Question

Consider a population with a uniform distribution on the range 0 to ?. Consider a sample of size 2 from this population and define the sample range ? = ???(?1,?2) − ???(?1,?2).Find ??(?),the PDF of ?.

We have n samples of ???????(0,?). ?? = ???{?1,?2,…,??}; ?1 = {?1,?2,…,??}. The joint PDF of ??,?1 is

??1,??(?1,??) =
? ? − 2
[??(??) − ??(??)]?−2??(?1)??(??) =
? ? − 2
[
?? ?
−
?1 ?
]
?−2
∙
1 ?
∙
1 ?


=
? ? − 2
(?? − ?1)?−2

To find distribution of ? = ?? − ?1 ??? ? = ?1;

⇒
?? = ? + ? ?1 = ?
}(?1,??) ⟶ (?,?)

Jacobian transformation,

|?| = |
??1 ??
??1 ??
??2 ??
??2 ??
| = |
0 1 1 1
| = −1

∴ ??,?(?,?) =
? ? − 2
??−2; −? < ? < ? ??? 0 < ? < ?.

So,

??(?) = ∫ ?(? − 1)??−2?? = ?(? − 1)???−2; −? < ? < ? ? 0


b. Use your answer to part (a) to find the constant c such that ? < ? < ?? is a 100(1 − ?)% confidence interval for ?.

Now, ?? − ?1 = ? ⇒
?? ?
−
?1 ?
=
? ?
?? ?ℎ? ???????????? ?? ?.

To find c, such that,

?(? < ? < ??) = 1 − ?

= ?(1 <
? ?
< ?) = 1 − ? ⇒ ?( ? ?
< ? < ?) = 1 − ? ⇒ ??(?) − ?? ( ? ?
) = 1 − ?

⇒ ?? ( ? ?
) = ? [∵ ??(?) = 1] ⇒ ∫ ?(? − 1)??−2??? ? ? −?
= ?
⇒ ?(? − 1)? ∙
1 ? − 1
∙ ??−1 |? ? ⁄ −?
= ?

⇒ ??{( ? ?
) ?−1
− (−?)?−1} = ? ⇒ ??? {( 1 ?
) ?−1
− (−1)?−1} = ? ⇒ ( 1 ?
) ?−1
=
? ???
+ (−1)?−1

⇒ ? = {
1
? ??? + (−1)?−1
}
1 ?−1


Repeat (a) and (b) for a sample of size n.

Answer 1