John e freunds mathematical statistics with applications 8th edition miller solutions manual

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Chapter 10 10.1

E 

 a x    a E ( x )   a μ  μ  a i i



i

i

i

i

n

a  1 i

i 1

10.2

E [k1θˆ1  k2θˆ2 ]  k1θ  k2θ  θ , k1  k2  1

10.3

x    (2m  1)!   h ( x )  f ( x ) dx f ( x )    f ( x ) dx  m ! m !     x 

m





m

x  θ (1/2)  (2m  1)!   h ( x )  dx   1   dx  m ! m ! θ (1/2)      x  



m



m



m

(2m  1)!  1  1   x  θ   θ   x  m! m!

2 2

m

m 1

1 1   h ( x )  6(  x  θ   θ   x      2 2 θ  (1/2)



1



1

x  x  θ   θ   x dx    2 2 θ

E ( x )  6

 (1/2)

let u = x  θ 

1 2

1

1   6  u  θ  u(1  u )du  θ  2

 0

10.4

h( x ) 

6 2 x /θ 1  e x /θ  e   8

E[ x ] 

6

θ 0



6





x e 2 x /θ 1  e  x /θ  dx

x e θ 0

2 x /θ

dx 

6



x e θ

3 x /θ

dx

0

5  θ  biased 6 Use gamma integrals.

139

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140

10.5

Mathematical Statistics, 8E

1 E n

1 n

2

i

i

i 1



10.6

E( x )  μ E( x 2 ) 

10.7

10.8

2

 ( x  μ )   n  E ( x  μ   n



1 1 σ 2   nσ 2  σ 2 n i 1 n

var( x ) 

σ2

σ2 n

 μ 2  μ 2 as n  

n

1 1 n 1  x 1   θ E E ( x  1)  ( nθ  1)   n2 n2 n2 n2 n2 E  θ when n   , so is asymptotically unbiased    ( x δ )   ( y1 δ ) g1 ( y1 )  n e dx   e  y1 

n 1



 n e  ( y1 δ )  e  ( n 1)( y2 δ )  n e  n ( y1 δ ) 

δ

E ( y1 )  n y1 e  n ( y1 δ ) dy1 



 n (u  δ )e  nu du  0

let u  y1  δ 1 δ n

The unbiased estimate is Y1 

10.9

β 1 1  g1 ( y1 )  n   dx  β y β  1 

E (Y1 ) 

E (Y1 )  δ as n  

n 1



n

1 n

β



n

β

n

(β  y1 )n 1

y ( β  y ) dy β  n 1

1

n

1

u

1

0



b

β

1

1

0

0

y1

β

du 

β u( β  β u ) n1β du  nβ  u(1  u ) n1du  n 

dy1

β β

n 1

Unbiased estimate is ( n  1)Y1  n xi2  1 n E xi2 10.10 E   n n i 1  i 1 

  





1 n 1 n 2 σ  σ2 (σ 2  μ 2 )  n i 1 n i 1





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Chapter 10

141

 x  x  1 10.11 E n   1     E ( x )  E ( x 2 )   n  n  n 1  nθ   nθ (1  θ )  n 2θ 2  n  ( n  1)θ (1  θ )  nθ (1  θ )

10.12 (a)

 yn  1 n  1 values before yn in  ways.  n  1   yn  1  n  1  f ( yn )  k   n 

for yn  n,, k

 yn  1  n  1  n k  yn  n  k  1 E (Yn )  yn      k   k  yn  n  n   k   n  1 yn  n  n   n   n  k

(b)

biased





n (k  1) n 1



see Exercise 1.15 or Theorem 1.11, respectively

 n 1  n  1 n (k  1) E  Yn  1   1  k n n 1  n 

QED

10.13 E (θˆ 2 )  var(θˆ )  E (θˆ )2  var(θˆ )  θ 2 E (θ 2 )  θ 2 since var(θ )  0 10.14 f ( x;θ )  θ x (1  θ )1 x E ( x)  θ ln f ( x;θ )  x ln θ  (1  x )ln(1  θ )  ln f ( x;θ ) x 1  x x θ    θ θ 1  θ θ (1  θ )

E( x2 )  θ

   ln f ( x;θ )2   1 1 E  E ( x  θ )2   2 2  θ (1  θ ) θ   θ (1  θ )   1 θ (1  θ ) x   var   when x is binomial random variable. u n E n x  is minimum variance estimator n  x  nθ θ E   n n  unbiased

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142

Mathematical Statistics, 8E

10.15 f ( x; λ ) 

λ x e λ x!

μλ

var( x ) 

σ2  λ

λ n

E ( x )  λ  unbiased ln f  x ln λ  λ  ln x !    ln f  2  E ( x 2 ) 2  E ( x)  1 E      λ λ2  λ 

 ln f x  1 λ λ



λ  λ2 2 1  λ 1  2 λ λ λ

1 λ   var( x ) nE n  x is minimum variance unbiased estimator

10.16 var(θˆ1 )  3var(θˆ2 ) E ( a θˆ  a θˆ )  a θ  a θ  θ  a  a  1 1 1

2 2 1 2 2 2 ˆ var  ai var( 1 )  a2 var( ˆ2 )

θ

1

2

θ

var  3ai2 var( ˆ2 )  a22 var( ˆ2 )  (3a12  a22 ) var( ˆ2 )

θ

θ

θ

 [3a12  (1  a1 ) 2 ]var(θˆ2 )   6a2  2(1  a1 )(1) a1  8a1  2  0

10.17 f ( x;θ ) 

1  x /θ e

θ

a1  E ( x)  θ

E ( x )  θ  unbiased ln f   ln θ 

1 4

a2 

3 4

E ( x 2 )  2θ 2

var( x ) 

σ2  θ2

θ2 n

x

θ

 ln f 1 x x θ   2  2 θ θ θ θ    ln f  2  1 1 2 E    4 E(x  θ )  2     θ θ  θ  1 θ2   var( x )  x is minimum variance unbiased estimator nE n n nβ 2 nβ 2 β , E (Yn2 )  , var(Yn )  n 1 n2 ( n  2)( n  1)2 n 1 let B   Yn n n 1 n   β  β  unbiased E ( B)  n n 1

10.18 E (Yn ) 

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Chapter 10

143

( n  1) 2 nβ 2 β2   2 2 n (n  2) n (n  2)( n  1)

var( B ) 

1 1 β2 β2     var( B ) n n (n  2)   ln f ( X )  n 1 nE   β2 β  so the Cramèr-Rao inequality is not satisfied.  ln f ( x ) 1 f ( x ) f ( x )  ln f ( x )   f ( x)  f ( x ) θ θ θ θ  ln f ( x )   f ( x )dx  0 θ 2 ln f ( x )  ln f ( x )  ln f ( x )  f ( x)    f ( x) 2 θ θ θ 2 2 ln f ( x )   ln f ( x )   f ( x )dx    f ( x )dx θ 2  θ 

10.19 (a)



(b)





2    ln f ( x )  2     ln f ( x )   E  E       θ       θ  

10.20

 ln f ( x ) 1  x  μ  from Example 10.5   μ σ  σ  2 ln f ( x ) 1  2 2 σ μ 1    ln f ( x )  2  nE       μ  



1 σ2  n  1  n  2   σ 

E [ wx1  (1  w) x2 ]  wμ  (1  w)μ  μ

10.21 (a)

var[ wx1  (1  w) x2 ]  w2

(b)

σ12 n

 (1  w)2

σ 22 n

σ σ d  2w  2(1  w)( 1) 0 dw n n



2 1

2 2



w

w σ12  σ 22  σ 22

10.22 var1  w2 w

1 2

σ12 n

 (1  w)2 var 

σ

2 1

4n

σ 22 σ12  σ 22

σ 22



n

σ 22 4n



1 2 (σ1  σ 22 ) 4n

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144

Mathematical Statistics, 8E

2

2

 σ2  σ2  σ2  σ2 var 2   2 2 2  1   2 1 2  2  σ1  σ 2  n  σ1  σ 2  n 





σ12σ 22 

2 2 σ 22 σ12   σ1 σ 2  n  σ12  σ 22  σ 2  σ 2 2  n (σ12  σ 22 ) 1 2



efficiency 



2 2 1  2 2 2 2 1  2)

σ σ n(σ σ

1 (σ12  σ 22 ) 4n



σ12σ 22 4n  2 2 2 n(σ1  σ 2 ) σ1  σ 22



10.23 var  w 2

σ2 n1

 (1  w) 2

4σ12σ 22 (σ 12  σ 22 ) 2

σ2 n2

d 2 wσ 2 2(1  w)σ 2   0 dw n1 n2 w 1 w n1  w n1 n2 n1  n 2

10.24 For w 

1 2

var 

2

 n1  σ 2  n2  σ 2 var      n1  n2  n1  n1  n 2  n2 2

n1 For w  n1  n2



σ

Efficiency =

1 σ2 1 σ2 σ2  1 1      4 n1 4 n2 4  n1 n2 

σ2 ( n1  n2 )

( n1  n2 )  2

σ2 n1  n2

2

4n1n2 n1  n2  σ  1 1  ( n1  n2 )2  4  n1 n2  2

 x  2 x2  x3  1 2 1 2 1 2 3 2 10.25 var  1   σ  σ  σ  σ  4 16 4 16 8

σ Efficiency =

var( x ) 

σ2 3

2

3 8 3 2 9 σ 8

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Chapter 10

145

10.26 μ  θ and σ 2  θ 2

var( x ) 

θ2

2 2 2 y1 /θ g1 ( y1 )  e

From Ex. 8.4

for y1  0

θ

θ  var(Y1 )    2

E (2Y1 )  θ unbiased 2

θ θ  var(Y1 )     2 4

2

var(2Y1 )  4 

θ2 4

 θ2

θ /2 1  2 θ2 2

Efficiency = n

10.27 g n ( yn ) 

β

n

ynn 1

n

y dy β 

E (Yn )  

n

n n

n

β n1

β

E (Yn )2 

 n

n

n 1 β



βn 0

0  yn  β

n



βn n 1

ynn 1dyn 

n

β n 2

β

n2

 n



nβ 2 n2

β  n( n  1)  n ( n  2) nβ 2 n2β 2 var(Yn )    n  2 ( n  1)2 ( n  2)( n  1)2 2

2

2

nβ 2 ( n  2)( n  1)2 n 1 n  1 nβ Z  Yn  E(Z )    β unbiased n n n 1 2 nβ 2 β2  n 1 var( Z )     QED  n  ( n  2)(n  1)2 n( n  2) 

10.28 Y  x  1

var(Y )  var( x ) 

1 Z  Y1  n

g1 ( y1 )  ne  n ( y1 δ ) E (Y1 )  

E (Y12 )  n



2  n ( y1 δ ) dy1 1

 n (u  δ )2 e  nu du  0

n



1 n

1 δ n

δ y e



θ2

u  y1  δ 2 2δ  δ2 n2 n

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146

Mathematical Statistics, 8E

2

2 2δ 1 1  δ2   δ   2 var(Y1 )  2  n  n n n 2

1 var( Z )  n  1 efficiency    1 var(Y ) n n

10.29 Continue from Exercise 10.12  yn  1 n( n  1) k  yn  1 1 k  E Yn (Yn  1)  yn ( yn  1)   k  yn  n  n  1   k  yn  n  n  1   n   n 







n( n  1)  k  2    k   n  2   n 

k

Exerxise 1.15 or

i

 k  1

  n    n  1 i n

n( k  1)( k  2) n2 n( k  1)(k  2)  E (Yn2 )  E (Yn )2 var(Yn )  n2 n( k  1)(k  2) n( k  1) n 2 ( k  1)2    n2 n 1 ( n  1)2 

 n 1  (n  1) var(   Yn  1  var(Yn )  n  n2 2

 

(k  1) ( k  2)( n  1)2  ( n  1)(n  2)  ( k  1)n (n  2) n( n  2)

(k  1)(k  n ) n (n  2)

k 1 (k  1)(2k  1) k2 1 , E( x2 )  , σ2  2 6 12 for population (k  1) E (2 x  1)  2 E ( x )  1  2 1  k unbiased 2 (k 2  1) k  n ( k  1)( k  n ) var( x )    12n k 1 12n ( k  1)(k  n ) ( k  1)( k  n ) 3 n (n  2)  var(2 x  1)  , efficiency  ( k  1)(k  n ) n  2 3n 3n 3 3 (a) efficiency = ; (b) efficiency = 4 5 E ( x) 

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Chapter 10

147

1

10.30 (a)

1



E ( x )  x dx  0

var( x ) 

(b)

1 1 1 1 1 , E ( x 2 )  x 2 dx  , var( x )    2 3 3 4 12 0



1 / 12 1  3 36

g1 ( y1 )  3(1  y1 )2

0  y1  1

g3 ( y3 )  3 y32

0  y3  1 0  y1  y3  1

f ( y1 , y3 )  6( y3  y1 ) 1

1 4



E (Y1 )  3 y1 (1  y1 ) 2 dy1  0

1

E (Y12 )  3

1

 y (1  y ) dy  10 2 1

2

1

1

0

1

1

3 3 E (Y3 )  3 y32 dy3  , E (Y32 )  3 y34 dy3  4 5 0 0





1 y3

E (Y1Y3 )  b

1

  y y ( y  y ) dy dy  5 1 3

3

1

1

3

0 0

1 1 3 3 9 3 var(Y1 )    , var(Y3 )    10 16 80 5 16 80 1 3 1 cov(Y1 , Y3 )    5 16 80

(c)

Y Y  1  1 3  1 E  1 3        unbiased  2  2 4 4 2 1 Y Y  1 3 1 3 1 1 var  1 3          2  4 80 4 80 2 80 40

Since

1 1 is less than midrange here is more efficient than the mean.

40 36

10.31 E (θˆ )  θ  b(θ ) E (θˆ  θ )2   E (θˆ 2 )  2θ E (θˆ )  θ 2  E (θˆ )2  2θ θ  b(θ )  θ 2 =E (θˆ 2 )  θ 2  2θ b(θ ) 2 2 var(θˆ )  E (θˆ 2 )  θ  b(θ )  E (θˆ 2 )  θ 2  2θ b(θ )   b(θ )

 E (θˆ  θ 2 )   b(θ )

2

2  E (θˆ  θ )2   var(θˆ )   b(θ )

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θ (1  θ ) 1  10.32 var(θˆ1 )  n 4n n θ  1 1 1 E (θˆ2 )  for θ  E (θˆ2 )   unbiased n2 2 2 nθ (1  θ ) n variance(θˆ2 )   = mean square error 2 (n  2) 4( n  2)2 2

1 1 1 E (θˆ2  θ )2       3 2 36 n 1 (a)  2 4n 4( n  2)

n 2  (n  2) 2

for all values of n (b)

1 1  , 4n  36, n  9 36 4n

α 1  10.33 g1 ( y1 )  n  f ( x ) dx  y   1 

n 1

f ( x)  1 0



elsewhere

for α  y1  α  1

= n (α  1  y1 ) n 1 P  Y1  α  c  

α  x  α 1

0

α c

elsewhere

 n(α  1  y ) dy α n 1

1

1

 1n  (1  c )n  1 when n   with c fixed. α 1

10.34 E (α  1  Y1 )  n

n

 (α  1  y ) dy  n  1 α n

n

α 1

E (α  1  Y1 ) 2  n E (Y1 )  α  1 

QED

n

 (α  1  y ) dy  n  2 α n 1

n 1 α  n2 n 1

1   E ( Y1    α  unbiased  n 1 2

n n  n  var(α  1  Y1 )      ( n  2) n 1 ( n  2)(n  1)2 1  n  var Y1   0 as n      n 1 ( n  2)(n  1)2

QED

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149

10.35 g n ( yn ) 

n

βn

ynn 1

P  Yn  β  c  

0  yn  β β

 β   β  c )  y dy   β β β  n

n

n 1 n

1

n

n

n

n

c

n

βc  1  1  β  when n   with c fixed.

10.36 x is consistent estimate of the mean of any population with a finite variance. Since θ is the mean and σ 2  θ 2 if follows that x is consistent estimate of θ .

10.37 For any single observation and for c  θ , P ( X  θ  θ )  1  e 2θ /θ  e 2 does not converge to 0, so X n is not consistent for θ .

10.38 Shown is (a) of 10.21 that it is unbiased. From 10.22 variance is

σ12σ 22 0 n σ12  σ 22



So it is consistent by Theorem 10.3.

1 nθ (1  θ )  X 1 10.39 Var   Var ( X )   2  n  2  (n  2) ( n  2)2

 θ as n  

asymptotically unbiased 1 nθ (1  θ )  X 1 Var   var( x )   n  2  ( n  2)2 n2 

θ (1  θ ) n

 0 as n  

QED

n β  β as n    asymptotically unbiased n 1 From Example 10.6 (see Exercise 10.27) n β2 β2 var(Yn )     0 as n   n  1 n (n  2) ( n  1)(n  2) consistent by Theorem 10.3

10.40 E (Yn ) 

10.41 (a)

(b)



n 1 1 P  x  μ  c  P n2  μ  c n n n 1 since x is known to be consistent and 1 n Let estimate be x n 1 2 1 n 1 E ( x)  μ  n   μ n  μ n n n not unbiased and not asymptotically unbiased.

P  x  μ  c

 1 0  1

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

150

Mathematical Statistics, 8E

10.42 f ( x1 , x2 ,, xn ) 

1

θ

e

1 n  xi    θ i 1 





1

e  (1/θ ) nx

n θ  g (θˆ,θ ) Since the joint density depends only on θ and x , x is a sufficient estimator of θ . n

n  n  10.43 f ( x1 , x2 )   1   2  θ x1  x2 (1  θ )( n1 n2 )( x1  x2 )  x1   x2  x x θˆ  1 2 n1  n2  n1   n2  ˆ ˆ f ( x1 , x2 )      θ ( n1 n2 )θ (1  θ )( n1 n2 )(1θ )    x1   x2      (θˆ ,θ ) h( x , x ) 1 2  by theorem, estimator is sufficient.

10.44 Try x1  0 and x2  1 2 2 f (0,1)      θ (1  θ )3  2θ (1  θ ) 2 0 1 x1  0 x2  1 x1  2 x2 2 1   only possibilities x1  2 x2  0 n1  2n2 6 3  by theorem, estimator is sufficient.  2  2 f (2,0)      θ 2 (1  θ )2  θ 2 (1  θ )2  2  0 Y

1 2θ (1  θ )3 2(1  θ )  f  0,1 Y     2 2 2   3 2(1  θ )  θ 2θ (1  θ )  θ (1  θ ) 2  2θ 2 θ  Y not sufficient 

10.45 f ( x1 , xn ) 

1

not independent of θ

g ( yn ) 

βn

n

βn

ynn 1

1 f ( x1 , xn Yn ) 

βn n

β 10.46 f ( x1 , x2 ) 

n 1 n yn



λ x1  x2 e 2 λ

x

x1 ! x2 !

λ 2 x e λ 

1 n ynn 1

1 x1 ! x2 !

   g ( x , λ ) ( x1 , x2 )

independent of β  sufficient

x1  x2 2

satisfies Theorem 10.3  sufficient

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151

10.47 Try x1  0, x2  1, x3  0, Y  2 The only possibility is x1  1, x2  0, x3  1 f (0,1,0)  θ (1  θ ) 2 f (1,0,1)  θ 2 (1  θ )

θ (1  θ ) 2  1θ θ (1  θ )2  θ 2 (1  θ ) not independent of θ  not sufficient f (0,1,0 Y  2) 

10.48 f ( x )  θ (1  θ ) x 1 f ( x1 , xn )  θ n (1  θ )  θ n (1  θ )nx  n Depends only on θ and x  sufficient xn

2 ˆ2  (1/2)  ( xi  μ )2  /σ 2 1 1 e  e  ( n /2σ )σ n /2 n n /2 n (2π ) σ (2π ) σ 2 2 Depends only on σ and σˆ  sufficient.

10.49 f ( x1  xn ) 

10.50 μˆ  m1, μ 2  σ 2  m2

σˆ 2  m2  (m1) 2 10.51 m1  μ  θ

θˆ  m1

p ˆ , β  2m1 2

10.52 μ 

λˆ  m1

10.53 μ  λ 10.54 β  1

μ

α α 1

α  m1 α 1

α  α m1  m1

α (1  m1)  m1, αˆ  10.55 μ 

θ

2

m1 1  m1

θ

x (θ  x )dx  , θˆ  3m  3 θ  1

2

0

10.56 μ 

1



θ δ

xe

 (1/θ )( x δ )

dx 

1



θ 0

(u  δ )e (1/θ )u du  θ  δ

u  x δ

μ2 

1



θ δ

2  (1/θ )( x δ )

x e

dx 

1



θ 0

(u  δ ) 2 e  (1/θ ) u du  2θ 2  2δθ  δ 2

m1  δ  θ , m2  2θ 2  2δθ  δ 2  θ 2  (θ  δ ) 2  θ 2  (m1) 2 θˆ  m   ( m  )2 and δˆ  m   m   ( m ) 2 2

1

1

2

1

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10.57

Mathematical Statistics, 8E

α β 1 1 (β  α )2  (α  β ) 2  m2  m1 12 4 2 1 m2  ( β  α ) 2  ( m1 )2 ( β  α )2  12[m2  ( m1 )2 ] 12 β  α  2 3[m2  ( m1 )2 ] β  α  2m1 βˆ  m   3[m   ( m )2 ]

add

1

2

1

subtract αˆ  m1  3[m2  ( m1 ) 2

10.58 μ  38

m1 

n0  0  n1  1  n2  2  n3  3  3θ N

n  2n2  3n2 θˆ  1 3N

λ  e  nλ x

10.59 L(λ ) 

ln L(λ ) 

 xi !

 x   (ln λ )  nλ  ln  x ! i



x d ln L( λ )  n 0 dλ λ x λˆ  x n



10.60 b( x;α ) 

Γ (α  1) α 1 x  α xα 1 Γ (α )Γ (1)

L(α )  α n (  xi )α 1

ln L(α )  n ln( n )  (α  1)

d ln L(α ) n   ln xi α dα n α n ln xi

 ln x

i



 i 1

1 xα 1e  x / β β Γ (α ) 1  2 x e x / β

10.61 f ( x ) 

α2

α

β

L( β ) 

1

  xi  e (1/ β ) x

β d ln L( β ) 2n 1   2 x  0 dβ β β 2n

β

ln L(β )  2n ln β  ln  xi 

x  x 2n

2

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β 1

x

Chapter 10

153

2 1 1  (1/2σ 2 )  ( x  μ )2 e  (1/2)[( x  μ )/σ ] L(σ )  e n n (2π ) σ 2πσ n 1 ln L(σ )   ln 2π  n ln σ  2 ( x  μ )2 2 2σ d ln L(σ ) n 1 ( x  μ )2  0   2 dσ σ σ2

10.62 f ( x ) 





 ( x  μ ) and σˆ   ( x  μ ) 2

σˆ 2 

n

μ

10.63 (a)

2

n

1  m1 8

θˆ 

1 1  m1 x

g ( x )  θ (1  θ ) x 1 L(θ )  θ n (1  θ )

(b)

xn

 x  n  ln(1  θ ) d ln L(θ ) n  1     x  n   0  1  θ  dθ θ

ln L(θ )  n ln θ 

10.64 f ( x )  2α xe α x

L(α )  2n α n   xi  e

2

ln L(α )  n ln 2  n ln α  ln  xi  α ln L(α ) n   dα α

10.65 f ( x ) 

α

x α 1

θˆ 

x  0 L( a ) 

x



1 x

 x  2

 x 

αˆ 

2

α

n

2

n x2



αn

  xi α 1

ln L(α )  n ln α  (α  1)ln(  xi ) dL(α ) n n   ln  xi   ln xi  0 dα α α n α ln xi





1 10.66 f ( x )  e  ( x δ )/θ 8 1  (1/θ ) ( x δ ) L(θ , δ )  n e

θ Maximized with respect to δ let δ̂ be y1 (smallest sample value)

δˆ  y1 θ 1

(x  δ)

d ln L(θ ,6) n 1   2 θ θ σ

(x  δ)

ln L(θ , δ )   n ln θ 

θˆ 

 x  δˆ n

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Mathematical Statistics, 8E

10.67 f ( x ) 

1 β α

L(α , β ) 

1 (β  α ) n

To maximize αˆ  y1 , and βˆ  yn 10.68 L  [(1  θ )3 ]n0 [3θ (1  θ )2 ]n1 [3θ 2 (1  θ )]n2 [θ 3 ]n3  3n1  n2 θ n1 2 n2 3n3 (1  θ )3n0  2 n1 n2 ln L  ( n1  n2 )ln 3  ( n1  2n2  3n3 ) ln θ  (3n0  2n1  n2 )ln(1  θ ) dL n1  2n2  3n3 3n0  2n1  n2   1 θ θ dθ ( n1  2n2  3n3 )(1  θ )  (3n0  2n1  n2 )θ θ (3n0  3n1  3n2  3n3 )  n1  2n2  3n3 n  2n2  3n3 θˆ  1 3N

1 xα 1e  x / β β Γ (α ) 1 α 1  (1/ β )  xi L(β )  nα  xi  e n  β [Γ (α )]

10.69 f ( x )  (a)

α

ln L(β )   nα ln β  n ln Γ (α )  (α  1) ln  xi  d ln L( β )  nα 1   2 dβ β β

(b)

 2x   1 α 

x

βˆ 

i

x  x

β i 1

x

i

nα

α

2

τ

10.70 L(α , β ) 

 2π 

ln L(α , β )  k 

2 n

1 2

e

 (1/2)

 [ v (α β )]2 (1/2) [ w(α β )]2

 [v  (α  β )]  2  [w  (α  β )] 2

1

2

 ln L  α

 (v  (α  β )]   ( w  (α  β )]  0 v  w  v  w αˆ   v   w  2nα  0  ln L  β

2n

2

 (v  (α  β )]   ( w  (α  β )]  0 v  w  v  w βˆ   v   w  2 nβ  0 2v

2

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Chapter 10

155

10.71 V n1 μ1 σ W n2 μ 2 σ L

 2π 

1 n1  n2

σ n1 n2

e

 (1/2σ 2 )

ln L  k  ( n1  n2 )ln σ 

 ( v  μ1 )2 (1/2σ 2 ) ( w μ2 )2

 (v  μ )  2σ  ( w  μ ) 2σ 1

2

2

 ln L 1   2  2 (v  μ1 )  0 μ1 2σ  ln L 1   2  2 ( w  μ2 )  0 μ 2 2σ L n n 1   1 2  3  ( v  μ1 ) 2  σ σ σ 2 ( v  v )  ( w  wˆ )2 2 σ  n1  n2



1

1

2



μˆ 1  v



μ2  w

 

2

2

 ( w  μ )  2

2

10.72 Any value θ̂ will do so long as 1 1 θˆ   y1 and yn  θˆ  2 2 1 1 θˆ  y2 + and θˆ  yn  2 2 1 ˆ 1 yn   θ  y1  2 2 1 1 1 10.73 (a) It is if Yn   (Y1  Yn )  Y1  2 2 2 make use of Y1  Yn  Y1  1 1 1 1 (Y1  Yn )  (Y1  Y1  1)  Y1  2 2 2 1 1 1 (Y1  Yn )  (Yn  Yn  1)  Yn  2 2 2

(b)

both conditions are satisfied

Suppose Y2  Y1  1 let n  2 1 1 2 1 (Y1  2Y2 )  (3Y1  2)  Y1   Y2  3 3 3 2 not max likelihood estimate

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Mathematical Statistics, 8E

10.74 E (θ x ) 

 θ (1  σ 2 )  x α where α  θ 0  0 2 0  1 α βn σ0  

 θ (1  θ )  β  (1  θ 0 )  0 2 0  1  σ0 

E (θ x ) 

α β 

θ 0 (1  θ 0 ) 1 σ 02

x n α   n α βn α βn

 θ (1  θ )  θ 0  0 2 0  1 σ0 x n      (1 ) (1 ) θ θ θ θ   n 0 0 0 0  1 n 1 n σ o2 σ 02 

x n  w  θ 0 (1  w) where w  θ (1  θ 0 ) n n 0 1

σ 02

40  40 1  2 α β 80  81 324 1 σ 18 1 Distribution is symmetrical about x  2 The function as well as its derivatives are 0 at x  0 and 1, and with k = 3 in Chebyshev's Theorem 8 1 1 1 2 of area under curve falls between   and 9 2 6 3 3

10.75 μ 

α

10.76 μ1  x  w



σ2 

nσ 02 σ2 μ    xw  μ0 (1  w) 0 nσ 02  σ 2 nσ 02  σ 2

nσ 02  nσ 02  σ 2

10.77 f ( x λ )  (a)

40 1  40  40 2

n n

QED

σ2 σ 02

λ x e λ x!

f ( x, λ ) 

λ x e λ x!



1 λ α 1e  λ / β β Γ (α ) α



xα 1e  β λ x e λ d λ g( x)  x!β α Γ (a ) 0





gamma distribution with α  x  1 and β  1

xα 1e  β xα 1e β x ! xα 1e  β ( 1)     x Γ x ! β α Γ (α ) x ! β α Γ (α ) β α Γ (α )

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Chapter 10

157

λ x α 1e  λ [1(1/ β )] x ! β α Γ (α )

f ( x, λ ) 



1 g( x)  λ x α 1e  λ ( β 1)/ β α x ! β Γ (α ) 0



gamma distribution with x  α and

β β 1

x α

 β   β  1  Γ ( x  α ) g ( x)  x !β α Γ (α )

φ (λ x) 



λ x α 1e  λ ( β 1)/β x ! β α Γ (α )  x α x ! β α Γ (α )  β   β  1  Γ ( x  α ) 1

 β   β  1 

.λ x α 1e  λ ( β 1)/ β

x α

Γ( x  α )

gamma distribution with parameters x  α and E (Λ x) 

(b)

25 50 (27.6)  (38.1)  34.6 75 75

10.79

9 4 (26.0)  (32.5)  28 13 13

10.80

4  210  1  279 3 nx 2 ( x  x )2

βˆ 

( m1 )2 m2  ( m1 )2

βˆ 



or αˆ 

β 1

( x  α )β from Theorem 6.3 β 1

10.78

10.81 αˆ 

β

(x  x)

2

nx m2  (m1) 2 m1

 x  86.4 and  x  756.52 2

86.4 756.52  7.2 and m2   63.0433 12 12 (7.2)2 51.84 αˆ    4.627 2 63.0433  51.84 63.0433  (7.2) m1 

βˆ 

63.0433  (7.2) 2  1.556 7.2

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158

Mathematical Statistics, 8E

10.82 θˆ  m1

 x  201,000

θˆ 

201,000  40,200 miles 5

 3  N  3  1   3  10.83 The likelihoods are N   4  N Likelihood  3  6   1   3  3  20 9   0.4762 126 9  4 

10

 3  7   1   3  3  35   0.5000 210 10   4 

11

 3  8   1   3 3  56   0.5091 330 11  4 

N

12

Likelihood  3  9   1   3  3  84   0.5091 495 12   4 

13

 3 10   1   3  3  120   0.5035 715 13  4 

14

 3 11  1   3  3  165   0.4945 1001 14   4 

Likelihood greatest for N  11 or N  12 10.84 θˆ  3m1 10.85

 x  5524,

 x  0.39

m1 

0.39  0.065 6

 x  2,570,176

n = 12

2

θˆ  3 

0.39  0.195 6

m1  460.3333 m2  214,181.3333 θˆ  214,181.3333  211,906.7471  47.69

δˆ  460.3333  47.69  412.64 10.86 δˆ  y1  403 10.87 n = 8

θˆ  460.33  403  57.33

 x  63.1  x  541.55 2

63.1  7.8875 8 541.55 m2   67.69375 8

m1 

αˆ  7.8875  3(67.69375  62.2126)  7.8875  4.0550  3.83 ˆ β  7.8875  4.0550  11.9427  11.95

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Chapter 10

159

10.88 αˆ  4.1 and βˆ  11.5 αˆ  y βˆ  y 1

10.89 αˆ 



n

log10 x  4.37840 n  15 4.33244 4.42160 4.39445 4.52634 4.38917 4.46538 4.55871 4.35025 4.33244 4.45179 4.42813 4.49693 4.35603 4.36361 66.24567

n n (0.4343)  ln xi log10 x



15(0.4343) αˆ   0.098 66.24567

n0  11

10.90 n = 3 N = 20 7  2  2  3  0 11  θˆ  3  20 60

n1  7

n2  2

10.91 1, 3, 5, 1, 2, 1, 3, 7, 2, 4, 4, 8, 1, 3, 6, 5, 2, 1, 6, 2

 x  67 10.92

θˆ 

 v  107.4  w  674

20  0.30 67

 v  116,108  w  76,246 2

n1  10

2

n2  6

1074 674  107.4 μˆ 2   112.3 10 6 116,108  115,347.6  76,246  75,712.7 1,293.7   80.86 σˆ 2  16 16

μˆ 1 

10.93 n = 100

θ 0  0.20

σ 0  0.04

x  38

38 w  0.20(1  w) 100 100 100 w   0.5025 (0.2)(0.8) 99  100 99  (0.04)2 E (θ 38) 

E (θ 38)  0.38(0.5025)  0.20(0.4975)  0.29

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n3  0

160

Mathematical Statistics, 8E

10.94 θ 0  0.74 (a) θˆ  0.74 (b) (c)

σ 0  0.03

n  30

x  18

x 18   0.60 n 30 30 30 30 w    0.1236 (0.74)(0.26) 29  213.8 242.8 29  (0.03)2 θˆ  (0.1236)(0.60)  (0.8764)(0.74)  0.72

θˆn 

712  715  0.32 9.5 725  715 z  1.05 9.5 p  0.1255  0.3531  0.4786

10.95 μ1  715

63  65.2  1.47 1.5 68  65.2 z  1.87 1.5 p  0.4292  0.4693  0.8985 40 40 w   0.62 μ1  (0.62)72.9  (0.38)65.2 2 64.34 7.4 40  2  69.97 1.5 1 40 1 σ12  0.92   2  0.730  0.444  1.174 2 2 σ1 7.4 1.5 63  70 z  7.6 0.92 68  70 z  2.18 0.92 p  0.5000  0.4854  0.0146

10.96 μ0  65.2

(a) (b)

10.97 (a) (b) (c)

z

σ1  9.5

z

σ 0  1.5

μˆ  αβ  50  2  100 μˆ  x  112 μˆ  μ1 

2(50  112)  108 3 2

z 2σ 2  2.575  4.2  10.98 n  2     467.9. Rounding up to the next integer, n = 468.  0.5  E

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Chapter 10

10.99 z 

161

E 6  15   9.0; yes.

1 σ/ n

10.100 The sample is more likely to include longer sections than shorter ones; They take more time to pass the inspection station.

10.101 Heads of households may tend to have somewhat different political opinions than other members of the household who are likely to be younger and/or of a different sex.

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