# MATH 448 confidence intervals discussion

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1. Let X1; X2; :::; Xn be iid with density

f (x; ) =  3x 23 0  x  0 otherwise where  > 0 is unknown.

a. Show that X(n) is a pivotal quantity.

b. Using (a), Önd the shortest and longest (1   ) 100% conÖdence intervals

for .

c. Using (a), Önd a (1   ) 100% lower conÖdence interval and a (1   ) 100%

upper conÖdence interval for . Compare with (b).

d. Refer to (b). For each of the Öve samples given below, Önd the shortest

and longest 40% conÖdence interv

Vishalc0
School: Boston College

Attached.

Question 1:

 3x 2
, 0  x 

a) Since X1 , X 2 ,..., X n are i.i.d. with the density f ( x; ) =   3
. It follows that
0, otherwise

X
X1 X 2
,
,..., n are i.i.d. with the following density:
 

  x 2
x
x 
3   , 0   1 d

f   ;1 =  f ( x; ) =    
~ U ( 0,1)

 
 0, otherwise

X 
X X
=  1 , 2 ,..., n  has the same distribution as the maximum of a random

 
 
X (n)
sample from the U ( 0,1) distribution. Therefore,
is a pivotal quantity.

Hence,

X ( n)

b) We select two numbers a and b satisfying the following conditions:

Pa 

Pa 

And since we have P  a 

 b  = 1−

X (n)
 b n −1
 b  =  nt dt = b n − a n

 a
X (n)

X (n) 

 X (n)
 b = P
 
 , it immediately implies that

 b

a 

X ( n)

 X (n) X (n) 
,

 is a (1 −  )100% confidence intervals for  , provided that 0  a, b  1 and
a 
 b
b n − a n = 1 −  . And it would be desirable to find an interval of the above form which has the

1

1

shortest possible length which is X ( n )  −  .
a b

1

1

We cannot control over X ( n) but we can pick a and b to minimize − subject to the
a b
constraint b n − a n = 1 −  . And it is straight forward that the solution is a =  1/ n and b = 1 .
Hence, the shortest (1 −  )100% confidence intervals for  is calculated to be

 X ( n) X ( n) 
shortest (1 −  )100% CI = 
,
=
a 
 b
Similarly,

X (n) 

 X ( n ) , 1/ n 
 


1 1
− is maximized when a = 0 and when b is equal to
a b

bn − a n = 1 − 
 bn − 0 = 1 − 
 bn = 1 − 
 b = (1 −  )

1/ n

Since a = 0 , we have

X (n)
a

=  . Hence, the longest (1 −  )100% confidence intervals for  is

calculated by

 X ( n) X ( n)   X ( n)
longest (1 −  )100% CI = 
,
,
=
1/ n

a   (1 −  )
 b

c) Let 1   2 be 2 distinct levels of significance where 0  1 ,  2  1 . We will use  1 for lower
confidence interval and  2 for upper confidence interval.
Hence, a (1 −  )100% lower confidence intervals for  can be found as follow

 X ( n) X ( n) 
lower (1 −  )100% CI = 
,

a 
 b
X ( n) 

 lower (1 −  )100% CI =  X ( n ) , 1/ n 
1 

where 2 numbers a and b satisfy 0  a, b  1 and b n − a n = 1 − 1 . In fact, we choose a = 11/ n
and b = 1 .
Similarly, a (1 −  )100% upper confidence intervals for  can be found as follow

 X (n) X ( n) 
upper (1 −  )100% CI = 
,

a 
 b
X (n) 

 upper (1 −  )100% CI =  X ( n ) , 1/ n 
 2 


where 2 numbers a and b satisfy 0  a, b  1 and b n − a n = 1 −  2 . In fact, we choose a =  21/ n
and b = 1 .
Hence, compared to the answer in part (b), both of the lower and upper (1 −  )100%
confidence intervals are wider than the shortest (1 −  )100% CI because of the 2 inequalities

X ( n)

 1/ n

− X ( n) 

X ( n)

11/ n

− X ( n ) and

X ( n)

 1/ n

− X ( n) 

X ( n)

 21/ n

− X ( n ) , and are less wider than the longest

(1 −  )100% CI whose right limit is  .
d) For 40% confidence interval, we can solve for  by

(1 −  )100% = 40%
40%
= 0.4
100%
  = 1 − 0.4
  = 0.6
 1− =

For sample 1, the shortest 40% confidence interval is calculated by

X ( n) 

shortest 40% CI =  X ( n ) , 1/ n 
 

0.978844113 

 shortest 40% CI = 0.978844113,

0.61/3

 shortest 40% CI =  0.979,1.161
For sample 1, the longest 40% confidence interval is calculated by

 X ( n)

longest 40% CI = 
,

1/ n

 (1 −  )

 0.978844113 
 longest 40% CI = 
,
1/3

 (1 − 0.6 )

 longest 40% CI = 1.3285,  )
For sample 2, the shortest 40% confidence interval is calculated by

X ( n) 

shortest 40% CI =  X ( n ) , 1/ n 
 

0.849384877 

 shortest 40% CI = 0.849384877,

0.61/3

 shortest 40% CI =  0.849385,1.007057 
For sample 2, the longest 40% confidence interval is calculated by

 X (n)

longest 40% CI = 
,

1/ n

 (1 −  )

 0.84938487

 longest 40% CI = 
,

1/3

 (1 − 0.6 )

 longest 40% CI = 1.152793,  )
For sample 3, the shortest 40% confidence interval is calculated by

X ( n) 

shortest 40% CI =  X ( n ) , 1/ n 
 

0.940479844 

 shortest 40% CI = 0.940479844,

0.61/3

 shortest 40% CI =  0.940479844, 1.115062
For sample 3, the longest 40% confidence interval is calculated by

 X ( n)

longest 40% CI = 
,

1/ n

 (1 −  )

 0.940479844

 longest 40% CI = 
,

1/3

 (1 − 0.6 )

 longest 40% CI = 1.276428,  )
For sample 4, the shortest 40% confidence interval is calculated by

X (n) 

shortest 40% CI =  X ( n ) , 1/ n 
 

0.90503358 

 shortest 40% CI = 0.90503358,

0.61/3

 shortest 40% CI =  0.90503358, 1.073036 
For sample 4, the longest 40% confidence interval is calculated by

 X ( n)

longest 40% CI = 
,

1/ n

 (1 −  )

 0.90503358

 longest 40% CI = 
,

1/3

 (1 − 0.6 )

 longest 40% CI = 1.22832,  )
For sample 5, the shortest 40% confidence interval is calculated by

X ( n) 

shortest 40% CI =  X ( n ) , 1/ n 
 

0.868601028 

 shortest 40% CI = 0.868601028,

0.61/3

 shortest 40% CI =  0.868601028, 1.02984
For sample 5, the longest 40% confidence interval is calculated by

 X ( n)

longest 40% CI = 
,

1/ n

 (1 −  )

 0.868601028

 longest 40% CI = 
,

1/3

 (1 − 0.6 )

 longest 40% CI = 1.178873,  )
e) It can be checked that all of the five shortest confidence intervals in part (d) contained  = 1,
and none of the five longest confidence intervals in part (d) contained  = 1. The expected
numbers show that half of the number of confidence intervals should contain  = 1.
Question 2:

 2x
 x2 
exp

 − , x  0
a) It is given that X1 , X 2 ,..., X n are i.i.d. with the density f ( x; ) =  
  
0,
x0

where   0 is unknown.
Hence, X 12 has the following density function (ignore the zero-part):

 x2  2x
 x2 
exp  −   exp  − 

   
  

( f ( x; ) ) =

2x

2

 ( f ( x; ) ) =
2

 2x2 
exp
−

2
  

4x2

 ( f ( x; ) ) = exp ( )
2 d

d

 X 12 = exp ( )
b) Summing up the result obtained in part (a) for the remaining n − 1 variables, it follows that
d

X 12 = exp ( )
n

d

  X i2 = n exp ( )
i =1

But we have

2n

2

n

d

 X i2 =
i =1

d

2n

exp ( )

exp ( ) =  22n . Hence, it immediately follows that

2

n

X

i =1

2
i

=  22n , as desired

result.

(

)

c) Let a and b be such that P a   22n  b = 1 −  . Then, we will have
n
n

 n 2

2
2
X
2
X
2
X i2 

i
i

P ( a   22n  b ) = P  a  i =1
 b  = P  i =1
   i =1

a 

 b...

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