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©2015 Pearson Education, Inc.

Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 2 1 ____________________________________ ____________________________________ _____________________________

2.1. (a) Probability distribution function for Y Outcome (number of heads) Probability

Y=0 0.25

(b) Cumulative probability distribution function for Y Outcome (number of Y<0 0 ! Y < 1 heads) Probability 0 0.25

(c) µ Y = E (Y ) = (0 ! 0.25) + (1! 0.5 0.50) + (2! 0.2 0.25) = 1.00 Using Key Concept 2.3: var(Y )

=

E (Y

2

) ! [ E (Y )] 2 ,

and 2

E (Y

) = (02 ! 0.25) + (12 ! 0.50) + (22 ! 0.25) = 1.50

so that

var(Y )

=

E(Y

2

) ! [ E(Y )] 2 1.50 ! (1.00) 2 =

=

0.50.


Y = = 1 0.50

Y = = 2 0.25

1 ! Y < 2

Y " 2

0.75

1.0


2.2. We know from Table 2.2 that Pr (Y = 0) = 0.22, Pr (Y = 1) = 0.78, Pr ( X = 0) = 0.30, Pr ( X = 1) = 0.70. So (a) µ Y

( ) = 0 ! Pr (Y

= E Y =

µ X

=

0 ! 0.22 + 1! 0.78 = 0. 78, ( ) = 0 ! Pr ( X

= E X =

0) + 1! Pr (Y = 1)

=

0) + 1! Pr ( X = 1)

0 ! 0.30 + 1! 0.70 = 0. 70.

(b) 2 !

X

= E[( X " µ X ) = (0 " 0.70) = ( "0 .70)

2

2

#

2

#

] Pr ( X

0) + (1" 0.70)2 # Pr (X = 1)

=

0. 30 + 0. 302 # 0. 70 = 0. 21,

2 ! = E

[(Y " µ Y )2 ]

Y

=

(0 " 0.78)2 # Pr (Y

=

0) + (1" 0.78)2 # Pr (Y = 1)

( 0 78)2 # 0. 22 + 0. 222 # 0. 78 = 0. 1716.

= " .

(c) ! XY

=

cov ( X , Y ) = E[( X

=

(0 " 0.70)(0 " 0.78) Pr( X

"

µ X )(Y " µ Y )] = 0, Y = 0)

+ (0 " 0.70)(1 " 0 .78) Pr ( X = 0 , Y = 1) + (1 " 0 .70)(0 " 0 .78) Pr ( X = 1, Y = 0) + (1 " 0.70)(1 " 0 .78) Pr ( X = 1, Y = 1) =

( "0.70) # ( "0.78) # 0 .15 + ( "0 .70) # 0 .22 # 0 .15 + 0.30 # ( "0 .78) # 0 .07 + 0 .30 # 0.22 # 0.63

=

0.084,

corr ( X , Y ) =

! XY ! X ! Y

=

0.084 0.21# 0.1716


= 0 .4425 .


2.3. For the two new random variables W

=

3 + 6 X

and

V

=

20 ! 7Y , we have:

(a) E (V ) = E (20 ! 7Y ) =

20 ! 7E (Y ) = 20 ! 7 " 0. 78 = 14. 54,

E (W ) = E (3 + 6 X ) = 3 + 6 E ( X ) = 3 + 6 " 0.70 =

7.2.

(b) !

2

W

!

2

V

=

2 = 36 " 0.21 = 7.56, var(3 + 6 X ) = 62! X

=

var(20 # 7Y ) = (#7) 2 $ ! Y 2 = 49 " 0.1716 = 8.4084.

(c) !

WV

=

cov(3 + 6 X , 20 " 7Y ) = 6("7)cov( X , Y ) = "42 # 0.084 = "3.52

corr(W , V ) =

3.528

!

WV

! ! W

V

" =

=

7.56 # 8.4084


0.4425.

"


2.4. (a) E ( X 3 ) = 03 ! (1 " p) + 13 ! p = p (b) E ( X k ) = 0k ! (1 " p) +1k ! p = p (c)

E ( X )

=

var ( X ) Thus,

0.3 =

E( X

! =

2

) ![ E ( X )] 2

0.21

=

=

0.3 !0.09

=

0.21

0.46.

To compute the skewness, use the formula from exercise 2.21: E ( X

Alternatively,

)3

! µ

E ( X ! µ )

Thus, skewness

=

3

3

) ! 3[ E( X 2 )][ E( X )] + 2[ E( X )]3

=

E( X

=

0.3 ! 3 " 0.32 + 2 " 0.33

=

0.084

= [(1 ! 0.3)3 " 0.3] + [(0 ! 0.3)3 " 0.7] = 0.084

E ( X " µ )

3

/! 3

=

.084/0.463

=

0.87.

To compute the kurtosis, use the formula from exercise 2.21: E ( X

)4

! µ

Alternatively,

) ! 4[ E( X )][ E( X 3)] + 6[ E( X)] 2[ E( X 2)] !3[ E( X )] 4

E( X

=

0.3 ! 4 " 0.32 + 6 " 0.33 ! 3 " 0.34

E ( X ! µ )

Thus, kurtosis is

4

=

4

=

0.0777

= [(1 ! 0.3) 4 " 0.3] + [(0 ! 0.3) 4 " 0.7] = 0.0777

E ( X " µ )

4

/! 4 = .0777/0.464

=

1.76



2.5.

Let X denote temperature in F and Y denote temperature in C. Recall that Y = 0 °

°

when X = 32 and Y =100 when X = 212. This implies

Y = (100/180) ! ( X

"

32) or Y

17.78 + (5/9) ! X.

="

o

Using Key Concept 2.3, µ X = 70 F implies that µ Y o

and ! X = 7 F implies

!

Y

=

(5/9) " 7 = 3.89°C.


= !17.78 + (5/9) " 70 = 21.11°C,


2.6.The table shows that Pr ( X = 0, Y = 0) = 0.053, Pr ( X = 0, Y = 1) = 0.586, Pr ( X = 1, Y = 0) = 0.015, Pr( X = 1, Y = 1) = 0.346, Pr ( X = 0) = 0.639, Pr ( X = 1) = 0.361, Pr (Y = 0) = 0.068, Pr (Y = 1) = 0.932. (a) E (Y ) = µ Y = =

0 ! Pr(Y = 0) + 1 ! Pr(Y = 1)

0 ! 0.068 + 1 ! 0.932 = 0.932.

(b)

Unemployment Rate = =

# (unemployed) # (labor

force)

Pr(Y = 0) = 1 ! Pr(Y = 1) = 1 ! E (Y ) = 1 ! 0.932 = 0.068.

(c) Calculate the conditional probabilities first:

Pr (Y = 0| X = 0) =

Pr (Y = 1| X = 0) =

Pr (Y = 0| X = 1) =

Pr (Y = 1| X = 1) =

Pr ( X = 0, Y = 0)

0.053 =

Pr ( X = 0) Pr ( X = 0, Y = 1)

0.586 =

Pr ( X = 0) Pr ( X = 1, Y = 0)

Pr( X = 1)

0.917,

=

0.639 0.015 =

Pr ( X = 1) Pr ( X = 1, Y = 1)

0.083,

=

0.639

0.042,

=

0.361 0.346 =

0.958.

=

0.361

The conditional expectations are E (Y |X = 1) = =

0 ! 0.042 + 1 ! 0.958 = 0.958,

E (Y |X = 0) = =

0 ! Pr (Y = 0| X = 1) + 1 ! Pr (Y = 1| X = 1)

0 ! Pr (Y = 0| X = 0) + 1 ! Pr (Y = 1| X = 0) 0 ! 0.083 + 1 ! 0.917 = 0.917.



(d) Use the solution to part (b), Unemployment rate for college graduates = 1 – E (Y |X =1) = 1!0.958 = 0.042 Unemployment rate for non-college graduates = 1 – E (Y |X =0) = 1!0.917 = 0.083

(e) The probability that a randomly selected worker who is reported being unemployed is a college graduate is

Pr ( X = 1|Y = 0) =

Pr ( X = 1, Y = 0)

0.015 =

Pr (Y = 0)

0.221.

=

0.068

The probability that this worker is a non-college graduate is Pr ( X = 0|Y = 0) = 1 ! Pr ( X = 1|Y = 0) = 1 ! 0.221 = 0.778.

(f) Educational achievement and employment status are not independent because they do not satisfy that, for all values of x and y,

Pr ( X

=

For example, from part (e) Pr ( X Pr( X = 0) = 0.639.

x|Y = y) = Pr ( X =

=

x).

0|Y 0) = 0.778, while from the table =



2.7. Using obvious notation, 2 2 2 ! C = ! M + ! F +

(a) µ C

=

2 cov (M ,

C

=

M

+

F ). This

F ; thus µC

=

+

µM

µ F and

implies

40 + 45 = $85,000 per year.

(b) corr( M , F )

=

cov( M , F ) !

!

M F

, so that cov ( M,

F)

=

!

M

!

corr ( M, F ). Thus

F

cov( M , F ) 12 !18 ! 0.80 172.80, where the units are squared thousands of =

=

dollars per year.

(c)

2 2 2 ! C = ! M + ! F +

! C

=

813.60

2 cov (M , F ), so that

=

28.524 thousand

2

! C =

12

2

2

+ 18 +

2 " 172.80 = 813.60, and

dollars per year.

(d) First you need to look up the current Euro/dollar exchange rate in the Wall Street Journal, the Federal Reserve web page, or other financial data outlet. Suppose that this exchange rate is e (say e = 0.75 Euros per Dollar or 1/e = 1.33 Dollars per Euro); each 1 Eollar is therefore with e Euros. The mean is therefore e! µ C (in units of thousands of euros per year), and the standard deviation is e!! C (in units of thousands of euros per year). The correlation is unit-free, and is unchanged.



2.8. µ Y

=

E (Y )

=

1,

2 ! Y

=

var (Y )

µ Z

=

2 ! Z

=

=

4. With

Z

=

1 2

(Y ! 1),

" 1 (Y $ 1) # 1 ( µ $1) 1 (1 $1) %2 & 2 2 ' ( "1 # 1 ! 2 1 ) 4 1 var % (Y $ 1) & 4 '2 ( 4

E

=

=

Y

=

Y

=


=

.

=

0,


H*I* Value of Y

Value of X

1 5 8 Probability distribution of Y

14 0.02 0.17 0.02 0.21

22 0.05 0.15 0.03 0.23

30 0.10 0.05 0.15 0.30

40 0.03 0.02 0.10 0.15

Probability Distribution 65 of X 0.01 0.21 0.01 0.40 0.09 0.39 0.11 1.00

(a) The probability distribution is given in the table above. E (Y ) = 14 ! 0.21 + 22 ! 0.23 + 30 ! 0.30 + 40 ! 0.15 + 65 ! 0.11 = 30.15 2

E (Y

) = 142 ! 0.21 + 222 ! 0.23 + 302 ! 0.30 + 402 ! 0.15 + 652 ! 0.11 = 1127.23

var(Y ) = E (Y 2 ) " [ E (Y )]2 = 218.21 #

=

Y

14.77

J"K LA' .-017,7-0%< ?3-"%"7<7,# -D ! M " N O 7( F7@'0 70 ,A' ,%"<' "'<-; Value of Y 14 22 30 40 0.02/0.39 0.03/0.39 0.15/0.39 0.10/0.39 E (Y |X = 8) = 14 ! (0.02/0.39) + 22 ! (0.03/0.39) + 30 ! (0.15/0.39) +

E (Y

40 ! (0.10/0.39) + 65! (0.09/0.39) = 39.21

2

| X = 8) = 142 ! (0.02/0.39) + 22 2 ! (0.03/0.39) + 30 2 !(0.15/0.39) +

402 ! (0.10/0.39) + 652 ! (0.09/0.39) = 1778.7

var(Y ) = 1778.7 " 39.212 = 241.65 ! Y | X =8

= 15.54

J.K E ( XY ) = (1!14 ! 0.02) + (1! 22: 0.05) + !

cov( X , Y ) corr( X, Y )

=

=

(8 ! 65 ! 0.09) =171.7 E ( XY ) ! E ( X ) E (Y ) 171.7 ! 5.33" 30.15 11.0 =

cov( X, Y )/(! X ! Y )

=

11.0 / (2.60 "14.77)


=

=

0.286

65 0.09/0.39


2.10. Using the fact that if

Y

!

Y " µ Y 2 # " N $ µ Y , ! Y % then & ' !

~ N (0,1) and Appendix Table 1,

Y

we have

" Y ! 1 $ 3 ! 1 # % (1) ' 2 2

(a) Pr (Y $ 3) = Pr &

=

=

0.8413.

(b)

Pr(Y

>

0) = 1 ! Pr(Y " 0)

=

1 ! Pr &

# Y ! 3 " 0 ! 3 $ = 1 ! %( !1) = %(1) = 0 .8413 . ' 3 ) ( 3

(c)

" 40 ! 50 $ Y ! 50 $ 52 ! 50 # & 5 5 ( ' 5 = ) (0.4) ! ) ( !2) = ) (0.4) ! [1 ! ) (2)] = 0.6554 ! 1 + 0.9772 = 0.6326 .

Pr (40 $ Y $ 52) = Pr %

(d)

" 6 ! 5 $ Y ! 5 $ 8 ! 5 # & 2 2 ( ' 2 ) (2 1213) ! ) (0 7071) 0 9831! 0 7602 0 2229

Pr (6 $ Y $ 8) = Pr % =

=

.

.

.

.


=

.

.


2.11. (a) 0.90 (b) 0.05 (c) 0.05 (d) When

Y

(e)

2

Y

=

Z

2

~ ! 10 , then

Y /10

~

F 10,

!

.

, where Z ~ N(0,1), thus Pr (Y

! 1)

=

Pr ("1 ! Z ! 1)


=

0.32.


2.12. (a) 0.05 (b) 0.950 (c) 0.953 (d) The t df distribution and N(0, 1) are approximately the same when df is large. (e) 0.10 (f) 0.01



2.13. (a) E (Y 2 ) = Var (Y ) + µY2

=

1 + 0 = 1; E(W 2 ) = Var (W ) + µ W 2 = 100 + 0 = 100.

(b) Y and W are symmetric around 0, thus skewness is equal to 0; because their mean is zero, this means that the third moment is zero.

(c) The kurtosis of the normal is 3, so 3

E (Y " µ Y )

=

4

4 ! Y

; solving yields E(Y 4 )

=

3; a

similar calculation yields the results for W .

(d) First, condition on E ( S |X

=

0)

=

X

=

0; E (S 2|X

0, so that =

S

=

W :

0) 100, E ( S 3| X =

=

0)

=

0, E ( S 4|X

=

0) 3 !100 2. =

Similarly, E (S |X

=

1) 0; E (S 2| X =

=

1) 1, E (S 3|X =

=

1) 0, E (S 4|X 1) 3. =

=

=

From the large of iterated expectations E (S )

=

E (S |X

=

0) ! Pr (X = 0) + E (S |X

=

1) ! Pr( X = 1) = 0

E ( S

2

) = E ( S 2 |X

=

0) ! Pr (X = 0) + E( S 2| X

=

1) ! Pr( X = 1) =100 ! 0.01 +1 ! 0.99 =1.99

E (S

3

) = E (S 3 |X

=

0) ! Pr (X = 0) + E (S 3 |X

=

1) ! Pr( X = 1) = 0

E ( S

4

) = E (S 4 |X

=

0) ! Pr (X = 0) + E (S 4 |X

=

1) ! Pr( X = 1)

=

(e) µ S

=

3 !1002 ! 0.01 + 3!1! 0.99 = 302.97

E (S )

Similarly,

=

0, thus E ( S ! µ S )3

2 ! S

=

E ( S

" µ S ) 2

=

3

E ( S

2

E (S

Thus, kurtosis 302.97 / (1.992 ) =

=

=

)

=

0 from part (d). Thus skewness = 0.

) 1.99, and E (S ! µ S )4 =

76.5


=

4

E (S

)

=

302.97.


2.14. The central limit theorem suggests that when the sample size (n) is large, the distribution of the sample average (Y ) is approximately Given µ Y (a)

n

=

2

100,

=

100,

2

!

Y

! Y

=

2

=

! Y

n

=

43 = 100

=

165,

Pr (Y

>

2

!

Y

2

=

! Y

n

=

43 = 165

0.43

$

101# 100 " 0.43

( % & (1 525) .

n

=

64,

! '

98) = 1 # Pr (Y $ 98) = 1 # Pr %

2

!

Y

2

=

! Y

64

=

2 ! Y

n

.

43 = 64

=

0. 9364.

0.2606, and Y

# 100

0.2606

$

" & 0.2606 (

98 # 100

) 1 # * (#3.9178) = * (3.9178) = 1.000

(c)

Y

=

0.43, and

! Y # 100

n

2

!

43.0,

Pr (Y $ 101) = Pr '

(b)

2 # " N $ µ Y , ! Y % with & '

(rounded to four decimal places).

0.6719, and

! 101 # 100 Y #100 103 #100 " $ $ & ' 0 6719 0 6719 0 6719 ( ) * (3 6599) # * (1 2200) 0 9999 # 0 8888

Pr (101 $ Y $ 103) = Pr %

.

.

.

.


.

=

.

.

=

0.1111.


2.15. (a) Pr (9.6 ! Y ! 10.4)

=

=

" 9.6 $ 10 Y $ 10 10.4 $ 10 # ! ! & 4/n 4/n ( ' 4/n " 9.6 $ 10 ! ! 10.4 $ 10 # Pr % Z & 4/n ( ' 4/n

Pr %

where Z ~ N (0, 1). Thus,

# 9.6 "10

(i) n = 20; Pr %

4/n

! Z !

10.4 "10 $

&

=

4/n

Pr ("0.89 ! Z ! 0.89)

10.4 " 10 $ # 9.6 " 10 ! Z ! & 4/n ( ' 4/n

(ii) n = 100; Pr %

=

# 9.6 " 10 ! 10.4 "10 $ ! Z & 4/ n 4/n ( '

(iii)n = 1000; Pr %

=

Pr("2.00 ! Z ! 2.00)

=

0.63

=

0.954

Pr("6.32 ! Z ! 6.32) 1.000 =

(b) Pr (10 $ c ! Y

As n get large

c

4/

" $c Y $ 10 c # ! 10 + c) = Pr % ! ! & 4/n 4/n ( ' 4/n " $c ! ! c # . = Pr % 4/n Z 4/n & ' (

gets large, and the probability converges to 1. n

(c) This follows from (b) and the definition of convergence in probability given in Key Concept 2.6.



2.16. There are several ways to do this. Here is one way. Generate n draws of Y , Y 1, Y 2, … Y n. Let X i = 1 if Y i < 3.6, otherwise set X i = 0. Notice that X i is a Bernoulli random variables with µ X = Pr( X = 1) = Pr(Y < 3.6). Compute probability to µ X = Pr( X = 1) = Pr(Y < 3.6),

X will

X . Because X converges

in

be an accurate approximation if n

is large.



2.17. µ Y = 0.4 and (a) (i) P ( Y

2

! Y

=

" 0.43)

(ii) P ( Y

0.4 " 0.6

!

= Pr %

! 0.37)

Y

=

0.24

# 0.4

0.24/n

!

= Pr %

Y

$

# 0.4

0.24/n

# 0.4 " & 0.24/n

0.43

$

=

# 0.4 " & 0.24/ n

Pr

0.37

=

! %

Pr

Y

# 0.4

0.24/n

! %

Y

" $ 0.6124&

# 0.4

0.24/ n

" $ #1.22 &

=

=

0.27

0.11

b) We know Pr(!1.96 ! Z ! 1.96) = 0.95, thus we want n to satisfy 0.41 =

0.41! 0.4 0.24/n

>

!1.96 and

0.39 !0.4 0.24/n

<

!1.96. Solving these inequalities yields n "

9220.



2.18. Pr (Y

=

$0) = 0.95, Pr (Y

(a) The mean of

=

$20000)

0.05.

Y is

µ Y

The variance of

=

0 ! Pr (Y

(Y ( µ ) $&

Y

=

$0) + 20, 000 ! Pr (Y

=

$20000) = $1000.

Y is

" 2 ! = E $

2#

% %'

Y

=

(0 ( 1000)2 ) Pr (Y

=

((1000)2 ) 0.95 + 190002 ) 0. 05 = 1. 9) 107 ,

so the standard deviation of Y is

(b) (i) E (Y ) µ Y =

=

=

$1000,

2

!

Y

=

2 ! Y

n

=

0 ) + (20000 ( 1000)2 ) Pr (Y = 20000)

1

! Y

=

(1.9 "107 ) 2

7

=

1. 9"10 100

=

$4359.

5

=

1.9 "10 .

(ii) Using the central limit theorem,

Pr (Y

>

2000) = 1 ! Pr (Y " 2000)

# Y ! 1000 2, 000 ! 1, 000 $ ! Pr % " & 5 1.9 ' 105 ) ( 1.9 '10 * 1 ! + (2.2942) = 1 ! 0.9891 = 0.0109. =1



2.19. (a) l

Pr (Y

y j )

=

=

! Pr ( X

=

xi , Y

=

yj)

i 1 =

l

=

! Pr (Y

=

y j | X xi )Pr ( X xi ) =

=

i 1 =

(b) k

E (Y ) =

k

' y Pr (Y

=

j

' y ' Pr (Y

yj ) =

l

! # # ## %

=

y j |X

y j |X

=

xi ) Pr ( X

=

i=1

" $ $ $ &

k

'' i =1

j =1

y j Pr (Y

=

j

j =1

=

l

=

xi ) $ Pr ( X xi )

j =1

=

l

=

' E (Y | X

=

xi )Pr ( X = xi ).

i =1

(c) When

X and Y are

independent,

Pr ( X

=

xi , Y

=

y j ) = Pr (X

=

xi )Pr (Y

=

xi , Y y j )

=

y j ),

so ! XY

=

E[( X l

" µ X )(Y " µ Y )]

k

** ( x "

=

"

µ X )( y j µ Y ) Pr ( X

i

=

i 1 j 1 =

=

l

k

**( x "

=

i

"

µ X )( y j µ Y ) Pr ( X

=

xi ) Pr (Y y j ) =

i 1 j 1 =

=

=

# l ( x " µ ) Pr ( X x ) $ # k ( y " µ ) Pr (Y i & % * j Y %* i X 'i1 ( ' j 1 E ( X " µ X ) E (Y " µ Y ) 0 ) 0 0, =

=

=

=

=

cor (X , Y )

=

! XY ! X ! Y

=

0 =

=

! X ! Y


0.

=

yj

$ & (

xi )


l

2.20. (a) Pr (Y

=

yi )

=

m

!! Pr (Y

=

yi| X

xj , Z

=

=

zh ) Pr ( X

=

xj , Z

=

z h )

j 1 h 1 =

=

(b) k

E (Y )

=

' y Pr (Y i

=

yi ) Pr (Y

yi )

=

i 1 =

k

=

l

' '' Pr (Y i 1

=

=

l

yi | X

=

xj , Z

=

zh ) Pr ( X

=

xj , Z

=

z h )

=

! k '' # ' yi Pr (Y j 1 h 1 % i 1 l

=

=

j 1 h 1

=

=

m

yi

"

m

=

=

yi | X

=

xj , Z

=

zh ) $ Pr ( X

&

=

=

xj , Z

=

z h )

m

'' E (Y| X

=

x j , Z

=

zh ) Pr ( X

=

xj , Z

=

z h )

j 1 h 1 =

=

where the first line in the definition of the mean, the second uses (a), the third is a rearrangement, and the final line uses the definition of the conditional expectation.



2. 21. (a) E ( X

! µ )3 = E[( X ! µ )2 ( X ! µ )] = E [X 3 ! 2X 2 µ + X µ 2 ! X 2 µ + 2X µ2 ! µ 3 ] = E

( X 3 ) ! 3E ( X 2 ) µ + 3E (X )µ 2 ! µ 3

= E

( X 3 ) ! 3E ( X 2 )E ( X ) + 2E (X )3

=

E (X

3

) ! 3E (X 2 )E (X ) + 3E (X )[E (X )]2 ! [E (X

(b) E ( X

! µ )4

=

E[( X

3

! 3 X 2 µ + 3 X µ 2 ! µ 3 )( X ! µ )]

[ X 4 ! 3 X 3 µ + 3 X 2 µ 2 ! X µ 3 ! X 3µ + 3 X 2µ 2 ! 3 X µ 3 + µ 4]

= E = E

( X 4 ) ! 4 E( X 3 ) E( X ) + 6 E( X 2) E( X ) 2 ! 4 E( X ) E( X ) 3 + E( X ) 4

= E

( X 4 ) ! 4[ E( X )][ E( X 3)] + 6[ E( X)] 2[ E( X 2)] ! 3[ E( X )] 4



2. 22. The mean and variance of R are given by µ =

w"

0.08 + (1 # w) " 0.05

2 2 ! = w "

0.07 2 + (1 # w) 2 " 0.042 + 2 " w " (1 # w) " [0.07 " 0.04 " 0.25]

where 0.07 ! 0.04 ! 0.25 Cov ( R s , Rb ) follows from the definition of the correlation =

between R s and Rb.

(a) µ 0.065; ! =

=

(b) µ 0.0725; ! =

0.044

=

0.056

(c) w = 1 maximizes µ ; !

=

0.07 for this value of w.

(d) The derivative of ! 2 with respect to w is 2

d !

dw

= =

2 w " .072 # 2(1 # w) " 0.042 + (2 # 4w) " [0.07 " 0.04 " 0.25] 0.0102w # 0.0018

Solving for w yields w 18/ 102 0.18. (Notice that the second derivative is positive, so that this is the global minimum.) With w 0.18, ! R .038. =

=

=


=


2. 23. X and Z are two independently distributed standard normal random variables, so µ X

=

µ Z

2

0, ! X

=

=

2

!Z

=

1, ! XZ

(a) Because of the independence between X and and E ( Z |X ) E ( Z ) 0. Thus =

2

) = ! X2

E ( X

(c)

E ( XY ) = E ( X

0.

Pr ( Z

=

z|X

=

+

2

| ) = E( X 2| X ) + E( Z| X ) =

+Z X

=

Pr ( Z

=

z ),

2

µ X

3

=

1, and µY

+ ZX

(

= E X

2

+Z

2

2

X + 0 = X .

) = E ( X 2 ) + µ Z

=

1 + 0 = 1.

) = E( X 3 ) + E ( ZX ). Using the fact that the odd moments of

a standard normal random variable are all zero, we have E ( X 3 ) independence between X and E ( XY ) = E( X

x)

=

E (Y| X ) = E( X

(b)

Z ,

=

3

Z , we

have

E ( ZX )

=

µ Z µ X

=

0. Using the 0. Thus =

) + E( ZX ) = 0.

(d) cov ( XY ) = E[( X " µ X )(Y " µ Y )] = E[( X " 0)(Y "1)] = E =

corr ( X , Y ) =

( XY " X ) = E ( XY ) " E ( X )

0 " 0 = 0. 0

!

XY

!

X

!

Y

=

=

!

X

0.

! Y



2.24. (a) E (Y i 2 ) = ! 2 + µ 2

2 = !

and the result follows directly.

(b) (Y i/! ) is distributed i.i.d. N (0,1),

W

"

=

n i

1

=

(Y i /! ) 2 , and the result follows from the

definition of a ! 2 random variable. n

n

(c)

E (W )

=

E

Yi

2

n

2

Y i

" ! 2 " E ! 2 i

=

1

i

=

=

n.

1

=

(d) Write V

=

Y1 n

"i

=

2 Yi

2

=

Y 1 /! n

"i

2 (Y /! )

2

=

n #1

n #1

which follows from dividing the numerator and den ominator by ! . Y 1/! ~ N (0,1),

"

n i

=

2

(Y i /! ) 2 ~ ! "1 , and Y 1/! and 2 n

"

n i

=

2

(Y i /! ) 2 are independent. The result then

follows from the definition of the t distribution.



n

n

! ax

2.25. (a)

i

(

= ax1 + ax2 + ax3 + ! + axn

) = a ( x1 + x2 + x3 + !

+ xn

) = a ! xi

i =1

i =1

(b) n

! ( x

i +

yi ) = ( x1 + y1 + x2 + y2 + ! xn + yn )

i =1

=

( x1 + x2 + ! xn ) + ( y1 + y2 + ! yn ) n

=

! xi

n

+

i =1

!y

i

i =1

n

(c)

! a = (a + a + a + !

+a

) = na

i =1

(d) n

! (a + bxi

n

+

cyi ) 2

=

i =1

! (a

2

+

b2 xi2 + c2 yi2 + 2 abxi + 2 acyi + 2 bcxi yi )

i =1 n

=

na

2

+

b

2

!x

2 i +

i =1

n

c

2

!y

2 i +

n

n

n

i=1

i=1

i=1

2 ab! xi + 2 ac! yi + 2bc! xi yi

i =1



cov(Yi , Y j )

2.26. (a) corr(Y i,Y j) =

"Y "Y i

cov(Yi , Y j )

=

cov(Yi , Y j )

=

2

" Y" Y

j

"Y

! , where the first equality

=

uses the definition of correlation, the second uses the fact that Y i and Y j have the same variance (and standard deviation), the third equality uses the definition of standard deviation, and the fourth uses the correlation given in the problem. Solving for cov(Y i, Y j) from the last equality gives the desired result.

(b)

Y =

1 2

1

Y1 +

2

1

var( Y ) =

(c)

Y

=

1 n

4

Y 2 ,

so that E ( Y ) =

var(Y1 ) +

1

1 2

var(Y2 ) +

4

n

! Y , so that E (Y )

1

=

i

n

i 1 =

E (Y )1 +

2 4

E (Y 2 ) = µ Y

2

2

!Y

cov(Y1 , Y2 ) =

n

! E (Y ) i

i

1

=

1

=

1 n

2

2

+

"! Y

2

n

! µ

Y

i

=

µ Y

1

=

1 n % $ var(Y ) = var & * Y i ' ( n i 1 ) =

=

=

n

1 n

2

i

n

2

*

! Y

n

+

2

=

2 !Y +

i =1

2

=

+

i =1

1 n

* var(Y )

! Y

n

+

2

n(n # 1) n

2

n

2

n #1

n

* * cov(Y ,Y ) i

j

i =1 j = i+1

n #1

2 n

2

n

* * "!

2 Y

i =1 j = i+1

2

"! Y

$1 # 1 % "! 2 & n ' Y ( ) n !1

where the fourth line uses

n

" " a = a(1 + 2 + 3 + ! i =1 j =i +1

+

n !1) =

an(n ! 1)

2

variable a. 2

(d) When n is large

! Y

n

"

0 and

1 !

0,

and the result follows from (c).

n


for any


2.27

! )| Z ] = E [ E ( X |Z ) – E ( X |Z ) ] = 0. (a) E (W ) = E [ E (W |Z ) ] = E [ E ( X " X

(b) E (WZ ) = E [ E (WZ |Z ) ] = E [ ZE (W ) |Z ] = E [ Z !0] = 0

2

2

2

(c) Using the hint: V = W – h( Z ), so that E (V ) = E (W ) + E [h( Z ) ] – 2" E [W !h( Z )]. 2

2

Using an argument like that in (b), E [W !h( Z )] = 0. Thus, E (V ) = E (W ) + 2

2

E [h( Z ) ], and the result follows by recognizing that E [h( Z ) ] # 0 because h( z ) for any value of z .


2

# 0

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