Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

6.6. The EM Algorithm 407

By expression (6.6.1), the conditional distributionZgivenXis the ratio of (6.6.12)

to (6.6.11); that is,

k(z|θ,x)=

∏n 1

i=1f(xi−θ)

∏n 2

i=1f(zi−θ)

[1−F(a−θ)]n^2

∏n 1

i=1f(xi−θ)

=[1−F(a−θ)]−n^2

∏n^2

i=1

f(zi−θ),a<zi,i=1,...,n 2 .(6.6.13)

Thus,ZandXare independent, andZ 1 ,...,Zn 2 are iid with the common pdf
f(z−θ)/[1−F(a−θ)], forz>a. Based on these observations and expression
(6.6.13), we have the following derivation:

Q(θ|θ 0 ,x)=Eθ 0 [logLc(θ|x,Z)]

= Eθ 0

[n 1

∑

i=1

logf(xi−θ)+

∑n^2

i=1

logf(Zi−θ)

]

=

∑n^1

i=1

logf(xi−θ)+n 2 Eθ 0 [logf(Z−θ)]

=

∑n^1

i=1

logf(xi−θ)

+n 2

∫∞

a

logf(z−θ)

f(z−θ 0 )

1 −F(a−θ 0 )

dz. (6.6.14)

This last result is the E step of the EM algorithm. For the M step, we need the
partial derivative ofQ(θ|θ 0 ,x) with respect toθ. This is easily found to be

∂Q

∂θ

=−

{n

∑^1

i=1

f′(xi−θ)

f(xi−θ)

+n 2

∫∞

a

f′(z−θ)

f(z−θ)

f(z−θ 0 )

1 −F(a−θ 0 )

dz

}

. (6.6.15)

Assuming thatθ 0 =θ̂ 0 , the first-step EM estimate would be the value ofθ,sayθ̂(1),
which solves∂Q∂θ = 0. In the next example, we obtain the solution for a normal
model.
Example 6.6.1.Assume the censoring model given above, but now assume that
Xhas aN(θ,1) distribution. Thenf(x)=φ(x)=(2π)−^1 /^2 exp{−x^2 / 2 }.Itiseasy
to show thatf′(x)/f(x)=−x. Letting Φ(z) denote, as usual, the cdf of a standard
normal random variable, by (6.6.15) the partial derivative ofQ(θ|θ 0 ,x) with respect
toθfor this model simplifies to

∂Q

∂θ

=

∑n^1

i=1

(xi−θ)+n 2

∫∞

a

(z−θ)

1

√

2 π

exp{−(z−θ 0 )^2 / 2 }

1 −Φ(a−θ 0 )

dz

= n 1 (x−θ)+n 2

∫∞

a

(z−θ 0 )

1

√

2 π

exp{−(z−θ 0 )^2 / 2 }

1 −Φ(a−θ 0 )

dz−n 2 (θ−θ 0 )

= n 1 (x−θ)+

n 2

1 −Φ(a−θ 0 )

φ(a−θ 0 )−n 2 (θ−θ 0 ).