北师大统计学基础习题答案2

Exercise 2

1. Draw the picture of b a f )1()(θθθ-=for 10≤≤θusing R software, where 0

)(θf ?

Solution:

Choose a=1/2 and b=1/2, then input “plot(function(x)(x^(1/2)*(1-x)^(1/2)),0,1)” in R.2.0.0, and we obtain the picture of the function 2

12

1)1()(θθθ-=f :

From the picture we can see that both the maximum point of )(θf and the

maximum value are 0.5.

For any 0

d erivativ

e o

f ))(ln(θf is

θ

θ--

1b

a

. Let the formula equal to zero and we get the

maximum point of )(θf is b a a + and the maximum value is b

a b

a b b a a )()(

++.

2. P266-267: 6.

3. Let n Y Y Y <<< 21be the order statistics of a random sample from a distribution with p.d.f ∞<<-∞+≤≤-

=θθθθ,2

1

21,1);(x x f , zero elsewhere. Show that every statistic ),,,(21n X X X u such that 2

1

),,,(21121+≤≤-

Y X X X u Y n n is a m.l.e. of θ.In particular,6/)124(1++n Y Y ,2/)(1n Y Y + , and 6/)142(1-+n Y Y are three such statistics. Thus uniqueness is not in general a property of a m.l.e. Solution:

The joint p.d.f of n X X X 21,is );();();(21θθθn x f x f x f

.2

1

21,1),,,;(21+≤≤-

==θθθi n x x x x L zero elsewhere. Since for any 2121,+≤≤-

θθi i X X , then 21

2121+≤<<≤-θθN Y Y Y , thus 2

1211+≤≤-Y Y n θ. Since 1)(=θL is a constant number as 2

1

211+≤≤-

Y Y n θ, every statistic ),,,(21n X X X u such that 2

1

),,,(21121+≤≤-Y X X X u Y n n is a m.l.e of θ.

From 2

1

211+≤≤-Y Y n θ we obtain that 011≤--Y Y n .

0)1(3

2

6124)21(11≤--=++--Y Y Y Y Y n n n ,0)1(31)21(6124111

≤--=+-++Y Y Y Y Y n n So 21

61242111+≤++≤-

Y Y Y Y n n and 6

1241++n Y Y is such a statistic. 0)1(2

1

2)21(11≤--=+--Y Y Y Y Y n n n , 0)1(21)21(2111

≤--=+-+Y Y Y Y Y n n , So 21

22111+≤+≤-

Y Y Y Y n n and 2

1n Y Y + is such a statistic. 0)1(3

1

6142)21(11≤--=-+--Y Y Y Y Y n n n ,0)1(32)21(6142111

≤--=+--+Y Y Y Y Y n n So 21

61-422111+≤+≤-

Y Y Y Y n n and 6

1421-+n Y Y is such a statistic. Therefore, 6/)124(1++n Y Y ,2/)(1n Y Y + , and 6/)142(1-+n Y Y are three such statistics.

3* P267: 6.8.If a random sample of size n is taken from a distribution having p.d.f θθθ≤<=x x x f 0,/2);(2,zero elsewhere, find: (a) The m.l.e θ? for θ.

(b) The constant c so that θθ

=)?(c E . (c) The m.l.e. for the median of the distribution. Solution:

(a) The joint p.d.f of a random sample of size n n X X X ,,,21 is ∏==

n

i i

n

n

n x

x x x f 1

2212),,;(θθ ,which can be regarded as a function of θ.

∏==

n

i i n

n

x L 1

22)(θ

θ is an ever-decreasing function. Since θ≤

x

0,

then )max(i x ≥θ and the value of L can be no larger than

∏=n

i i

n i n

x

x 1

2)max(2

So the m.l.e. θ? for θ is the nth order statistic n Y . (b) Since θθ

θ≤<=x x

x f 0,2);(2

, then the distribution function is

θθθ≤<=

x x x F 0,);(2

2

.∏=≤=≤≤≤=n

i i n x x x P x x x x x x P F i 1

21)max()(),,(

∏==

=n

i n

n

x x x F i 1

22)(θ

. n

n x nx x f i 21

2)max(2);(θ

θ-=

θθ?θθ

θ

=?+?

=+?

?

=?

=+?-1

221

222))max((1

20

20

21

2n n

c n n

c dx nx

x c x c E n x n

n

n i Therefore c=n

n 21

2+ (c) 212);(2

2

2

=

==?

θθ

θx ds s

x F x

, then the median of the distribution is

θ2

2

So the m.l.e for the median of the distribution is n Y 2

2

.