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

, For Instructors
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Solutions to End-of-Chapter Exercises
n n n

,Chapter 2 n



Review of Probability
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2.1. (a)n ProbabilityndistributionnfunctionnfornY
Outcomen(numbernofnheads) Ynn0 Ynn1 Ynn2
n Probability 0.25 0.50 0.25n

(b)n CumulativenprobabilityndistributionnfunctionnfornY

Outcomen(numbernofnheads) Ynn0 0nnYnn1 1nnYnn2 Ynn2
Probability 0 0.25 0.75 1.0

(c) nnY =nE(Yn)nn(0nn0.25)nn(1n0.50)nn(2nn0.25)nn1.00n.nFn
d
Fq,n.
Usingn Keyn Conceptn 2.3:n var(Yn)nn E(Yn2n)n[E(Yn)]2n,
and
(uin|Xin )
sonthat
var(Yn)nnE(Yn2n)n[E(Yn)]2n n1.50nn(1.00)2n n0.50.

2.2. WenknownfromnTablen2.2nthatn Prn(Yn n0)nn022, Prn(Yn n1)nn078, Prn(nXn n0)nn030,
Pr(nXn n1)nn070.n So
(a) Yn n E(Yn)nn0nnPrn(Yn n0)nn1nPrn(Yn n1)
n 0nn022nn1n078nn078,
Xn n E(nXn)nn0nnPrn(nXn n0)nn1nPrn(nXn n1)
n 0nn030nn1n070n n 070
(b) n n E[(nXn nn )2n]
2n

X X

n(0nn0.70)2n nPrn(nXn n0)nn(1nn0.70)2n nPrn(nXn n1)
n(070)2n n030nn0302n n070nn021,
nY2n nE[(Yn nnY )2n]
n(0nn0.78)2n nPrn(Yn n0)nn(1nn0.78)2n nPrn(Yn n1)
n(078)2n n022nn0222n n078nn01716




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, SolutionsntonEnd-of-ChapternExercises 3

(c)n nXYn ncovn(Xn,nYn)nn E[(nXn nXn)(Yn nYn)]
n(0nn0.70)(0nn0.78)nPr(nXn n0,nYn n0)
n(0nn070)(1nn078)nPrn(nXn n0nYn n1)
n(1nn070)(0nn078)nPrn(nXn n1nYn n0)
n(1nn070)(1nn078)nPrn(nXn n1nYn n1)
n(070)nn(078)nn015nn(070)nn022nn015
n030nn(078)nn007nn030nn022nn063
n0084,
nXY 0084
corrn(Xn,nYn)n  n04425
nXnY 021n01716

2.3. Fornthentwonnewnrandomnvariablesn Wn n3nn6nXn andn Vn n20nn7Yn,n wenhave:
(a) E(Vn)nnE(20nn7Yn)nn20nn7E(Yn)nn20nn7nn078nn1454,
E(Wn)nnE(3nn6Xn)nn3nn6E(nXn)nn3nn6nn070nn72
(b)n n2n nvar(3nn6Xn)nn62nnn2n n36nn021nn756,
W X

Vn n nvar(20nn7Yn)nn(7) nnn2Yn n49nn01716nn84084
2 2


(c)n WVnnnncov(3nn6Xn,n20nn7Yn)nn6nn(7)ncov(Xn,nYn)nn42nn0084nn3528
WV 3528
corrn(Wn,n Vn)n  n04425
WnV 756nn84084

2.4.nnn(a)n(b E(nXn3 ) nn03nn(1n p)n13nn pnn pnE(n
)n( Xnkn)nn0kn n(1n p)n1kn n pnn p
c) E(nXn)nn0.3n,nandnvar(X)n=nE(X2)−[E(X)]2n=n0.3n−0.09n=n0.21.nThusnn= 0.21 =n0.46.
var(nXn)nnE(nXn2n)n[E(nXn)]2n n0.3nn0.09nn0.21nn  0.21 n0.46.n Toncomputenthenskewness,nuse
thenformulanfromnexercisen2.21:
E(nXn n )3n n E(nXn3n)nn3[E(nXn2n )][E(nXn)]nn 2[E(nXn)]3
n0.3nn3n0.32n n2nn0.33n n0.084
Alternatively,n E(nXn n)3n [(1n0.3)3nn0.3]n[(0nn0.3)3nn0.7]nn0.084
Thus,n skewnessn nE(nXn n)3/n3n n0.084/0.463n n0.87.
Toncomputenthenkurtosis,nusenthenformulanfromnexercisen2.21:
E(nXn n)4n n E(nXn4n)nn4[E(nXn)][E(nXn3n)]nn6[E(nXn)]2n[E(nXn2n)]nn3[E(nXn)]4
n0.3nn4nn0.32n n6nn0.33nn3n0.34n n0.0777
Alternatively,n E(nXn n)4n [(1n0.3)4n n0.3]n[(0nn0.3)4n n0.7]nn0.0777
Thus,n kurtosisn isn E(nXn n)4/n4n n0.0777/0.464n 1.76




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