Homework 3 - Solution
(p.338: 6.75)
Refer to Exercise 6.74. Suppose that the number of minutes that you need to wait for a bus is
uniformly distributed on the interval [0,15]. If you take the bus five times, what is the
probability that your longest wait less than 10 minutes.
Solution:
The distribution function for a random variable that is uniformly distributed on the interval [0,15]
is F(y) = y/(15-0) = y/15 for 0 y 15. The density of Y(5) is given by
G5 ( y ) [ F ( y )]5 ( y / 15) 5 , 0 y 15.
Therefore, P(Y( 5) 10) G5 (10) (10 / 15)
.1317
15
(p.338: 6.76)
Let Y1, ..., Yn be independent, uniformly distributed random variables on the interval [0, ].
a. Find the density function of Y(k), the kth-order statistic, where k is an integer between 1
and n.
b. Use the result from part (a) to find E(Y(k)).
c. Find V(Y(k))
d. Use the result from (c) to find E(Y(k) - Y(k-1)), the mean difference between two successive
order statistics. Interpret the result.
Solution:
a.
By Thm 6.5, g k ( y )
y k 1 y n k 1
n!
( k 1)! ( n k )!
b.
E (Y( k ) ) y ( k 1)!n(!n k )!
0
z y /
y k 1 ( y ) nk
n!
( k 1)! ( n k )!
n
1
y k 1
y nk 1
(z ) z 1 z
1
n!
( k 1)!( n k )! 0
z 1 z
dz
z 1 z
dz
1
n!
( k 1)!( n k )! 0
1
n!
( k 1)!( n k )! 0
n!
( k 1)!( n k )!
k !( n k )!
n!
( k 1)!( n k )! ( n 1)!
k
n 1
nk
nk
dy
nk 1
k 1
(k 1, n k 1)
, 0 y .
dz
c.
Using the same techniques in (b) above, we have
E (Y(2k ) ) y 2
0
z y /
y k 1
n!
( k 1)!( n k )!
1
n!
( k 1)!( n k )! 0
(z )
y nk 1
z k 1 1 z
dz
z 1 z
dz
1
n! 2
( k 1)!( n k )! 0
1
n! 2
( k 1)!( n k )! 0
n! 2
( k 1)!( n k )!
( k 1)!( n k )!
n!
( k 1)!( n k )!
( n 2 )!
k ( k 1)
( n 2 )( n 1)
nk
k 1
nk
k 1
dy
nk 1
z 1 z
dz
(k 2, n k 1)
so that the variance is V (Y( k ) ) E (Y( k ) ) [ E (Y( k ) )]
2
( n k 1) k
( n 1) 2 ( n 2 )
d.
E (Y( k ) Y( k 1) ) E (Y( k ) ) E (Y( k 1) )
k
n 1
nk 11
1
n 1
. Note that this is constant for all k,
so that the expected order statistics are equally spaced.
(p.394: 8.8)
Suppose that Y1, Y2, Y3 denote a random sample from an exponential distribution with
density function
1 e y / , y 0,
f ( y)
0, elsewhere.
Consider the following five estimators of :
1 Y1 , 2 Y 2Y , 3 Y 32Y , 4 min(Y1 , Y2 , Y3 ) , 5 Y
1
a.
b.
Which of these estimators are unbiased?
Among the unbiased estimators, which has the smallest variance?
Solution:
a.
First, we have E(Yi) = , Var (Yi) = 2, for i = 1, 2, 3.
E (1 ) E (Y1 ) , E (2 ) E ( Y1 2Y2 ) 12 E (Y1 ) 12 E (Y2 ) 12 12 ,
E (3 ) E ( Y1 32Y2 ) E ( Y31 ) E ( 23Y2 ) 3 23 ,
3
3
E (5 ) E (Y ) E ( 13 i 1 Yi ) 13 i 1 E (Yi ) 13 3 . So these four estimators are
unbiased.
For 4 min(Y1 , Y2 , Y3 ) :
We know that f ( y ) 1 e
y /
and F ( y ) 1 e y / , y 0.
Therefore, the density function of Y(1) is given by
g1 ( y ) 3[1 F ( y )]31 f ( y ) 3[e y / ]31 1 e y / 3 e 3 y / , y 0.
This is the exponential density with mean /3. E (4 ) / 3 . So this estimator is biased.
b.
var(1 ) var(Y1 ) 2 , var(2 ) var( Y1 2Y2 ) 14 E (Y1 ) 14 E (Y2 ) 14 2 14 2 12 2 .
var(3 ) var( Y1 32Y2 ) 19 var(Y1 ) 94 var(Y2 ) 9 49
2
var(5 ) var(Y ) var( 13 i 1 Yi )
3
1
9
3
i 1
5 2
9
var(Yi ) 19 3 2 3 .
2
So 5 has the smallest variance among those unbiased estimators.
(p.395 8.9)
Suppose that Y1, ..., Yn constitute a random sample from a population with probability
density function
11 e y /( 1) , y 0, 1
f ( y)
0, elsewhere.
Suggest a suitable statistics to use as an unbiased estimator for .
The density is in the form of the exponential with mean + 1. We know that E( Y ) = + 1, so
Y is unbiased for the mean + 1. Thus an unbiased estimator for is simply Y - 1.
