6.4. Multiparameter Case: Estimation 391Thus, 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 byf(x,p)=⎛
⎝k∏− 1j=1pxjj⎞
⎠⎛
⎝ 1 −k∑− 1j=1pj⎞
⎠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
pk1
pk ···1
pk
1
pk1
p 2 +1
pk ···1
pk
..
...
...
.
1
pk1
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)