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 H 0 against an alternative composite
hypothesis H1 if the set C is a best critical region of size
 for testing H 0 against each simple hypothesis in H1 . A
test defined by this critical region C is called a uniformly
most powerful (UMP) test , with significance level  for
testing the simple hypothesis H 0 against the alternative
composite hypothesis H1 .
Example: For X 1 , , X n iid N (  ,1) , suppose we want to
test H 0 :   0 versus the compositive alternative
H1 :   0 . Consider first testing H 0 :   0 versus the
simple alternative H1 :   1 for 1  0 . By the NeymanPearson Lemma, the most powerful level  test rejects for
small values of
  (X   ) 
 1 

exp  
n
n
i 1
f ( X1,
f ( X1,
2
i
0




2
, X n ; 0 )  2 



, X n ; 1 )  1 n
  n ( X i  1 ) 2 
i 1


 exp  

2
 2 


1 n
 1 n

 exp    i 1 ( X i2  2 X i 0  02 )   i 1  X i2  2 X i 1  12  
2
 2

1
n


 exp  ( 0  1 ) i 1 X i  n( 02  12 ) 
2


Because 1  0 , this is equivalent to rejecting for large
values of


n
i 1
X i . Under H 0 :   0 , we have
X i ~ N (n0 , n) . Thus, the most powerful level  test
Of H 0 :   0 versus the simple alternative H1 :   1 for
n
i 1
1  0 has critical region

1
X

n


n

(1   )
i
0
i 1
n
where  is the standard normal CDF. Because the critical
region of the most powerful test is the same for all simple
alternatives H1 :   1 for 1  0 , the test with this
1
critical region ( i 1 X i  n0  n (1   ) ) is the
uniformly most powerful test for testing H 0 :   0 versus
n
the composite alternative H1 :   0 .
Comment: The test with critical region

1
X

n


n

(1   ) is also the uniformly most
i
0
i 1
n
powerful level  test for testing H 0 :   0 versus the
alternative H1 :   0 . Recall from Section 5.5 that the
size of a test with a composite null hypothesis is
max H 0 P   X 1 , , X n   C;  . We have
max  0 P


1
X

n


n

(1   );    so that
i
0
i 1
n
the test with critical region

X i  n0  n1 (1   ) has size  and is
consequently the uniformly most powerful level  test for
testing H 0 :   0 versus the alternative H1 :   0
n
i 1
Monotone Likelihood Ratios:
Suppose X 1 , , X n iid f ( x; ) and we want to test
H 0 :   0 versus the alternative H1 :   0 . When can we
find a UMP test?
Definition: We say that the likelihood
L( ;( X1 , , X n )) has monotone likelihood ratio in the
statistic Y  u ( X1 , , X n ) if for 1  2 , the ratio
L(1 ;( X 1 , , X n ))
L( 2 ;( X 1 , , X n )) (*)
is a monotone function of Y  u ( X1 , , X n ) .
Assume then that our likelihood function
L( ;( X1 , , X n )) has monotone likelihood ratio in the
statistic Y  u ( X1 , , X n ) . 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 H 0 :   0 versus the alternative
H1 :   0 :
Reject H 0 if Y  cY , (**)
where cY is determined by   P(Y  cY ;0 ) .
'
Proof: First consider the simple null hypothesis H 0 :    0 .
Let  ''  0 be arbitrary but fixed. Let C denote the most
'
powerful critical region for testing H 0 :    0 versus
H1 :    '' . By the Neyman-Pearson Theorem, C is
defined by
L(0 ;( X 1 , , X n ))
 k if and only if ( X 1 , , X n )  C ,
L( '';( X 1 , , X n ))
where k is determined by   P[( X1 , , X n )  C;0 ] . But
by the definition of monotone likelihood ratio, because
 ''  0 ,
L( 0 ;( X 1 , , X n ))
 g (Y )  k  Y  g -1 (k ) ,
L( '';( X 1 , , X n ))
1
1
where g (k ) satisfies   P[Y  g (k ); ] ; i.e.,
cY  g 1 (k ) . Hence, the Neyman-Pearson test is equivalent
'
to (**). Furthermore, the test is UMP for H 0 :    0 versus
H1 :   0 because we assumed only that  ''  0 and the
test is uniquely determined by 0 .
Finally, to show that the test is UMP for H 0 :   0
versus H1 :   0 , we need to show that the test has size
 for testing H 0 :   0 , i.e.,
max  P(Y  g 1 (k ); )   .
Consider testing H 0 :    ''' versus H1 :   0 for
 '''  0 using the critical region Reject H 0 if Y  g 1 (k ) .
0
By the above argument, this test is the most powerful test
1
of level 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,
P(Y  g 1 (k ); ''')  P(Y  g 1 (k ); ))  
for  '''   0 . This proves that
max 0 P(Y  g 1 (k ); )  

Example of monotone likelihood ratio: Let X 1 , , X n be iid
with probability of success  . Let  '   '' . Consider the
ratio of likelihoods
n
x
n x
 xi
L( '; x1 , , xn ) ( ') i (1   ')  i  '(1   '')   1   ' 




L( ''; x1 , , xn ) ( '') xi (1   '')n  xi  ''(1   ')   1   '' 
Since  '/  ''  1 and (1   '') /(1   ')  1 so that
 '(1   '') /  ''(1   ')  1 , the ratio is a decreasing function
of y   xi . Thus we have a monotone likelihood ratio in
the statistic Y   X i .
Consider the hypotheses
H 0 :    ' versus H1 :    ' .
The UMP level  decision rule for testing H 0 versus H1 is
given by
Reject H 0 if Y  i 1 X i  c ,
where c is such that   P[Y  c; '] assuming that such a c
exists for the given  level.
n
Other examples of families with monotone likelihood ratio:
Family
Y
 X
Normal (  ,1)
n
i 1
i
Poisson(  )

n
i 1
Xi
Example of family that does not have monotone likelihood
ratio:
1
1
f
(
x
)

Cauchy (  ) family:
 1  ( x   )2 .
Two more examples where there is no UMP test:
(1) Let X 1 , , X n be iid N (  ,1) . Suppose we want to test
H 0 :   0 versus the alternative H1 :   0 . The most
powerful test for alternatives    '  0 is to reject if
1
X

n


n

(1   ) . The most powerful test for
i1 i 0
alternatives    '  0 is to reject if
n

1
X

n


n

(1   ) . For    '  0 , the power
i
0
i 1
n
1
of the test that rejects if i 1 X i  n0  n (1   ) is
greater than the power of the test that rejects if
n

1
X

n


n

(1   ) ; thus, there is no UMP test.
i
0
i 1
n
2
(2) Let X 1 , , X n be iid N (  ,  ) . Suppose we want to test
H 0 :   0 versus the alternative H1 :   0 . 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 X 1 , , X n iid N (  ,1) , H 0 :   0 versus the
alternative H1 :   0 , the test that rejects if

1
X

n


n

(1   ) is not unbiased because it has
i
0
i 1
n
power   for   0 . The UMP unbiased test rejects if

n
|  i 1 X i  n0 | n 1 (1  ) .
2
The t-test is UMP unbiased for testing X 1 , , X n iid
N (  ,1) , H 0 :   0 versus the alternative H1 :   0 ,
 2 unknown.
(2) Use Generalized Likelihood Ratio test. The generalized
likelihood ratio test that we studied in Chapter 6 for testing
H 0 :    versus H1 :     C
generally has good power properties. The test rejects for
small values of
max  L( )


Generalized Likelihood ratio:
max  L( )
(3) Use Decision Theory or Bayesian methods (rest of
course).