北师大统计学基础习题答案3

Exercise 3

1. Show that )(1111

n n n x n x n x -++-

++=;2122

1)(1

1)11(n n n n x x n s n s -++-++-=;and use Splus (or R) language to write programme of computing n x -

and 2

n s , where ∑=-

=n i i n x n x 11; and ∑=---=n i n i

n x x n s 1

22)(11. Solution ：Since ∑=-

=n

i i n x n x 11, then ∑=-=n i i n x x n 1

and

1`

1

11111)(11)(11+-+==+-+=+=++=++∑∑n n i i n i i n n n x x n x x n x n x n So )(1

1

11

n n n x n x n x -++-

++= Since ∑=-

--=

n

i n i n x x n s 1

22

)(11

, then ∑+=+-+-=1

1

212

1

)(1n i n i n x x n s 211

2

11211121)1

(1)(1)(1)(1++-+-+-=-+-=

-

+++-

-

-

=+-

+=+-

∑∑n x n x x n x x x x n x x n x x n n n n n n n n

i i n n n

i n i 2

12

1211112)()

1()(1))((2)(1n n n i n n n n n i n i n i n i x x n n x x n x x x x n x x n -+=+--+--==---++-+--+-=∑∑∑=2

1

2

21112)()1()1(1)(1n n n n n n i n i n i x x n n n x n x x n x x n -+-

+-==--++++-+-∑∑ 212

1212)()

1()1(1)11(n n n

i n n n x x n n n x x n s n -+=-

+-+++-+-=∑ 2

12

2122)()1()()1(1)11(n n n n n x x n n x x n s n -+-+-++-++-= 212)()

1(1)11(n n n x x n s n -+-++-=

So 21221

)(1

1

)11(n n n n x x n s n s -++-++

-= Program: We give the program by generating 100 standard normal random numbers. The program is: >x=rnorm(100,0,1)

> mean(x)

[1] -0.08212821 > var(x)

[1] 0.819366

2. P267 6.7. For each of the distributions in Exercise 6.1, find an estimator of θ by the method of moments and show that it is consistent.

(a)

,zero

elsewhere,

where

Solution: θθθθθθθθθθθ

θ

=??=?=?-=?

=--∞

=∞

=--∞

=-∑

∑

∑e e e x e

x e

x x x E x x

x x x x

1

1

0!

)!

1(!

)(

So the estimator of θ by the method of moments is -

=x θ?.

According to the Law of Large Numbers, ∑=?→?-n k P

k k x E x n 10)]([1, which

shows θ?→?

-

P x . So -

=x θ? is consistent. (b)

,zero elsewhere.

Solution: 1

)(1

1

1+=

=?=??-θθθθθθdx x dx x x x E

Let -

=+x 1??θ

θ, then -

-

-=

x

x 1?θ and θ? is consistent.

(c)

,zero elsewhere.

Solution: θθ

θθ

θθ=+?-=-=??

=???∞-

∞

-

∞-

-

∞

dx e e

x e xd dx e x x E x

x

x

x

|)(1

)(

So -

=x θ? and θ? is consistent.

(d)

Solution: dx e x dx e x dx e x x E x x x -∞-∞---∞∞-???+=?=θθθθθ222

1

)(||

θθθθθθθθθ=----=----)(2

)(2e e e e e e . So -

=x θ? and θ? is consistent. (e)

zero elsewhere.

Solution: ???∞

∞

--∞

--+=-==?=θ

θ

θθ

θθθ1)()()(x x x e xd e dx xe e dx e x x E

Let -=+x 1?θ and 1?-=-x θ. 1?-=-

n

n x θ. (|lim P n ∞

→n

θ?-0)|)((|)|1(||=≥-=≥--=≥-

∞

→-∞→εεθεθx E x P im l x P im l n n n n . So θ? is consistent.

3. p217-218

4.97. Find the mean and variance of n X X S n

i i /)(212

∑=-

-= ,

where n X X X ,,,21 , is a random sample from ),(2σμN .

Solution: Since )1(~2

2

2

-n nS χσ

,

1)(

2

2

-=n nS E σ

, )1(2)(

2

2

-=n nS Var σ

So 221)(σn n S E -=, 4

22)1(2)(σn

n S Var -=.

4. P218 4.101. Let 521,,X X X be a random sample of size n=5 from

),0(2σN

(a) Find the constant c so that 252

4

2321/)(X X X X X c ++- has a t-distribution.

(b) How many degrees of freedom are associated with this T? Solution: According to Theorem 1 of Section 4.7, since 1X ,2X ~),0(2σN .

)2,0(~2

211σN X X Y -=, then

)1(~2

2

2

χσ

i X . )3(~22

2

52

4232χσ

X X X Y ++=

.

(

)

25

24

2

3

212524232

121263

/23/2X

X X X X X X X X X Y Y ++-=

++-=σσ has a t-distribution.

So 2

6

=

c an

d th

e degrees are 3.