𝑖 in (𝑟 + 1)𝑡ℎ step can be obtained by
(𝑟)
(𝑟)
𝜏𝑖𝑘 (𝑟+1) = Pr(trio 𝑘 ′ s genotype vector is 𝑮𝑖 |𝝁(𝑟) , 𝑟1 , 𝑟2 , 𝜀0 (𝑟) , 𝜀1 (𝑟) , 𝒙𝑘 , 𝒗𝑘 )
(𝑟)
=
(𝑟)
𝜋𝑖 (𝝁(𝑟) ,𝑟1 ,𝑟2 )𝑓𝑖 (𝒙𝑘 ;𝒗𝑘 ,𝑮𝑖 ,𝜀0 (𝑟) ,𝜀1 (𝑟) )
(𝑟)
(𝑟)
(𝑟)
(𝑟) ,𝜀 (𝑟) )
∑15
1
𝑠=1 𝜋𝑠 (𝝁 ,𝑟1 ,𝑟2 )𝑓𝑠 (𝒙𝑘 ;𝒗𝑘 ,𝑮𝑗 ,𝜀0
(𝑟)
=
(𝑟) (𝑟)
(𝑟) (𝑟)
(𝑟) 1−𝜀0 −𝜀1
𝑔𝑖𝑗 )
2
𝜇𝑔 𝑔 Pr(𝑔𝑖3 |𝑔𝑖1 ,𝑔𝑖2 ,𝑟1 ,𝑟2 ) ∏3𝑗=1 𝑓𝐵 (𝑥𝑘𝑗 |𝑣𝑘𝑗 ,𝜀0 +
𝑖1 𝑖2
(𝑟)
(𝑟) (𝑟)
(𝑟)
(𝑟)
.
(𝑟) (𝑟)
(𝑟) 1−𝜀0 −𝜀1
𝑔𝑠𝑗 )
2
3
∑15
𝑠=1 𝜇𝑔𝑠1 𝑔𝑠2 Pr(𝑔𝑠3 |𝑔𝑠1 ,𝑔𝑠2 ,𝑟1 ,𝑟2 ) ∏𝑗=1 𝑓𝐵 (𝑥𝑘𝑗 |𝑣𝑘𝑗 ,𝜀0 +
Therefore, 𝐸(𝑧𝑖𝑘 |𝑟1 , 𝑟2 , 𝝁(𝑟) , 𝜀0 (𝑟) , 𝜀1 (𝑟) , 𝒙𝑘 , 𝒗𝑘 ) = 𝜏𝑖𝑘 (𝑟+1) .
M-step
Let 𝑄 (𝑟+1) be the conditional expectation of the complete-data log-likelihood
function:
(𝑟)
(𝑟)
(𝑟)
(𝑟)
𝑄 (𝑟+1) = 𝐸[𝑙𝐶 |𝝁(𝑟) , 𝑟1 , 𝑟2 , 𝜀0 , 𝜀1 , 𝒙, 𝒗]
(𝑟+1)
15
= ∑𝑁
𝑘=1 ∑𝑠=1 𝜏𝑠𝑘
1−𝜀0 −𝜀1
2
[log 𝜇𝑔𝑠1 𝑔𝑠2 + log Pr(𝑔𝑠3 |𝑔𝑠1 , 𝑔𝑠2 , 𝑟1 , 𝑟2 ) + ∑3𝑗=1 log 𝑓𝐵 (𝑥𝑘𝑗 |𝑣𝑘𝑗 , 𝜀0 +
𝑔𝑠𝑗 )]
M-step for mating frequencies
To derive the (𝑟 + 1)𝑡ℎ step estimates of mating frequencies, differentiating 𝑄 (𝑟)
with respect to 𝜇𝑖𝑗 . It is obtained that
𝜕𝑄 (𝑟+1)
𝜕𝜇𝑖𝑗
𝜕
15
= ∑𝑁
𝑘=1 ∑𝑠=1 𝜏𝑠𝑘 𝜕𝜇 [log 𝜇𝑔𝑠1 𝑔𝑠2 ] =
∑15
𝑠=1 𝜏𝑠+ 𝐼(𝑔𝑠1 =𝑖,𝑔𝑠2 =𝑗)
𝜇𝑖𝑗
𝑖𝑗
Since ∑ 𝜇𝑖𝑗 = 1, using the Lagrange multiplier 𝜆,
𝜇𝑖𝑗 =
∑15
𝑠=1 𝜏𝑠+ 𝐼(𝑔𝑠1 =𝑖,𝑔𝑠2 =𝑗)
𝜆
𝜕𝑄 (𝑟)
𝜕𝜇𝑖𝑗
.
− 𝜆 = 0.
. From ∑ 𝜇𝑖𝑗 = 1, the Lagrange multiplier is equal to the
total number of trios 𝜆 = 𝑁 and the (𝑟 + 1)𝑡ℎ step estimates of the mating
frequencies are
(𝑟+1)
𝜇𝑖𝑗
(𝑟+1)
∑15
𝐼(𝑔𝑠1 = 𝑖, 𝑔𝑠2 = 𝑗)
𝑠=1 𝜏𝑠+
=
.
𝑁
When mating symmetric 𝜇𝑖𝑗 = 𝜇𝑗𝑖 ,
𝜕𝑄 (𝑟+1)
𝜕𝜇𝑖𝑗
=
∑15
𝑠=1 𝜏𝑠+ {𝐼(𝑔𝑠1 =𝑖,𝑔𝑠2 =𝑗)+𝐼(𝑔𝑠1 =𝑗,𝑔𝑠2 =𝑖)}
𝜇𝑖𝑗
for 𝑖 ≠ 𝑗 and
𝜕𝑄 (𝑟+1)
𝜕𝜇𝑖𝑖
=
∑15
𝑠=1 𝜏𝑠+ 𝐼(𝑔𝑠1 =𝑖,𝑔𝑠2 =𝑖)
𝜇𝑖𝑖
.
Since ∑ 𝜇𝑖𝑗 = ∑3𝑖=1 𝜇𝑖𝑖 + 2 ∑𝑖<𝑗 𝜇𝑖𝑗 = 1,
𝜕𝑄 (𝑟+1)
𝜕𝜇𝑖𝑖
𝜇𝑖𝑖 =
= 𝜆 and
𝜕𝑄 (𝑟)
𝜕𝜇𝑖𝑗
= 2𝜆. This implies that
∑15
𝑠=1 𝜏𝑠+ 𝐼(𝑔𝑠1 =𝑖,𝑔𝑠2 =𝑖)
𝜆
=
∑15
𝑠=1 𝜏𝑠+ {𝐼(𝑔𝑠1 =𝑖,𝑔𝑠2 =𝑖)+𝐼(𝑔𝑠1 =𝑖,𝑔𝑠2 =𝑖)}
∑15
𝑠=1 𝜏𝑠+ {𝐼(𝑔𝑠1 =𝑖,𝑔𝑠2 =𝑗)+𝐼(𝑔𝑠1 =𝑗,𝑔𝑠2 =𝑖)}
2𝜆
2𝜆
for 𝑖 ≠ 𝑗.
and 𝜇𝑖𝑗 =
Similarly, 𝜆 = 𝑁 and
(𝑟+1)
∑15
{𝐼(𝑔𝑠1 = 𝑖, 𝑔𝑠2 = 𝑗) + 𝐼(𝑔𝑠1 = 𝑗, 𝑔𝑠2 = 𝑖)}
𝑠=1 𝜏𝑠+
=
.
2𝑁
(𝑟+1)
𝜇𝑖𝑗
M-step for error parameters
For symmetric error model 𝜀 = 𝜀0 = 𝜀1 ,
𝜕𝑄 (𝑟+1)
𝜕𝜀
𝜕
𝜕
(𝑟+1)
15
= 𝜕𝜀 ∑𝑁
𝑘=1 ∑𝑠=1 𝜏𝑠𝑘
(𝑟+1)
=
1
(𝑟+1)
[∑3𝑗=1 log 𝑓𝐵 (𝑥𝑘𝑗 |𝑣𝑘𝑗 , 𝜀 + (2 − 𝜀)𝑔𝑠𝑗 )]
15
= 𝜕𝜀 ∑𝑁
𝑘=1 ∑𝑠=1 𝜏𝑠𝑘
15
∑𝑁
𝑘=1 ∑𝑠=1 𝜏𝑠𝑘
[∑3𝑗=1 𝑥𝑘𝑗 𝐼(𝑔𝑠𝑗 = 0) log 𝜀 + (𝑣𝑘𝑗 − 𝑥𝑘𝑗 )𝐼(𝑔𝑠𝑗 = 2) log(1 − 𝜀)]
[∑3𝑗=1 𝑥𝑘𝑗 𝐼(𝑔𝑠𝑗 =0)]
𝜀
(𝑟+1)
−
15
∑𝑁
𝑘=1 ∑𝑠=1 𝜏𝑠𝑘
[∑3𝑗=1(𝑣𝑘𝑗 −𝑥𝑘𝑗 )𝐼(𝑔𝑠𝑗 =2)]
1−𝜀
=0
Therefore,
𝜀
(𝑟+1)
=
15
3
(𝑟+1)
∑𝑁
{𝑥𝑘𝑗 𝐼(𝑔𝑖𝑗 = 0) + (𝑣𝑘𝑗 − 𝑥𝑘𝑗 )𝐼(𝑔𝑖𝑗 = 2)}
𝑘=1 ∑𝑖=1 ∑𝑗=1 𝜏𝑖𝑘
15
3
(𝑟+1) 𝑣 {𝐼(𝑔 = 0) + 𝐼(𝑔 = 2)}
∑𝑁
𝑘𝑗
𝑖𝑗
𝑖𝑗
𝑘=1 ∑𝑖=1 ∑𝑗=1 𝜏𝑖𝑘
.
For asymmetric error model, it is necessary to introduce another missing random
vector 𝒘 = (𝑤1 , 𝑤2 ) for genotype 1. For genotype 1, define 𝑤1 to be the number of
correct reads of allele A and 𝑤2 to be the number of correct reads of allele B. The
total number of reads 𝑣𝑘 = 𝑤2𝑘 + (𝑥𝑘 − 𝑤1𝑘 ) + (𝑣𝑘 − 𝑥𝑘 − 𝑤2𝑘 ) + 𝑤1𝑘 . The random
vector (𝑤2𝑘 , 𝑥𝑘 − 𝑤1𝑘 , 𝑣𝑘 − 𝑥𝑘 − 𝑤2𝑘 , 𝑤1𝑘 ) follows a multinomial distribution with
the probability vector (
1−𝜀0 𝜀0 𝜀1 1−𝜀1
,
2
2
,
2
,
2
) unconditionally. The complete-data log-
likelihood for asymmetric error parameters can be written as
𝑙𝐶∗ (𝝁, 𝑟1 , 𝑟2 , 𝜀0 , 𝜀1 |𝑿, 𝑽)
15
3
𝑥𝑘𝑗
(1 − 𝜀0 )𝑣𝑘𝑗 −𝑥𝑘𝑗 } +
= ∑𝑁
𝑘=1 𝑧𝑖𝑘 {∑𝑖=1 [log 𝜋𝑖 (𝝁, 𝑟1 , 𝑟2 ) + ∑𝑗=1 (𝐼(𝑔𝑖𝑗 = 0) log{(𝜀0 )
1−𝜀0 𝑤2𝑘𝑗
𝐼(𝑔𝑖𝑗 = 1) log {(
2
)
𝜀
( 20 )
𝑥𝑘𝑗 −𝑤1𝑘𝑗
𝜀
𝑣𝑘𝑗 −𝑥𝑘𝑗 −𝑤2𝑘𝑗
( 21 )
(
1−𝜀1 𝑤1𝑘𝑗
2
)
2) log{(1 − 𝜀1 )𝑥𝑘𝑗 (𝜀1 )𝑣𝑘𝑗−𝑥𝑘𝑗 })]}.
} + 𝐼(𝑔𝑖𝑗 =
(A2)
The conditional expectation of 𝑤1𝑘𝑗 given the 𝑟 𝑡ℎ estimates of the parameters and
(𝑟)
(𝑟)
(𝑟)
(𝑟)
the data is 𝐸(𝑤1𝑘𝑗 |𝝁(𝑟) , 𝑟1 , 𝑟2 , 𝜀0 , 𝜀1 , 𝒙, 𝒗) = 𝑥𝑘𝑗
(𝑟)
1−𝜀1
(𝑟)
(𝑟)
𝜀0 +1−𝜀1
and that of 𝑤2𝑘𝑗 is
(𝑟)
(𝑟)
(𝑟)
(𝑟)
1−𝜀0
(𝑟)
𝐸(𝑤2𝑘𝑗 |𝝁(𝑟) , 𝑟1 , 𝑟2 , 𝜀0 , 𝜀1 , 𝒙, 𝒗) = (𝑣𝑘𝑗 − 𝑥𝑘𝑗 )
(𝑟)
(𝑟)
𝜀1 +1−𝜀0
. The conditional
expectation of Equation (A2) can be given by
(𝑟+1)
𝑄 ∗(𝑟+1) = ∑𝑁
𝑘=1 𝜏𝑖𝑘
3
{∑15
𝑖=1 [log 𝜋𝑖 (𝝁, 𝑟1 , 𝑟2 ) + ∑𝑗=1 (𝐼(𝑔𝑖𝑗 = 0){𝑥𝑘𝑗 log 𝜀0 +
(𝑟)
(𝑣𝑘𝑗 − 𝑥𝑘𝑗 ) log(1 − 𝜀0 )} + 𝐼(𝑔𝑖𝑗 = 1) {(𝑣𝑘𝑗 − 𝑥𝑘𝑗 )
(𝑟)
𝑥𝑘𝑗
𝜀0
(𝑟)
1−𝜀0
(𝑟)
(𝑟)
𝜀1 +1−𝜀0
(𝑟)
(𝑟)
(𝑟) log 𝜀0 + (𝑣𝑘𝑗 − 𝑥𝑘𝑗 )
𝜀0 +1−𝜀1
𝜀1
(𝑟)
log(1 − 𝜀0 ) +
(𝑟) log 𝜀1 + 𝑥𝑘𝑗
𝜀1 +1−𝜀0
1−𝜀1
(𝑟)
(𝑟)
𝜀0 +1−𝜀1
log(1 − 𝜀1 )} +
𝐼(𝑔𝑖𝑗 = 2){𝑥𝑘𝑗 log(1 − 𝜀1 ) + (𝑣𝑘𝑗 − 𝑥𝑘𝑗 ) log 𝜀1 })]}.
By differentiating 𝑄 ∗(𝑟+1) with respect to the error parameters,
𝜕𝑄 ∗(𝑟+1)
𝜕𝜀1
𝜕𝑄 ∗(𝑟+1)
𝜕𝜀0
= 0 and
= 0 yield
𝜀0 (𝑟+1)
𝜀0 (𝑟)
𝐼(𝑔𝑖𝑗 = 1)}
𝜀0 (𝑟) + 1 − 𝜀1 (𝑟)
=
,
𝜀0 (𝑟)
𝑥𝑘𝑗 {𝐼(𝑔𝑖𝑗 = 0) + (𝑟)
𝐼(𝑔𝑖𝑗 = 1)}
𝜀0 + 1 − 𝜀1 (𝑟)
15 ∑3
(𝑟+1)
∑𝑁
∑
𝜏
𝑘=1 𝑖=1 𝑗=1 𝑖𝑘
1 − 𝜀0 (𝑟)
+(𝑣𝑘𝑗 − 𝑥𝑘𝑗 ) {𝐼(𝑔𝑖𝑗 = 0) + ( (𝑟)
)𝐼(𝑔𝑖𝑗 = 1)}
𝜀1 + 1 − 𝜀0 (𝑟)
[
]
15
3
(𝑟+1)
∑𝑁
𝑥𝑘𝑗 {𝐼(𝑔𝑖𝑗 = 0) +
𝑘=1 ∑𝑖=1 ∑𝑗=1 𝜏𝑖𝑘
𝜀1 (𝑟+1)
𝜀1 (𝑟)
𝐼(𝑔𝑖𝑗 = 1)}
𝜀1 (𝑟) + 1 − 𝜀0 (𝑟)
=
.
𝜀 (𝑟)
(𝑣𝑘𝑗 − 𝑥𝑘𝑗 ) {𝐼(𝑔𝑖𝑗 = 2) + (𝑟) 1
𝐼(𝑔
=
1)}
𝑖𝑗
𝜀1 + 1 − 𝜀0 (𝑟)
15 ∑3
(𝑟+1)
∑𝑁
∑
𝜏
𝑘=1 𝑖=1 𝑗=1 𝑖𝑘
1 − 𝜀1 (𝑟)
+𝑥𝑘𝑗 {𝐼(𝑔𝑖𝑗 = 2) + ( (𝑟)
)𝐼(𝑔𝑖𝑗 = 1)}
𝜀0 + 1 − 𝜀1 (𝑟)
[
]
15
3
(𝑟+1)
∑𝑁
(𝑣𝑘𝑗 − 𝑥𝑘𝑗 ) {𝐼(𝑔𝑖𝑗 = 2) +
𝑘=1 ∑𝑖=1 ∑𝑗=1 𝜏𝑖𝑘
M-step for transmission parameters
𝑟
For multiplicative model 𝑟1 = 𝑟 and 𝑟2 = 𝑟 2 , let 𝑡 = 1+𝑟.
log Pr(𝑔𝑖3 |𝑔𝑖1 , 𝑔𝑖2 , 𝑟1 , 𝑟2 ) =
{
𝐼(𝑮𝑖 = (0,1,1)) + 𝐼(𝑮𝑖 = (1,0,1)) + 𝐼(𝑮𝑖 = (1,1,1))
} log 𝑡 +
+2𝐼(𝑮𝑖 = (1,1,2)) + 𝐼(𝑮𝑖 = (1,2,2)) + 𝐼(𝑮𝑖 = (2,1,2))
{
𝐼(𝑮𝑖 = (0,1,0)) + 𝐼(𝑮𝑖 = (1,0,0)) + 2𝐼(𝑮𝑖 = (1,1,0))
} log(1 − 𝑡) + 𝐶𝑜𝑛𝑠𝑡.
+𝐼(𝑮𝑖 = (1,1,1)) + 𝐼(𝑮𝑖 = (1,2,1)) + 𝐼(𝑮𝑖 = (2,1,1))
𝜕𝑄 (𝑟+1)
𝜕𝑡
15
= ∑ 𝜏𝑖+ (𝑟+1) {
𝑖=1
15
− ∑ 𝜏𝑖+ (𝑟+1) {
𝑖=1
𝐼(𝑮𝑖 = (0,1,1)) + 𝐼(𝑮𝑖 = (1,0,1)) + 𝐼(𝑮𝑖 = (1,1,1)) 1
}
+2𝐼(𝑮𝑖 = (1,1,2)) + 𝐼(𝑮𝑖 = (1,2,2)) + 𝐼(𝑮𝑖 = (2,1,2)) 𝑡
𝐼(𝑮𝑖 = (0,1,0)) + 𝐼(𝑮𝑖 = (1,0,0)) + 2𝐼(𝑮𝑖 = (1,1,0))
1
}
= 0.
+𝐼(𝑮𝑖 = (1,1,1)) + 𝐼(𝑮𝑖 = (1,2,1)) + 𝐼(𝑮𝑖 = (2,1,1)) 1 − 𝑡
𝐼(𝑮𝑖 = (0,1,1)) + 𝐼(𝑮𝑖 = (1,0,1)) + 𝐼(𝑮𝑖 = (1,1,1))
}
+2𝐼(𝑮𝑖 = (1,1,2)) + 𝐼(𝑮𝑖 = (1,2,2)) + 𝐼(𝑮𝑖 = (2,1,2))
.
𝐼(𝑮𝑖 = (0,1,1)) + 𝐼(𝑮𝑖 = (1,0,1)) + 𝐼(𝑮𝑖 = (1,1,1))
+2𝐼(𝑮𝑖 = (1,1,2)) + 𝐼(𝑮𝑖 = (1,2,2)) + 𝐼(𝑮𝑖 = (2,1,2))
(𝑟+1)
∑15
{
𝑖=1 𝜏𝑖+
𝑡 (𝑟+1) =
(𝑟+1)
∑15
𝑖=1 𝜏𝑖+
+𝐼(𝑮𝑖 = (0,1,0)) + 𝐼(𝑮𝑖 = (1,0,0)) + 2𝐼(𝑮𝑖 = (1,1,0))
{ +𝐼(𝑮𝑖 = (1,1,1)) + 𝐼(𝑮𝑖 = (1,2,1)) + 𝐼(𝑮𝑖 = (2,1,1)) }
Therefore,
(𝑟+1)
𝑟1
𝐼(𝑮𝑖 = (0,1,1)) + 𝐼(𝑮𝑖 = (1,0,1)) + 𝐼(𝑮𝑖 = (1,1,1))
}
+2𝐼(𝑮𝑖 = (1,1,2)) + 𝐼(𝑮𝑖 = (1,2,2)) + 𝐼(𝑮𝑖 = (2,1,2))
=
,
𝐼(𝑮𝑖 = (0,1,0)) + 𝐼(𝑮𝑖 = (1,0,0)) + 2𝐼(𝑮𝑖 = (1,1,0))
15
(𝑟+1)
∑𝑖=1 𝜏𝑖+
{
}
+𝐼(𝑮𝑖 = (1,1,1)) + 𝐼(𝑮𝑖 = (1,2,1)) + 𝐼(𝑮𝑖 = (2,1,1))
(𝑟+1)
∑15
{
𝑖=1 𝜏𝑖+
(𝑟+1)
𝑟2
(𝑟+1) 2
= (𝑟1
) .
𝑟
For genetic model-free TDT2-NGS, let 𝛼 = log 𝑟1 and 𝛽 = log 𝑟2.
1
log Pr(𝑔𝑖3 |𝑔𝑖1 , 𝑔𝑖2 , 𝑟1 , 𝑟2 ) = {𝐼(𝑮𝑖 = (0,1,1)) + 𝐼(𝑮𝑖 = (1,0,1)) + 𝐼(𝑮𝑖 = (1,1,1)) +
𝐼(𝑮𝑖 = (1,1,2))}𝛼 − {𝐼(𝑮𝑖 = (0,1,1)) + 𝐼(𝑮𝑖 = (1,0,1)) + 𝐼(𝑮𝑖 = (0,1,0)) +
𝐼(𝑮𝑖 = (1,0,0))} log(1 + 𝑒 𝛼 ) + {𝐼(𝑮𝑖 = (1,2,2)) + 𝐼(𝑮𝑖 = (2,1,2)) + 𝐼(𝑮𝑖 =
(1,1,2))}𝛽 − {𝐼(𝑮𝑖 = (1,2,2)) + 𝐼(𝑮𝑖 = (2,1,2)) + 𝐼(𝑮𝑖 = (1,2,1)) + 𝐼(𝑮𝑖 =
(2,1,1))} log(1 + 𝑒 𝛽 ) − {𝐼(𝑮𝑖 = (1,1,0)) + 𝐼(𝑮𝑖 = (1,1,1)) + 𝐼(𝑮𝑖 =
(1,1,2))} log(1 + 2𝑒 𝛼 + 𝑒 𝛼+𝛽 )
There is no closed iteration formula in this case. To find the (𝑟 + 1)𝑡ℎ step estimates
of the transmission parameters, the Fisher’s scoring method is applied. For
(𝑟+1)
simplicity, let 𝑎 = (∑15
{𝐼(𝑮𝑖 = (0,1,0)) + 𝐼(𝑮𝑖 =
𝑖=1 𝜏𝑖+
𝑇
(𝑟+1)
(1,0,0))} , ∑15
{𝐼(𝑮𝑖 = (0,1,1)) + 𝐼(𝑮𝑖 = (1,0,1))}) , 𝑎+ = 𝑎1 + 𝑎2,
𝑖=1 𝜏𝑖+
(𝑟+1)
(𝑟+1)
(𝑟+1)
𝑏 = (∑15
𝐼(𝑮𝑖 = (1,1,0)) , ∑15
𝐼(𝑮𝑖 = (1,1,1)) , ∑15
𝐼(𝑮𝑖 =
𝑖=1 𝜏𝑖+
𝑖=1 𝜏𝑖+
𝑖=1 𝜏𝑖+
𝑇
(1,1,2))) , 𝑏+ = 𝑏1 + 𝑏2 + 𝑏3,
(𝑟+1)
(𝑟+1)
𝑐 = (∑15
{𝐼(𝑮𝑖 = (1,2,1)) + 𝐼(𝑮𝑖 = (2,1,1))} , ∑15
{𝐼(𝑮𝑖 =
𝑖=1 𝜏𝑖+
𝑖=1 𝜏𝑖+
(1,2,2)) + 𝐼(𝑮𝑖 = (2,1,2))}) , 𝑐+ = 𝑐1 + 𝑐2 .
The first and second derivatives of 𝑄 (𝑟+1) in 𝛼 and 𝛽 can be obtained as
𝑇
𝜕𝑄 (𝑟+1) 𝜕𝑄 (𝑟+1)
(
,
)
𝜕𝑟1
𝜕𝑟2
𝑒𝛼
2𝑒 𝛼 + 𝑒 𝛼+𝛽
𝑒𝛽
= (𝑎2 − 𝑎+
+ 𝑏2 + 𝑏3 − 𝑏+
, 𝑐 − 𝑐+
1 + 𝑒𝛼
1 + 2𝑒 𝛼 + 𝑒 𝛼+𝛽 1
1 + 𝑒𝛽
𝑇
𝑒 𝛼+𝛽
+ 𝑏3 − 𝑏+
)
1 + 2𝑒 𝛼 + 𝑒 𝛼+𝛽
𝜕2 𝑄 (𝑟+1)
𝜕2 𝑄(𝑟+1)
𝜕𝛼2
𝜕𝛼𝜕𝛽
𝜕2 𝑄 (𝑟+1)
𝜕2 𝑄(𝑟+1)
𝜕𝛼𝜕𝛽
𝜕𝛽 2
(
)=
𝑒𝛼
𝑎+ (1+𝑒 𝛼)2 + 𝑏+
𝑏+
(
(𝑟),𝑠
Let 𝑟1
2𝑒 𝛼 +𝑒 𝛼+𝛽
𝑏+
2
(1+2𝑒 𝛼 +𝑒 𝛼+𝛽 )
𝑒 𝛼+𝛽
𝑐+
2
(1+2𝑒 𝛼 +𝑒 𝛼+𝛽 )
(𝑟),𝑠
= 𝑒 𝛼 and 𝑟2
scoring algorithm.
(𝑟),𝑠 𝛽
= 𝑟1
𝑒 𝛼+𝛽
(1+2𝑒 𝛼 +𝑒 𝛼+𝛽 )
𝑒 𝛼 𝑒 𝛼+𝛽
(𝑟),𝑠
(𝑟),𝑠 2
(𝑟1 +𝑟2 )
+ 𝑏+
2
,
(1+2𝑒 𝛼 )𝑒 𝛼+𝛽
2
(1+2𝑒 𝛼 +𝑒 𝛼+𝛽 )
)
𝑒 for the 𝑠 𝑡ℎ step iteration in the following Fisher’s
𝛿 (𝑟),𝑠 = (𝑎2 − 𝑎+
𝑟2
(𝑟),𝑠
1+2𝑟1
(𝑟),𝑠
(𝑟),𝑠 + 𝑏2 + 𝑏3 − 𝑏+
1+𝑟1
(𝑟),𝑠
+𝑟2
(𝑟),𝑠
1+2𝑟1
(𝑟),𝑠
+𝑟2
(𝑟),𝑠
(𝑟),𝑠 , 𝑐1 − 𝑐+
+𝑟2
𝑟2
(𝑟),𝑠
𝑟1
(𝑟),𝑠
+𝑟2
+ 𝑏3 −
)
(𝑟),𝑠
𝑎+
𝐻 (𝑟),𝑠 =
2𝑟1
𝑇
(𝑟),𝑠
𝑏+
(𝑟),𝑠
𝑟1
(𝑟),𝑠
𝑟1
(𝑟),𝑠 2
(1+𝑟1 )
+ 𝑏+
2𝑟1
(𝑟),𝑠
(𝑟),𝑠
+𝑟2
𝑏+
(𝑟),𝑠
(𝑟),𝑠 2
(1+2𝑟1 +𝑟2 )
(𝑟),𝑠
(
𝑏+
𝑟2
(𝑟),𝑠
(1+2𝑟1
𝑐+
(𝑟),𝑠 2
+𝑟2
)
𝑟2
(𝑟),𝑠
(1+2𝑟1
(𝑟),𝑠 (𝑟),𝑠
𝑟2
(𝑟),𝑠
(𝑟),𝑠 2
(𝑟1 +𝑟2 )
𝑟1
(𝑟),𝑠 2
+𝑟2
)
(𝑟),𝑠
+ 𝑏+
(1+2𝑟1
(𝑟),𝑠
(1+2𝑟1
,
(𝑟),𝑠
)𝑟2
(𝑟),𝑠 2
+𝑟2
)
)
−1
𝛿 (𝑟),𝑠+1 = (𝐻 (𝑟),𝑠 ) 𝛿 (𝑟),𝑠
To recover the transmission parameters,
(𝑟),𝑠+1
𝑟𝑖
= exp(𝛿 (𝑟),𝑠+1 ), for 𝑖 = 1,2.
(𝑟),𝑠 ∗ +1
Iterate these until |𝑟1
(𝑟),𝑠∗
− 𝑟1
(𝑟),𝑠 ∗ +1
| + |𝑟2
(𝑟),𝑠∗
− 𝑟2
| is less than the specified
tolerance for some 𝑠 ∗ or 𝑠 ∗ reaches the pre-specified maximum number of
iterations.
(𝑟+1)
𝑟1
(𝑟),𝑠∗ +1
= 𝑟1
(𝑟+1)
and 𝑟2
(𝑟),𝑠∗ +1
= 𝑟2
.