6.4. Multiparameter Case: Estimation 391
Thus, the information matrix can be written as (1/b)^2 times a matrix whose entries
are free of the parametersaandb. As Exercise 6.4.12 shows, the off-diagonal entries
of the information matrix are 0 if the pdff(z) is symmetric about 0.
Example 6.4.5(Multinomial Distribution).Consider a random trial which can re-
sult in one, and only one, ofkoutcomes or categories. LetXjbe 1 or 0 depending
on whether thejth outcome occurs or does not, forj=1,...,k. Suppose the prob-
ability that outcomejoccurs ispj; hence,
∑k
j=1pj=1. LetX=(X^1 ,...,Xk−^1 )
′
andp=(p 1 ,...,pk− 1 )′. The distribution ofXis multinomial; see Section 3.1.
Recall that the pmf is given by
f(x,p)=
⎛
⎝
k∏− 1
j=1
pxjj
⎞
⎠
⎛
⎝ 1 −
k∑− 1
j=1
pj
⎞
⎠
1 −Pkj−=1^1 xj
, (6.4.18)
where the parameter space is Ω ={p:0<pj< 1 ,j=1,...,k−1;
∑k− 1
j=1pj<^1 }.
We first obtain the information matrix. The first partial of the log offwith
respect topisimplifies to
∂logf
∂pi
=
xi
pi
−
1 −
∑k− 1
j=1xj
1 −
∑k− 1
j=1pj
.
The second partial derivatives are given by
∂^2 logf
∂p^2 i
= −
xi
p^2 i
−
1 −
∑k− 1
j=1xj
(1−
∑k− 1
j=1pj)^2
∂^2 logf
∂pi∂ph
= −
1 −
∑k− 1
j=1xj
(1−
∑k− 1
j=1pj)
2
,i =h<k.
Recall that for this distribution the marginal distribution ofXjis Bernoulli with
meanpj. Recalling thatpk=1−(p 1 +···+pk− 1 ), the expectations of the negatives
of the second partial derivatives are straightforward and result in the information
matrix
I(p)=
⎡
⎢
⎢
⎢
⎣
1
p 1 +
1
pk
1
pk ···
1
pk
1
pk
1
p 2 +
1
pk ···
1
pk
..
.
..
.
..
.
1
pk
1
pk ···
1
pk− 1 +
1
pk
⎤
⎥
⎥
⎥
⎦
. (6.4.19)
This is a patterned matrix with inverse [see page 170 of Graybill (1969)],
I−^1 (p)=
⎡
⎢
⎢
⎢
⎣
p 1 (1−p 1 ) −p 1 p 2 ··· −p 1 pk− 1
−p 1 p 2 p 2 (1−p 2 ) ··· −p 2 pk− 1
..
.
..
.
..
.
−p 1 pk− 1 −p 2 pk− 1 ··· pk− 1 (1−pk− 1 )
⎤
⎥
⎥
⎥
⎦
. (6.4.20)