Transcript PowerPoint

COMPOSITE TESTS AND THE
LIKELIHOOD RATIO TEST
Lecture XXIV
Simple Tests Against a Composite

Mathematically, we now can express the tests as
testing between H0:q = q0 against H1:q  Q1,
where Q1 is a subset of the parameter space.

Given this specification, we must modify our
definition of the power of the test because the b
value (the probability of accepting the null
hypothesis when it is false) is not unique. In this
regard, it is useful to develop the power function.

Definition 9.4.1. If the distribution of the sample X
depends on a vector of parameters q, we define
the power function of the test based on the critical
region R by
Qq   PX  R q 

Definition 9.4.2. Let Q1(q) and Q2(q) be the power
functions of two tests respectively. Then we say that
the first test is uniformly better (or uniformly most
powerful) than the second in testing H0:q = q0
against H1:q  Q1 if Q1(q0)=Q2(q0) and
Q1 q   Q2 q  for all q  Q1
and
Q1 q   Q2 q  for at least one q  Q1

Definition 9.4.3. A test R is the uniformly most power
(UMP) test of size (level) a for testing H0:q = q0
against H1:q  Q1 if P(R|q0)= (≤) a and any other
test R1 such that P(R1|q0)= (≤) a, we have P(R|q) ≥
P(R1|q) for any qQ1.

The likelihood ratio test usually gives the UMP test if
an UMP test exists. Even if the UMP does not exist,
the likelihood ratio test has good properties.

Definition 9.4.4. Let L(x|q) be the likelihood function
and let the null and alternative hypotheses be H0:q
= q0 and H1:q  Q1, where Q1 is a subset of the
parameter space Q. Then the likelihood ratio test
of H0 against H1 is defined by the critical region

Lq 0 x 
sup Lq x 
q0 Q
c
where c is chosen to satisfy
P( < c|H0) = a for a certain value of a.

Example 9.4.3. Let the sample be Xi~N(m,s2),
i=1,2,…n, where s2 is assumed to be known. Let xi
be the observed value of Xi. Testing H0:m=m0
against H1:m>m0. The likelihood ratio test is to
reject H0 if
 1
2
xi  m 0  
exp 
2 
2s i 1



c
n
 1
2
xi  m  
sup exp 
2 
m m0
 2s i 1

n

Assume that we had the sample X={6,7,8,9,10}
from the preceding example and wanted to
construct a likelihood ratio test for m>7.5:


 1
2
2
2 
exp  2 6  7.5  7  7.5  10  7.5 
2s



 1
2
2
2 
exp  2 6  8  7  8  10  8 
 2s



where 8 is the maximum likelihood estimate of m.
Assuming a standard deviation of 1.5 yields a
likelihood ratio of 1.284.

Theorem 9.4.1. Let  be the likelihood ratio test
statistic. Then, -2  is asymptotically distributed as
chi-squared with the degrees of freedom equal to
the number of exact restrictions implied by H0.
Composite Against Composite

Definition 9.5.1. A test R is the uniformly most power
test of size (level) a if
sup qQP(R|q) = (≤) a and for any other test R1
such that sup qQP(R|q) = (≤) a we have P(R|q) ≥
P(R1|q) for any q  Q.

Definition 9.5.2. Let L(x|q) be the likelihood
function. Then the likelihood ratio test of H0 against
H1 is defined by the critical region
sup Lq x 

Q0
sup Lq x 
q 0  Q1
c
where c is chosen to satisfy supQP(<c|q)=a for a
certain specified value of a.

Example 9.5.2. Let the sample Xi~N(m,s2) with
unknown s2, i=1,2,…n. We want to test H0:m=m0
and 0<s2<∞ against H1:m>m0 and 0<s2<∞.
Lq   2 
n
2
s 
n
2  2
 1
exp 
2
 2s

i 1 xi  m  
n
2

Using the concentrated likelihood function at the null
hypothesis:
sup Lq   2 
n
Q0
2
s 
n
2  2
 n
exp  
 2
1 n
2
s  i 1 xi  m 0 
n
2

This likelihood value can be compared with the
maximum likelihood value of
sup Lq   2 
n
Q0
2
sˆ 
n
2  2
 n
exp  
 2
1 n
2
sˆ  i 1 xi  x 
n
1 n
x  i 1 xi
n
2

The critical region then becomes
s 
 2 
 sˆ 
2
n
2
c

Turning back to the Bayesian model, the Bayesian
would solve the problem testing H0:q≥q0 against
H1: q>q0. Let L2(q) the loss incurred by choosing H0
and L1(q) by choosing H1. The Bayesian rejects H0 if



L1 q  f q x dq   L2 q  f q x dq


where f(q|x) is the post distribution of q.