Statistics 522 Midterm 2
April 16, 2014
Name: _________________________________ (please print)

There are 3 questions.

Show your work for full credit.

You may assume well known theorems for exponential families that were covered in class.
The general form of exponential families is given below. If using such theorems, you need
to show that the conditions (if any) hold.

Confidence intervals should be in a form where the parameter is between two random
quantities.
For your information:
1. The Cramer-Rao inequality for the variance of an unbiased estimator of    is
VarW  X  
    
2
2
 
 
E 
log f  X    
 
 
.
2. The general form of an exponential family is
f  x    h  x  c   exp

k
i 1

wi   ti  x  ,
where h  x   0 and c    0 .
3. Definition: A family of pdfs or pmfs  f  x   with real-valued parameter  has monotone
likelihood ratio (MLR) in the statistic T  x  if for every 1   0 , whenever f  x 1  and f  x 0 
are distinct, the ratio
f  x 1 
f  x 0 
is a monotone function of T  x  for the set of values x for which at
least one of the pdfs or pmfs is positive.
1
, X n be a random sample from N   ,  2  .
1. (30 points total) Let X 1 ,



2


(a) (10 points) Show that X , S 2 is a complete sufficient statistic for  ,  2 . We have
f x ,

n
 1 


 exp  
2




n
i 1
 xi   
2

/ 2 2 

n
  n x 2  n x n 2  
 1 
 i1 i  i1 i   .

 exp    2 2   2
2 2  

 2 




1
1 
 , w1     2 , w2     2 2 ,
 2 

This is an exponential family, with h  x   1 , c    
n
T1  x   i 1 xi , and T2  x   i 1 xi2 . Then since  w1   , w2    contains an open set in
n
n
    ,  2  varies,
function of

n

n
2
as

x , i 1 xi2 is complete and sufficient. Since  X , S 2  is a one-to-one
n
i 1 i

x , i 1 xi2 , it is complete and sufficient as well.
i 1 i
n
2
2
We now show that S is UMVUE for  (give details). Since S 2
 2  2
2
2

  n1 , E  S    . Since
 n 1 
S 2 is unbiased for  2 and a function of the complete sufficient statistic, it is UMVUE.
(b) (10 points) Determine the UMVUE for  2 .
E  X 2   Var  X    E  X   
2
2
n
  2 . Then
2
n


Xi  X 
 2  n  1 2

 2 S2 

2
2
2
i

1

E X    E X 
S S  E X 
S
n 
n
n







  n X i2

 E  i 1
 S 2    2 . This statistic is unbiased and a function of the complete sufficient statistic,


n


and is thus the UMVUE for  2 .
2
(c) (10 points) The maximum likelihood estimator of  is W  X  
2
1 n
X

X
. Assume that



i
n i 1
D
n W  X    2  
 N  0, ???? holds. Determine
regularity conditions are satisfied so that the result
the asymptotic variance, i.e. replace the ????. Completely justify your answer.
2
When the regularity conditions hold, MLEs are asymptotically efficient. Thus ???? is 1/ I   , the Fisher
information from one observation.
2
 
 
I    E  2 log f  X       E
 
 
2


2
 
log f  X     .
   2 2



Now,
l   ,
2
  log f  x  , 
2

 2
l  ,
2
  2 
2
2
x   .
1
  log  2 2  
2
2 2
2
   2
1
l  ,
2
2
  2
1
4
x  

2  2 
2
2
x  

 2 
3
2
.
  2

1
2
1
.
 E  2 log f  X       4 

3
4
2


2

2






 
4
Thus ????  2 .
2. (50 points total) Let X1 ,
X n be a random sample from the probability density function
f x   
2x
2
, 0  x  .
(a) (10 points) Show that X  n  max  X1 , , X n  is a consistent estimator sufficient for  .
x
FX  x   
0
 
2t
x2

2

E X n  
0
dt 
2
2nx 2 n
 2n
, so that FX  n  x  
dx 
x2n
 2n
and
f X  n  x  
2nx 2 n 1
 2n
2n 2 n 1
2n

   as n   . Also
2n
 2n  1
 2n  1
3
, 0  x  .
Then

    2nx
E X
2
 n
2 n 1
2n
dx 
0
 
Var X
2
 n
2n 2 n  2
2n
n 2

2 

2n
n 1
 2n  2  
 2n  2 
so
that



n 2  2n
n

 
 2 
 0 as n   . Thus the bias and
2
n 1
  n  1 2n  1 
  2n  1 
2
variance go to zero as n   , a sufficient condition for consistency.
(b) (5 points) Show that
Let Y 
X

X  n

is a pivotal quantity.
y
. Then X   Y , J   , and fY  y   f X  y   2 y , 0  y  1 . FY  y    2tdt  y 2 and
thus FY   y   y . Differentiating we get fY   y   2ny
2n
n
Y n 
X  n

n
0
2 n 1
for 0  y  1 . Thus
is then a pivotal quantity.
(c) (10 points) Using the pivotal quantity
for  with lower limit L  X   X  n .

X  n

above, derive a 1   100% confidence interval

 b   1   for some 0  a  b  1 . The confidence interval is then of the form




X  n 
 X  n
C x n     :
 
 . Then b  1 since the lower bound of the interval has been specified

b
a 

1
X  n


 1  1   . Thus  2ny 2 n 1dx  1  a 2 n  1   . Then a   1/2n and
as X  n  . Therefore Pr  a 


a


X  n 

the interval is C X  n     : X  n     1/ 2 n  .
 

We set Pr  a 
X  n
 
 
(d) (10 points) Derive a 1   100% confidence interval by pivoting using the cumulative
distribution function of X  n  .


n
x2n
FX  n  x   P X  n  x   P  X 1  x   2 n . The interval is then

2n
2n
X  n
X n
X 2nn  
X n
X  n  

 
2n
 
 
. Here 1   2   .
 : 1  2 n  1   2    :
   :
1/ 2 n
1/ 2 n 

1   1   2 
1  

  1   2
4
(e) (10 points) Obtain a likelihood ratio test of H 0 :   0 versus H1 :    0 with level of
significance  .
The likelihood function is not differentiable, but graphically we see that X  n  is the MLE of  .
The likelihood ratio statistic is
2n x1 x2

  x 
xn 
I  X  n    0 
2n
0
n
2 x1 x2 xn
x2nn
we always reject if X  n 


0

  x n 

  0
if x n    0



 X  n
  0 or if 
 
 0

2n
if x n    0
. We reject for small values of   x  , i.e.
2n

2n
2n
1/2 n
  c  X  n    0 c , i.e. X  n  0 c .




 x
2n
Then PH X  n  0c1/2n  PH X  n  0   . The cdf of X  n  is P X  n   x   FX  x      .
 
0

Then PH X  n  0c
0
1/2 n
0
  P  X    
H0
0
n
n
2n
  c1/2 n 
 0
  0  c   , and thus we reject if X  n    0 or if
 0 
X  n  0 1/2n .
(f) (5 points) Invert your test from part (e) to obtain a 1   100% confidence interval for  .


The non-rejection region is of the form x : 0 1/2n  X  n  0 . The confidence interval is then of



the form  :  1/ 2 n  X  n      : X  n   
3. (20 points total) Let X1 ,

X  n  
.
 1/ 2 n 
X n be a random sample from a distribution with pdf
f  x     x 1 , 0  x  1 ,   0 .
(a) (10 points) Use the Neyman-Pearson Lemma to derive the form of the most powerful test for
testing H 0 :    0 vs. H1 :   1 ,  0  1 . (Your rejection region should depend on the sufficient
statistic.) Is your test most powerful for testing H 0 :   0 vs. H1 :    0 ? Why or why not?
5
We calculate the ratio of the pdfs:
function of T  x   x1 x2
0  x1 x2
xn  0
1  x1 x2
xn 
 1
1 1
 
  0   x1 x2
 1 
xn  0
 1
. This is an increasing
xn since  0  1 . We then reject when T  x   k . Since the form of the
test does not change, it is UMP.
(b) (5 points) Find explicitly the critical value k for testing H 0 :    0 vs. H1 :   1 when n  1 ,
 0  0.5 , 1  0.25 , and   0.05 .
 x 1
k
We reject if 0  1  2 x 0.50.25  k , i.e. x    
1 x
2
4
0
1
rejection region is then reject if x 
 k / 2 4

0
1 1/ 2
k2
x dx  0.05 
 0.05  k  0.20 . The
2
4
4
k
0.04

 0.0025 .
16 16
(c) (5 points) Under the conditions of the test from part (b), determine the power function the test
that you derived. Is your test unbiased? Why or why not?
We reject if x  0.0025 . The power function is P  x  0.0025  
0.0025

 x 1dx   0.0025  . Since

0
this is a decreasing function of  , it is larger than 0.05 (the power under the null hypothesis) when
  0.5  0 . Thus the test is unbiased.
6