Test Bank - for Introduction to Econometrics 3rd Edition by H STOCK JAMES & W. WATSON MARK, All Chapters 1-18 |Complete Guide A+

TEST BANK FOR
Introduction to Econometrics 3rd Edition by H STOCK JAMES & W. WATSON
MARK




FULL TEST BANK!!!

,Chapter 2
Review of Probability

2.1. (a) Probability distribution function for Y
Outcome (number of heads) Y=0 Y=1 Y=2
Probability 0.25 0.50 0.25

(b) Cumulative probability distribution function for Y

Outcome (number of heads) Y0 0Y1 1Y2 Y2
Probability 0 0.25 0.75 1.0

(c) Y = E(Y ) = (0  0.25) + (1 0.50) + (2  0.25) = 1.00 . F →
d
Fq, .
Using Key Concept 2.3: var(Y ) = E(Y 2 ) −[E(Y )]2 ,
and
(ui |Xi )
so that
var(Y ) = E(Y 2 ) −[E(Y )]2 = 1.50 − (1.00)2 = 0.50.

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) Y = E(Y ) = 0  Pr (Y = 0) + 1 Pr (Y = 1)
= 0  022 + 1 078 = 078,
X = E( X ) = 0  Pr ( X = 0) + 1 Pr ( X = 1)
= 0  030 + 1 070 = 070
(b)  2 = E[( X −  )2 ]
X X

= (0 − 0.70)  Pr ( X = 0) + (1 − 0.70)2  Pr ( X = 1)
2


= (−070)2  030 + 0302  070 = 021,
 Y2 = E[(Y −  Y )2 ]
= (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




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(c)  XY = cov (X , Y ) = E[( X − X )(Y − Y )]
= (0 − 0.70)(0 − 0.78) Pr( X = 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,
 XY 0084
corr (X , Y ) = = = 04425
 XY 021 01716

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 + 6X ) = 3 + 6E( X ) = 3 + 6  070 = 72
(b)  2 = var(3 + 6X ) = 62   2 = 36  021 = 756,
W X

V = var(20 − 7Y ) = (−7)   Y2 = 49  01716 = 84084
2 2


(c) WV = cov(3 + 6X , 20 − 7Y ) = 6  (−7) cov(X , Y ) = −42  0084 = −3528
WV −3528
corr (W , V ) = = = −04425
WV 756  84084

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 ) = 0.3 , and var(X) = E(X2)−[E(X)]2 = 0.3 −0.09 = 0.21. Thus  = 0.21 = 0.46.
var( X ) = E( X 2 ) −[E( X )]2 = 0.3 − 0.09 = 0.21  = 0.21 = 0.46. To compute the skewness, use
the formula from exercise 2.21:
E( X − )3 = E( X 3 ) − 3[E( X 2 )][E( X )] + 2[E( X )]3
= 0.3 − 3 0.32 + 2  0.33 = 0.084
Alternatively, E( X − )3 =[(1− 0.3)3  0.3] +[(0 − 0.3)3  0.7] = 0.084
Thus, skewness = E( X − )3/ 3 = 0.084/0.463 = 0.87.
To compute the kurtosis, use the formula from exercise 2.21:
E( X − )4 = E( X 4 ) − 4[E( X )][E( X 3 )] + 6[E( X )]2 [E( X 2 )] − 3[E( X )]4
= 0.3 − 4  0.32 + 6  0.33 − 3 0.34 = 0.0777
Alternatively, E( X − )4 =[(1− 0.3)4  0.3] +[(0 − 0.3)4  0.7] = 0.0777
Thus, kurtosis is E( X − )4/ 4 = 0.0777/0.464 =1.76

,4 Stock/Watson • Introduction to Econometrics, Third Edition


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. Using Key
Concept 2.3, X = 70oF implies that  Y = −17.78 + (5/9)  70 = 21.11C, and X = 7oF implies
Y = (5/9)  7 = 3.89C.

2.6. The table shows that Pr ( X = 0, Y = 0) = 0037, Pr ( X = 0, Y = 1) = 0622,
Pr ( X = 1, Y = 0) = 0009, Pr ( X = 1, Y = 1) = 0332, Pr ( X = 0) = 0659, Pr ( X = 1) = 0341,
Pr(Y = 0) = 0046, Pr (Y = 1) = 0954.
(a) E(Y ) = Y = 0  Pr(Y = 0) + 1 Pr (Y = 1)
= 0  0046 +1 0954 = 0954
#(unemployed)
(b) Unemployment Rate =
#(labor force)
= Pr (Y = 0) = 1 − Pr(Y = 1) = 1 − E(Y ) = 1 − 0954 = 0.046
(c) Calculate the conditional probabilities first:
Pr ( X = 0, Y = 0) 0037
Pr (Y = 0| X = 0) = = = 0056,
Pr ( X = 0) 0659
Pr ( X = 0, Y = 1) 0622
Pr (Y = 1| X = 0) = = = 0944,
Pr ( X = 0) 0659
Pr ( X = 1, Y = 0) 0009
Pr (Y = 0| X = 1) = = = 0026,
Pr ( X = 1) 0341
Pr ( X = 1, Y = 1) 0332
Pr (Y = 1| X = 1) = = = 0974
Pr ( X = 1) 0341
The conditional expectations are
E(Y|X = 1) = 0  Pr (Y = 0| X = 1) +1 Pr (Y = 1| X = 1)
= 0  0026 + 1 0974 = 0974,
E(Y|X = 0) = 0  Pr (Y = 0| X = 0) + 1 Pr (Y = 1|X = 0)
= 0  0056 +1 0944 = 0944
(d) Use the solution to part (b),
Unemployment rate for college graduates = 1 − E(Y|X = 1) = 1 − 0.974 = 0.026
Unemployment rate for non-college graduates = 1 − E(Y|X = 0) = 1 − 0.944 = 0.056
(e) The probability that a randomly selected worker who is reported being unemployed is a
college graduate is
Pr ( X = 1, Y = 0) 0009
Pr ( X = 1|Y = 0) = = = 0196
Pr (Y = 0) 0046

The probability that this worker is a non-college graduate is
Pr ( X = 0|Y = 0) = 1 − Pr ( X = 1|Y = 0) = 1 − 0196 = 0804




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(f) Educational achievement and employment status are not independent because they do not
satisfy that, for all values of x and y,
Pr ( X = x|Y = y) = Pr ( X = x)

For example, from part (e) Pr ( X = 0|Y = 0) = 0.804, while from the table Pr(X = 0) = 0.659.

Using obvious notation, C = M + F; thus  =  +  and  =  +  2 + 2cov(M, F). This
2 2
2.7.
C M F C M F
implies
(a) C = 40 + 45 = $85, 000 per year.
cov(M , F )
(b) corr (M, F ) = , so that cov (M, F ) =  M  F corr (M, F). Thus cov (M, F ) =
MF
12 18  0.80 = 172.80, where the units are squared thousands of dollars per year.
(c)  =  +  + 2cov(M, F), so that  =122 +182 + 2172.80 = 813.60, and
2 2 2 2

C M F C

 C = 813.60 = 28.524 thousand 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 = 0.80 Euros per dollar); each 1 dollar is therefore with e Euros. The mean is therefore
e  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.  = E(Y ) = 1,  2 = var (Y ) = 4. With Z = 1 (Y −1),
Y Y 2

1  1 1
Z = E  (Y −1) = ( Y−1) = (1 −1) = 0,
2 2 2
 
1  1 1
 Z2 = var  (Y −1)  =  Y2 =  4 = 1
2 4 4
 


2.9. Value of Y Probability
Distribution of
14 22 30 40 65 X
Value of X 1 0.02 0.05 0.10 0.03 0.01 0.21
5 0.17 0.15 0.05 0.02 0.01 0.40
8 0.02 0.03 0.15 0.10 0.09 0.39
Probability distribution of Y 0.21 0.23 0.30 0.15 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
E(Y 2 ) = 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




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,6 Stock/Watson • Introduction to Econometrics, Third Edition


(b) The conditional probability of Y|X = 8 is given in the table below

Value of Y
14 22 30 40 65
0.02/0.39 0.03/0.39 0.15/0.39 0.10/0.39 0.09/0.39

E(Y|X = 8) = 14  (0.02/0.39) + 22  (0.03/0.39) + 30  (0.15/0.39)
+ 40  (0.10/0.39) + 65  (0.09/0.39) = 39.21
E(Y 2|X = 8) = 142  (0.02/0.39) + 222  (0.03/0.39) + 302  (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

(c) E( XY ) = (114  0.02) + (1 22 : 0.05) + + (8  65  0.09) = 171.7
cov( X, Y ) = E( XY ) − E( X )E(Y ) = 171.7 − 5.33 30.15 = 11.0
corr( X, Y ) = cov( X, Y )/( X  Y ) = 11.0 / (2.60 14.77) = 0.286
Y − Y
2.10. Using the fact that if Y N 

Y 2  then
,Y ~ N (0, 1) and Appendix Table 1, we have
 
Y
 Y −1 3 −1 
(a) Pr (Y  3) = Pr   =  (1) = 08413
2 2 
 
Y −3 0−3
(b) Pr(Y  0) = 1 − Pr(Y  0) = 1 − Pr 
3 3 
 
= 1 − (−1) = (1) = 08413
 40 − 50 Y − 50 52 − 50 
(c) Pr (40  Y  52) = Pr  5  5  5
 
=(04) − (−2) =(04) −[1 − (2)]
= 06554 −1 + 09772 = 06326
(d)(d) Pr (6  Y  8) = Pr 6 − 5  Y − 5  8 − 5 
 
 2 2 2 
=(21213) − (07071)
= 09831 − 07602 = 02229

2.11. (a) 0.90
(b) 0.05
(c) 0.05
(d) When Y ~ 10 , then Y /10 ~ F10, .
2




(e) Y = Z 2 , where Z ~ N (0,1), thus Pr (Y  1) = Pr (−1  Z  1) = 0.32.




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2.12. (a) 0.05
(b) 0.950
(c) 0.953
(d) The tdf 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 ) + 2Y =1+ 0 =1; E(W 2 ) = Var (W ) + W2 =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 −  )4 / 4 ; solving yields E(Y 4 ) = 3; a similar
Y Y
calculation yields the results for W.
(d) First, condition on X = 0, so that S = W:
E(S | X = 0) = 0; E(S 2 | X = 0) =100, E(S3|X = 0) = 0, E(S 4 | X = 0) = 31002 .
Similarly,
E(S | X = 1) = 0; E(S 2 | X =1) =1, E(S3 | X =1) = 0, E(S 4 | X =1) = 3.
From the law 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)
= 31002  0.01 + 31 0.99 = 302.97
(e) S = E(S) = 0, thus E(S − S )3 = E(S3 ) = 0 from part (d). Thus skewness = 0. Similarly,
 S2 = E(S −  S )2 = E(S 2 ) =1.99, and E(S −  S )4 = E(S 4 ) = 302.97. Thus,
kurtosis = 302.97 / (1.992 ) = 76.5

2.14. The central limit theorem suggests that when the sample size (n) is large, the distribution of the
sample average (Y ) is approximately N   Y ,  Y2  with  2Y = nY2 . Given  Y = 100,  Y = 430,
2

 
2
(a) n = 100,  = 2 
Y = 43
= 043, and
Y n 100

 Y −100 101 −100 
Pr (Y  101) = Pr    (1525) = 09364
 043 043 
2
(b) n = 165,  = Y = = 02606, and
2  43
Y n 165

 −100 
Pr (Y  98) = 1 − Pr(Y  98) = 1− Pr Y −100  9802606

 02606 
 1 − (−39178) =  (39178) = 1000 (rounded to four decimal places)




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,8 Stock/Watson • Introduction to Econometrics, Third Edition


 Y2 43
(c) n = 64,  = 2
= = 06719, and
Y
64 64
 101 −100 Y −100 103 −100 
Pr (101  Y  103) = Pr   
 06719 06719 06719 
  (36599) − (12200) = 09999 − 08888 = 01111
 9.6 −10
2.15. (a) Pr (9.6  Y  10.4) = Pr  Y −10 10.4 −10 
  
 4/n 4/n 4/n 
 9.6 −10 10.4 −10 
= Pr  Z 
 4/n 4/n 
where Z ~ N(0, 1). Thus,
 9.6 −10 10.4 −10 
(i) n = 20; Pr Z  = Pr (−0.89  Z  0.89) = 0.63
 
 4/n 4/n 
 9.6 −10 10.4 −10 
(ii) n = 100; Pr Z = Pr(−2.00  Z  2.00) = 0.954
 
 4/n 4/n 
 9.6 −10 10.4 −10 
(iii) n = 1000; Pr  Z

= Pr(−6.32  Z  6.32) = 1.000
 4/n 4/n 
 −c c 
(b) Pr (10 − c  Y  10 + c) = Pr   Y −10  
4/n
 4/n 4/n 
 −c c 
= Pr  Z .
 4/n 4/n 
c
As n get large gets large, and the probability converges to 1.
4/ 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, Y1, Y2, … Yn. Let Xi = 1
if Yi  3.6, otherwise set Xi = 0. Notice that Xi is a Bernoulli random variables with X = Pr(X = 1)
= Pr(Y  3.6). Compute X . Because X converges in probability to X = Pr(X = 1) = Pr(Y  3.6),
X will be an accurate approximation if n is large.

2.17. Y = 0.4 and  Y2 = 0.4 0.6 = 0.24
 
(a) (i) P( Y  0.43) = Pr Y − 0.4 0.43 − 0.4    0.6124 = 0.27
 = Pr Y − 0.4
   
 0.24/n 0.24/n   0.24/n 




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 
(ii) P( Y  0.37) = Pr Y − 0.4 0.37 − 0.4   Y − 0.4  −1.22 = 0.11
   = Pr  
 0.24/n 0.24/n   0.24/n 
0.41 − 0.40
(b) We know Pr(−1.96  Z  1.96) = 0.95, thus we want n to satisfy 0.41 =  −1.96
24 / n
0.39 − 0.40
and  −1.96. Solving these inequalities yields n  9220.
24 / n
2.18. Pr (Y = $0) = 095, Pr(Y = $20000) = 005.
(a) The mean of Y is
Y = 0  Pr (Y = $0) + 20,000  Pr (Y = $20000) = $1000.
The variance of Y is
 2 = E (Y −  )2
Y Y
 

= (0 −1000)2  Pr(Y = 0) + (20000 −1000)2  Pr (Y = 20000)
= (−1000)2  095 + 190002  005 = 19 107,
so the standard deviation of Y is  = (19 107 ) 2 = $4359
1


Y

(b) (i) E(Y ) =  = $1000,  2 = Y2 1.9 107
= = 19 105.
Y Y
n 100
(ii) Using the central limit theorem,
Pr (Y  2000) = 1 − Pr (Y  2000)
 Y −1000 2 000 −1 000 
= 1 − Pr   
 19 10 5
19 10 5

 1 − (22942) = 1 − 09891 = 00109

l
2.19. (a) Pr (Y = y j ) =  Pr ( X = xi , Y = y j )
i =1
l

=  Pr (Y = yj |X =xi )Pr ( X =xi )
i =1
k k l
(b) E (Y ) =  y j Pr (Y = yj ) =  yj  Pr (Y = yj |X = xi ) Pr ( X = xi )
j =1 j =1 i =1


l  k

j j i


 i

= y Pr (Y = y |X = x ) Pr ( X = x )
i =1  j =1
 
l

= E(Y|X =xi )Pr ( X =xi )
i =1

(c) When X and Y are independent,
Pr (X = xi , Y = yj ) = Pr (X = xi )Pr (Y = yj )

so



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, 10 Stock/Watson • Introduction to Econometrics, Third Edition


 XY = E[( X − X )(Y − Y )]
l k
=  (xi − X )( y j − Y ) Pr ( X =xi , Y = y j )
i =1 j =1
l k
= (xi − X )( y j − Y ) Pr ( X =xi ) Pr (Y = y j )
i =1 j =1

 l
 k 
= (x −  ) Pr ( X = x ) ( y −  ) Pr (Y = y
 i X i 
 j Y j
 i =1   j =1 
= E( X − X )E(Y − Y ) = 0  0 = 0,
 XY 0
corr(X , Y ) = = = 0
 X Y  XY

l m
2.20. (a) Pr (Y = yi ) =  Pr (Y = yi |X = xj , Z = zh ) Pr (X = xj , Z = zh )
j =1 h=1

k
(b) E(Y ) =  yi Pr (Y = yi ) Pr (Y = yi )
i =1
k l m
=  yi  Pr (Y = yi |X = xj , Z = zh ) Pr (X = xj , Z = zh )
i =1 j =1 h=1
=  
l m
 k y Pr (Y = y |X = x , Z = =x,Z=z )
 i i j zh )  Pr (X j h
j =1 h=1  i =1 
l m

=  E(Y|X = xj , Z = zh ) Pr (X = xj , Z = zh )
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 − )3 = E[( X − )2 ( X − )] = E[ X 3 − 2 X 2  + X  2 − X 2  + 2 X  2 −  3 ]
= E( X 3 ) − 3E( X 2 ) + 3E( X ) 2 −  3 = E( X 3 ) − 3E( X 2 )E( X )
+ 3E( X )[E( X )]2 −[E( X )]3
= E( X 3 ) − 3E( X 2 )E( X ) + 2E( X )3
(b) E( X − )4 = E[( X 3 − 3X 2  + 3X  2 −  3 )( X − )]
= E[ X 4 − 3X 3  + 3X 2  2 − X  3 − X 3  + 3X 2  2 − 3X  3 +  4 ]
= E( X 4 ) − 4E( X 3 )E( X ) + 6E( X 2 )E( X )2 − 4E( 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




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