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profile likelihood 的理解
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profile likelihood 的理解
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Chapter 3
The Profile Likelihood
3.1 The Profile Likelihood
3.1.1 The method of profiling
Let us suppose that the unknown parameters ✓ can be parti t i on ed as ✓
0
=(
0
,
0
), where
are the p-dimensional parameters of interest (eg. mean) and are the q-dimensional
nuisance parameters (eg. v ariance). We will need to estimate both and ,butour
interest lies only in the parameter .Toachievethisoneoftenprofilesoutthenuisance
parameters. To motivate the profile likelihood, we first describe a method to estimate the
parameters ( , )intwostagesandconsidersomeexamples.
Let us suppse that {X
i
} ar e iid random variables, with density f(x; , ) where o ur
objective is to estimate and .Inthiscasethelog-likelihoodis
L
n
( , )=
n
X
i=1
log f(X
i
; , ).
To estimate and one can use (
ˆ
n
,
ˆ
n
)=argmax
,
L
n
( , ). However, this can be
difficult to directly maximise. Instead let us consider a di↵erent method, which may,
sometimes, be easier to eval u ate. Sup pose, for now, is known, then we rewrite the
likelihood as L
n
( , )=L
()(toshowthat is fixed but varies). To estimate we
maximise L
()withrespectto,i.e.
ˆ
=argmax
L
().
99
In reality is unknown, hence for each we can evaluate
ˆ
. Note that for each ,we
have a new curve L
()over. Now to estimate , we evaluate the maximum L
(),
over ,andchoosethe ,whichisthemaximumoverallthesecurves. Inotherwords,
we evaluate
ˆ
n
=argmax
L
(
ˆ
)=argmax
L
n
( ,
ˆ
).
Abitoflogicaldeductionshowsthat
ˆ
n
and
ˆ
n
are the maximum likelihood estimators
(
ˆ
n
,
ˆ
n
)=argmax
,
L
n
( , ).
We note that we have profiled out nuisance parameter ,andthelikelihoodL
(
ˆ
)=
L
n
( ,
ˆ
)isintermsoftheparameterofinterest .
The advantage of this procedure is best illustrated through some examples.
Example 3.1.1 (The Weibull distribution) Let us suppose that {X
i
} are iid random
variables from a Weibull distribution with density f(x; ↵, ✓)=
↵y
↵1
✓
↵
exp((y/✓)
↵
).We
know from Example 2.2.2, that if ↵, were known an explicit expression for the MLE can
be derived, it is
ˆ
✓
↵
=argmax
✓
L
↵
(✓)
=argmax
✓
n
X
i=1
✓
log ↵ +(↵ 1) log Y
i
↵ log ✓
Y
i
✓
↵
◆
=argmax
✓
n
X
i=1
✓
↵ log ✓
Y
i
✓
↵
◆
=(
1
n
n
X
i=1
Y
↵
i
)
1/↵
,
where L
↵
(X; ✓)=
P
n
i=1
✓
log ↵ +(↵ 1) log Y
i
↵ log ✓
Y
i
✓
↵
◆
. Thus for a given ↵,
the maximum likelihood estimator of ✓ can be derived. The maximum likelihood estimator
of ↵ is
ˆ↵
n
=argmax
↵
n
X
i=1
✓
log ↵ +(↵ 1) log Y
i
↵ log(
1
n
n
X
i=1
Y
↵
i
)
1/↵
Y
i
(
1
n
P
n
i=1
Y
↵
i
)
1/↵
↵
◆
.
Therefore, the maximum likelihood estimator of ✓ is (
1
n
P
n
i=1
Y
ˆ↵
n
i
)
1/ˆ↵
n
. We observe that
evaluating ˆ↵
n
can be tricky but no worse than maximising the likelihood L
n
(↵, ✓) over ↵
and ✓.
100
As we mentioned above, we are not interest in the nuisance parameters and are only
interesting in testing and constructing CIs for .Inthiscase,weareinterestedinthe
limiting distribution of the MLE
ˆ
n
. Using Theorem 2.6.2(ii) we have
p
n
ˆ
n
ˆ
n
!
D
!N
✓
0,
I
I
I
I
!
1
◆
.
where
I
I
I
I
!
=
E
@
2
log f (X
i
; ,)
@
2
E
@
2
log f (X
i
; ,)
@ @
E
@
2
log f (X
i
; ,)
@ @
0
E
@
2
log f (X
i
; ,)
@
2
!
. (3.1)
To derive an exact expression for the limiting variance of
p
n(
ˆ
n
), we use the block
inverse matrix identity.
Remark 3.1.1 (Inverse of a block matrix) Suppose that
AB
CD
!
is a square matrix. Then
AB
CD
!
1
=
(A BD
1
C)
1
A
1
B(D CA
1
B)
1
D
1
CB(A BD
1
C)
1
(D CA
1
B)
1
!
. (3.2)
Using (3.2) we have
p
n(
b
n
)
D
!N(0, (I
,
I
,
I
1
I
,
)
1
). (3.3)
Thus if is a scalar we can use the above to construct confidence intervals for .
Example 3.1.2 (Block diagonal information matrix ) If
I( , )=
I
,
0
0 I
,
!
,
then using (3.3) we have
p
n(
b
n
)
D
!N(0,I
1
,
).
101
3.1.2 The score and the log-likelihood ratio for the profile like-
lihood
To ease notation, let us suppose that
0
and
0
are the true parameters in the distribution.
We now consider the log-likelihood ratio
2
⇢
max
,
L
n
( , ) max
L
n
(
0
,)
, (3.4)
where
0
is the true parameter. However, to derive the limiti n g distribution in this case
for this statistic is a little more compli cated than the log-l i kelihood ratio test that does
not involve nuisance parameters. This is because di r ect l y applying Taylor expansion does
not work since this is usually expanded about the true parameters. We observe that
2
⇢
max
,
L
n
( , ) max
L
n
(
0
,)
=2
⇢
max
,
L
n
( , ) L
n
(
0
,
0
)
| {z }
2
p+q
2
n
max
L
n
(
0
,) max
L
n
(
0
,
0
)
o
| {z }
2
q
.
It seems reasonable that the di↵erence m ay be a
2
p
but it is really not clear by. Below,
we show that by using a few Taylor expansions why this is true.
In the theorem below we will derive the distribution of the score and the nested log-
likelihood.
Theorem 3.1.1 Suppose Assumption 2.6.1 holds. Suppose that (
0
,
0
) are the true
parameters. Then we have
@L
n
( , )
@
c
ˆ
0
,
0
⇡
@L
n
( , )
@
c
0
,
0
@L
n
( ,)
@
c
0
,
0
I
1
0
0
I
0
0
(3.5)
1
p
n
@L
n
( , )
@
c
0
,
ˆ
0
D
!N(0, (I
0
0
I
0
0
I
1
0
0
I
0
,
0
)) (3.6)
where I is defined as in (3.1) and
2
⇢
L
n
(
ˆ
n
,
ˆ
n
) L
n
(
0
,
ˆ
0
)
D
!
2
p
, (3.7)
where p denotes the dimension of . This result is often called Wilks Theorem.
102
PROOF. We first prove (3.5) which is the b a si s of the proofs of (3.6). To avoid, not a t i on a l
difficulties we will assume that
@L
n
( ,)
@
c
ˆ
0
,
0
and
@L
n
( ,)
@
c
=
0
,
0
are univariate random
variables.
Our objective is to find an expression for
@L
n
( ,)
@
c
ˆ
0
,
0
in terms of
@L
n
( ,)
@
c
=
0
,
0
and
@L
n
( ,)
@
c
=
0
,
0
which will allow us t o obtain its vari an ce and asymptotic distribution .
Making a Taylor expansi on of
@L
n
( ,)
@
c
ˆ
0
,
0
about
@L
n
( ,)
@
c
0
,
0
gives
@L
n
( , )
@
c
ˆ
0
,
0
⇡
@L
n
( , )
@
c
0
,
0
+(
ˆ
0
0
)
@
2
L
n
( , )
@@
c
0
,
0
.
Notice that we have used ⇡ instead of = because we replace the second derivative
with its true pa r am e te rs . If the sample size is large enough then
@
2
L
n
( ,)
@@
c
0
,
0
⇡
E
@
2
L
n
( ,)
@@
c
0
,
0
;eg.intheiidcasewehave
1
n
@
2
L
n
( , )
@@
c
0
,
0
=
1
n
n
X
i=1
@
2
log f(X
i
; , )
@@
c
0
,
0
⇡ E
✓
@
2
log f(X
i
; , )
@@
c
0
,
0
◆
= I
,
Therefore
@L
n
( , )
@
c
ˆ
0
,
0
⇡
@L
n
( , )
@
c
0
,
0
n(
ˆ
0
0
)I
. (3.8)
Next we make a decomposition of (
ˆ
0
0
). We recall that since L
n
(
0
,
ˆ
0
)=argmax
L
n
(
0
,)
then
@L
n
( , )
@
c
ˆ
0
,
0
=0
(if the maximum is not on the boundary). Therefore making a Taylor expansion of
@L
n
(
0
,)
@
c
ˆ
0
,
0
about
@L
n
(
0
,)
@
c
0
,
0
gives
@L
n
(
0
,)
@
c
ˆ
0
,
0
| {z }
=0
⇡
@L
n
(
0
,)
@
c
0
,
0
+
@
2
L
n
(
0
,)
@
2
c
0
,
0
(
ˆ
0
0
).
Replacing
@
2
L
n
(
0
,)
@
2
c
0
,
0
with I
gives
@L
n
(
0
,)
@
c
0
,
0
nI
(
ˆ
0
0
) ⇡ 0,
103
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