# MATH 448 confidence intervals discussion

*label*Mathematics

*timer*Asked: Feb 25th, 2019

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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

## Tutor Answer

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:

Pa

Pa

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,

x0

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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