Statistics 512 Notes 24: Uniformly Most
Powerful Tests
Definition 8.2.1: The critical region C is a uniformly most
powerful (UMP) critical region of size
for testing the
simple hypothesis
0
H
against an alternative composite
hypothesis
1
H
if the set C is a best critical region of size
for testing
0
H
against each simple hypothesis in
1
H
. A test
defined by this critical region C is called a uniformly most
powerful (UMP) test , with significance level
for testing
the simple hypothesis
0
H
against the alternative composite
hypothesis
1
H
.
Example: For
1
, ,
n
X X K
iid
( ,1) N
, suppose we want to
test
0 0
: H
versus the compositive alternative
1 0
: H >
. Consider first testing
0 0
: H
versus the
simple alternative
1 1
: H
for
1 0
>
. By the Neyman-
Pearson Lemma, the most powerful level
test rejects for
small values of
( )
2
0
1
1 0
2
1 1
1
1
2 2 2 2
0 0 1 1
1 1
( )
1
exp
2 2
( , , ; )
( , , ; )
( )
1
exp
2 2
1 1
exp ( 2 ) 2
2 2
n
n
i
i
n
n
n
n
i
i
n n
i i i i
i i
X
f X X
f X X
X
X X X X
,
,
,
,
_
+ + +
,
K
K
2 2
0 1 0 1
1
1
exp ( ) ( )
2
n
i
i
X n
_
,
Because
1 0
>
, this is equivalent to rejecting for large
values of
1
n
i
i
X
. Under
0 0
: H
, we have
0
1
~ ( , )
n
i
i
X N n n
. Thus, the most powerful level
test
Of
0 0
: H
versus the simple alternative
1 1
: H
for
1 0
>
has critical region
1
0
1
(1 )
n
i
i
X n n
> +
where is the standard normal CDF. Because the critical
region of the most powerful test is the same for all simple
alternatives
1 1
: H
for
1 0
>
, the test with this
critical region (
1
0
1
(1 )
n
i
i
X n n
> +
) is the
uniformly most powerful test for testing
0 0
: H
versus
the composite alternative
1 0
: H >
.
Comment: The test with critical region
1
0
1
(1 )
n
i
i
X n n
> +
is also the uniformly most
powerful level
test for testing
0 0
: H
versus the
alternative
1 0
: H >
. Recall from Section 5.5 that the
size of a test with a composite null hypothesis is
( ) ( )
0
1
max , , ;
H n
P X X C
K
. We have
( )
0
1
0
1
max (1 );
n
i
i
P X n n
> +
so that
the test with critical region
1
0
1
(1 )
n
i
i
X n n
> +
has size
and is consequently the uniformly most
powerful level
test for testing
0 0
: H
versus the
alternative
1 0
: H >
Monotone Likelihood Ratios:
Suppose
1
, ,
n
X X K
iid
( ; ) f x
and we want to test
0 0
: H
versus the alternative
1 0
: H >
. When can we
find a UMP test?
Definition: We say that the likelihood
1
( ; ( , , ))
n
L X X K
has monotone likelihood ratio in the statistic
1
( , , )
n
Y u X X K
if for
1 2
<
, the ratio
1 1
2 1
( ; ( , , ))
( ; ( , , ))
n
n
L X X
L X X
K
K
(*)
is a monotone function of
1
( , , )
n
Y u X X K
.
Assume then that our likelihood function
1
( ; ( , , ))
n
L X X K
has monotone likelihood ratio in the statistic
1
( , , )
n
Y u X X K
. Then the ratio in (*) is equal to
( ) g y
where
g
is a decreasing function (the case where the
likelihood function has a monotone increasing likelihood
ratio follows similarly by changing the sense of the
inequalities below). We claim that the following test is
UMP level
for testing
0 0
: H
versus the alternative
1 0
: H >
:
Reject
0
H
if
Y
Y c
, (**)
where
Y
c
is determined by
0
( ; )
Y
P Y c
.
Proof: First consider the simple null hypothesis
'
0 0
: H
.
Let
0
'' >
be arbitrary but fixed. Let C denote the most
powerful critical region for testing
'
0 0
: H
versus
1
: '' H
. By the Neyman-Pearson Theorem, C is
defined by
0 1
1
1
( ; ( , , ))
if and only if ( , , )
( ''; ( , , ))
n
n
n
L X X
k X X C
L X X
K
K
K
,
where k is determined by
1 0
[( , , ) ; ]
n
P X X C K
. But
by the definition of monotone likelihood ratio, because
0
'' >
,
-1
0 1
1
( ; ( , , ))
( ) Y g ( )
( ''; ( , , ))
n
n
L X X
g Y k k
L X X
K
K
,
where
1
( ) g k
satisfies
1
[ ( ); ] P Y g k
; i.e.,
1
( )
Y
c g k
. Hence, the Neyman-Pearson test is equivalent
to (**). Furthermore, the test is UMP for
'
0 0
: H
versus
1 0
: H >
because we assumed only that
0
'' >
and the
test is uniquely determined by
0
.
Finally, to show that the test is UMP for
0 0
: H
versus
1 0
: H >
, we need to show that the test has size
for testing
0 0
: H
, i.e.,
0
1
max ( ( ); ) P Y g k
.
Consider testing
0
: ''' H
versus
1 0
: H
for
0
''' <
using the critical region Reject
0
H
if
1
( ) Y g k
. By the
above argument, this test is the most powerful test of level
1
( ( ); ''') P Y g k
. By Corollary 8.1.1, the level of this
most powerful test must be less than or equal to its power
versus the alternative, i.e,
1 1
( ( ); ''') ( ( ); )) P Y g k P Y g k
for
0
''' <
. This proves that
0
1
max ( ( ); ) P Y g k
Example of monotone likelihood ratio: Let
1
, ,
n
X X K
be iid
with probability of success . Let ' '' < . Consider the
ratio of likelihoods
1
1
( '; , , ) ( ') (1 ') '(1 '') 1 '
( ''; , , ) ''(1 ') 1 ''
( '') (1 '')
i
i i
i i
x
n x n x
n
x n x
n
L x x
L x x
1
_
1
,
]
K
K
Since '/ '' 1 < and
(1 '') /(1 ') 1 <
so that
[ ]
'(1 '') / ''(1 ') 1 <
, the ratio is a decreasing function
of
i
y x
. Thus we have a monotone likelihood ratio in
the statistic
i
Y X
.
Consider the hypotheses
0 1
: ' versus : ' H H >
.
The UMP level
decision rule for testing
0
H
versus
1
H
is
given by
0
1
Reject if
n
i
i
H Y X c
,
where
c
is such that
[ ; '] P Y c
assuming that such a c
exists for the given
level.
Other examples of families with monotone likelihood ratio:
Family Y
Normal
( ,1)
1
n
i
i
X
Poisson( )
1
n
i
i
X
Example of family that does not have monotone likelihood
ratio:
Cauchy ( ) family:
2
1 1
( )
1 ( )
f x
x
+
.
Two more examples where there is no UMP test:
(1) Let
1
, ,
n
X X K
be iid
( ,1) N
. Suppose we want to test
0 0
: H
versus the alternative
1 0
: H
. The most
powerful test for alternatives
0
' <
is to reject if
1
0
1
(1 )
n
i
i
X n n
<
. The most powerful test for
alternatives
0
' >
is to reject if
1
0
1
(1 )
n
i
i
X n n
> +
. For
0
' <
, the power
of the test that rejects if
1
0
1
(1 )
n
i
i
X n n
<
is
greater than the power of the test that rejects if
1
0
1
(1 )
n
i
i
X n n
> +
; thus, there is no UMP test.
(2) Let
1
, ,
n
X X K
be iid
2
( , ) N . Suppose we want to test
0 0
: H
versus the alternative
1 0
: H >
. Here both
the null hypothesis and the alternative hypothesis is
composite (the null hypothesis is composite because
can
take on any positive value). There is no UMP test because
the z-test for known
is more powerful than the t-test for
unknown
.
What to do when there is no UMP test?
(1) Look for UMP unbiased test. A test is said to be
unbiased if its power never falls below its significance
level. By Corollary 8.1.1, the most powerful test of a
simple null versus simple alternative is unbiased. For
testing
1
, ,
n
X X K
iid
( ,1) N
,
0 0
: H
versus the
alternative
1 0
: H
, the test that rejects if
1
0
1
(1 )
n
i
i
X n n
> +
is not unbiased because it has
power
<
for
0
<
. The UMP unbiased test rejects if
1
0
1
| | (1 )
2
n
i
i
X n n
>
.
The t-test is UMP unbiased for testing
1
, ,
n
X X K
iid
( ,1) N
,
0 0
: H
versus the alternative
1 0
: H
,
2
unknown.
(2) Use Generalized Likelihood Ratio test. The generalized
likelihood ratio test that we studied in Chapter 6 for testing
0
: H
versus
1
:
C
H I
generally has good power properties. The test rejects for
small values of
Generalized Likelihood ratio:
max ( )
max ( )
L
L
(3) Use Decision Theory or Bayesian methods (rest of
course).
