1. Let X 1 , X 2 ,, X n be a random sample from the Gamma( ,  ) distribution
f (x |  ,  ) 
(a)
(b)
(c)
(d)

1
( ) 

x
x 1e  , x  0 ,   0 ,   0 .
( 8 %) Find the method of moment estimates of  and  .
( 7 %) Find the MLE of  , assuming  is known.
( 7 %) Giving   0 , find the Cramer-Rao lower bound of estimates of  .
( 8 %) Giving   0 , find the UMVUE of  .
2. Suppose that X 1 , X 2 ,, X n are iid ~ B ( 2, p ) , p  (0,1) .
Let  ( p)  2 p(1  p) .
n
(a) ( 5 %) Show that T   X i is a sufficient statistic for p .
i 1
1, if X 1  1
(b) ( 5 %) Let Y  
. Show that Y is an unbiased estimate of  ( p ) .
0
,
if
X

1
1

(c) (10%) Find the UMVUE W of  ( p ) .
3. Let X 1 , X 2 , , X n be a random sample from a Poisson ( ) ,   0 , distribution.
Consider testing H 0 :   1 vs H1 :   3 .
(a) (10%) Find a UMP level  test, 0    1.
(b) ( 7 %) For n  3 , the test rejects H 0 , if X 1  X 2  X 3  5 .
Find the power function  ( ) of the test.
(c) ( 8 %) For n  3 , the test rejects H 0 , if X 1  X 2  X 3  5 .
Evaluate the size and the power of the test.
4. (10%) Let X 1 , X 2 , , X n be iid Poisson ( ) distribution, and let the prior
distribution of  be a Gamma( ,  ) distribution,   0 ,   0 .
posterior distribution of  .
Find the
5. Let X 1 , X 2 , , X n be a random sample from an exponential distribution with mean
 ,  0 .
n
(a) ( 5 %) Show that T   X i is a sufficient statistic n for  .
i 1
(b) ( 5 %) Show that the Poisson family has a monotone likelihood ratio, MLR.
(c) ( 5 %) Find a UMP level  test of H 0 : 0    1 vs H1 :   1 by the
Karlin-Rubin Theorem shown below.
[Definition] A family of pdfs or pmfs {g (t |  ) |   } has a monotone likelihood ratio,
g (t |  2 )
MLR, if for every  2  1 ,
is a monotone function of t .
g (t | 1 )
[Karlin-Rubin Theorem] Suppose that T is a sufficient statistic for  and the pdfs or
pmfs {g (t |  ) |   } has a non-decreasing monotone likelihood ratio.
Consider testing H 0 :    0 vs H1 :    0 . A UMP level  test rejects
H 0 if and only if T  t 0 , where   P 0 (T  t0 ) .
1
數理統計期末考試試題答案

x
(  1)   1
 
x
e
dx

  and
 ( ) 


(

)

0
1. (a) Since E ( X ) 


1
x
(  2)    2
 1 
x
e
dx

  (  1)  2 ,
 ( ) 

( ) 
0
E( X 2 ) 

1
Let m1   and m2   (  1)  2 
 ~ 
m2
m12

 1

m1 m2  m12
,   ~ 
.

m1
m2  m12
m12
~
Furthermore, m1  X , m2  m12 
The MME of  .and  are ~ 
1 n 2
1 n
(n  1) 2
X i  X 2   ( X i2  X ) 2 
S ,

n i 1
n i 1
n
~ (n  1) S 2
,  
nX
(n  1) S 2
nX 2
n
n
x)  [
(b) L(  |  , ~
1
i 1 ( )  
 1
xi
 xi
x
 i
e  ]
n
1
 1
(  xi
[( )   ]n i 1
 i 1
)e

n
n
 x1
i 1

 ln L(  |  , ~
x )  n ln ( )  n ln   (  1)  ln xi  i 1
Let
n

n
1 n
x
ˆ 1
ln L(  |  , ~
x)  

x

0

xi  .



i

  2 i 1
n i 1

Furthermore,
n
2 n
n 2nx
ln L(  |  , ~
x) 

ln xi 


 2
 2  3 i 1
2 3
2
n 2nx nx  2nx  n
ln L( ˆ |  , ~
x) 



 0,
 2
ˆ 2 ˆ 3
x3
x2
X
So, ˆ 
is the MLE of  .

2

(c)  E  [
n
2 n
n 2n n
ln L(  |  , ~
x )]  E  (

Xi )  



 2
 2  3 i 1
2
3
2

2
1
 CRLB =
 E [
2

2

ln L(  |  , ~
x )]
2
n

X
is an unbiased estimate of  , and
  , ˆ 



X
1  2  2
X
is the UMVUE of  .
Var ( ) 

 CRLB, ˆ 
2

n
n


(d) Since E (
X
)
2
[Or] f ( x |  ,  ) 
1

I (0, ) ( x) x
 1

x
e  
1

1
I (0, ) ( x) x 1 exp[ x( )]

( ) 
( ) 
 Given  , { f ( x |  )} is an exponential family in  .
n
 T   X i is a sufficient statistic for  .
i 1
X
T
Since ˆ 
is an unbiased estimate of  and a function of sufficient

 n
X
statistics T , by Rao-Blackwell Theorem, ˆ 
is the UMVUE of  .

n
n 2
2. (a) f ( x1 , x 2 , , x n | p)   f ( xi | p )   [  I{0,1,2} ( xi ) p xi (1  p) 2  xi ]
x
i 1
i 1  i 
n
n 2
xi
2
p xi
p i
 [  I{0,1,2} ( xi )(
) (1  p) 2 ]  [   I{0,1,2} ( xi )](
) 1 (1  p) 2n
x
x
1 p
1 p
i 1  i 
i 1  i 
n
n 2
p T ( ~x )
Let g (T ( ~
x ), p)  (
)
(1  p) 2n and h( x)    I{0,1,2} ( xi ) . By
x
1 p
i 1  i 
n
factorization theorem, T   X i is a sufficient statistic for p .
i 1
 2
 2
p
[Or] f ( x | p)    I{0,1,2} ( x) p x (1  p) 2  x    I{0,1,2} ( x)(1  p) 2 exp[ x ln(
)]
1 p
 x
 x
n
 { f ( x | p)} is an exponential family  T   X i is a sufficient statistic.
i 1
 2
(b) E (Y )  1  P( X 1  1)  0  P( X 1  1)    p1 (1  p) 2 1  2 p(1  p) , so Y is an
1
unbiased estimate of  ( p ) .
n
(c) If X 1 , X 2 ,, X n , n  N , are iid ~ B ( 2, p ) , then T   X i ~ B (2n, p ) .
i 1
n
 E (Y | T  t ) 
P(Y  1 & T  t ) P( X 1  1 & T  t )


P(T  t )
P(T  t )
P( X 1  1 &  X i  t  1)
i2
P(T  t )
n
 2n  2  t 1
 p (1  p ) 2n  t 1
2 p (1  p )
t

1


i2


P(T  t )
 2n  t
  p (1  p ) 2n  t
 t 
2(2n  2)!
t!(2n  t )! t (2n  t )
, t  0,1,2,,2n .


(t  1)!(2n  t  1)! (2n)!
n(2n  1)
T (n  T )
By Rao-Blackwell Theorem, W  E (Y | T ) 
is the UMVUE of  ( )  e  .
2(2n  1)
P( X 1  1) P(  X i  t  1)
3. (a) By Neyman-Pearson Lemma, a UMP level  test rejects H 0 if and only if
3
f ( x1 , x2 ,, xn |   3)  kf ( x1 , x2 ,, xn |   1) .
n
n 1x i
3 xi  3
e ]  k [
e 1 ]  3 xi  ke2n  (  xi ) ln 3  2n  ln k
i 1 ( xi )!
i 1 ( xi )!
i 1
n
2  ln k
c
  xi 
ln
3
i 1

n
[
n
Since
 X i ~ Poisson (n ) , a UMP level
 test rejects H 0 if and only if
i 1
n
 X i  c , where c is the smallest integer satisfying
i 1

ni  n
 i! e   .
i  c 1
n
[Or] T   X i is sufficient for  and T ~ Poisson (n ) .
i 1
By the corollary of Neyman-Pearson Lemma, a UMP level  test rejects H 0
if and only if g (t |   3)  kg(t |   1) .

n
3t  3
1t
e  k e 1  3t  ke2  (  x1 ) ln 3  2  ln k
t!
t!
i 1
(b)  ( )  P ( X 1  X 2  X 3  5)  1  P ( X 1  X 2  X 3  4)
(3 ) 0 (3 )1 (3 ) 2 (3 ) 3 (3 ) 4  3
,  0




]e
0!
1!
2!
3!
4!
30 31 32 33 34  3
(c) The size of this test is  (1)  1  [




]e
 0.1847
0!
1!
2!
3!
4!
9 0 91 9 2 93 9 4  9
The power of this test is  (3)  1  [  

 ]e  0.9450
0! 1! 2! 3! 4!
 1[
n
4. Since T   X i is sufficient for  and T ~ Poisson (n ) .
i 1


1
(n ) t  n
 1 
 e
; and f  ( ) 
fT |  (t |  ) 
e
t!
( )  

1

 ( n  )
(n ) t  n
1
nt
 ,
e
 1e  
t   1e
 f (t ,  ) 
 0


t!
( ) 
t!( ) 

 f T (t ) 
nt
t   1
 t!( )  
1
 ( n  )

e
d 
0
nt
f (t ,  )
t!( ) 

 f  | t ( | t ) 
f T (t )
nt

t   1

e
nt
t!( ) 
 (n 
1


(t   )(
 t 
)
n  1
)

t   1

e
 (n 
1

)
 t   (t   )(  ) t  
(t   )(
)
n  1
n  1
t!( )  

The posterior distribution of  is Gamma(t   ,
).
n  1
4
,  0
 1
i
1 
1   n
5. (a) f ( x1 , x2 , , xn |  )   f ( xi |  )   ( e
I (0, ) ( xi )) 
e
 I (0, ) ( xi )
n
x
n
i 1
T (~
x)

1
x ),  ) 
e  and
Let g (T ( ~
n

x

i 1
n
i 1
n
h( ~
x )   I (0, ) ( xi ) . By factorization theorem,
i 1
n
T   X i is a sufficient statistic for  .
i 1
x
1 
1
1
[Or] f ( x |  )  e  I (0, ) ( x)  I (0, ) ( x) exp[ x( )]



n
 { f ( x |  )} is an exponential family.  T   X i is a sufficient statistic.
i 1
X
T
Since ˆ 
is an unbiased estimate of  and a function of sufficient

 n
X
statistics T , by Rao-Blackwell Theorem, ˆ 
is the UMVUE of  .

n
(b) T   X i ~ Poisson ( n )  g (t |  ) 
i 1
(n ) t  n
, t  0,1,2,
e
t!
(n2 ) t  n 2
t
e
 2   n( 2  1 )
g (t | 2 )
t
!


   e
g (t | 1 )
(n1 ) t  n1  1 
e
t!

g ( t | 2 )
If 2  1  2  1 
is an increasing function of t ,
1
g (t | 1 )
Hence {g (t |  ) |   0} of T has MLR.
n
1
i 1
(n) n
(c) T   X i ~ Gamma(n,  )  g (t |  ) 

1
g (t |  2 )


g (t | 1 )
(n) 2n
1
(n)1n
If  2  1   (
1
2

t
n 1

t
e  , t 0
t
t n 1e  2
1
1
    (  )t
  1  e  2 1 , t  0
t
 2 

n
t n 1e 1
1
1
 
) 2 1 0
1 2

Hence {g (t |  ) |   0} of T has an MLR.
g (t |  2 )
is increasing in t .
g (t | 1 )
n
By Karlin-Rubin Theorem, the UMP size  test rejecting H 0 if T   X i  c , where c
i 1
n
satisfies that P{ X i  c |   1}   ; i.e.,
i 1

c
5
1
  ( n) x
n 1  x
e
dx   .