PROBABILITY PRIMER. Exercise Solutions

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PROBABILITY PRIMER Exercse Solutons 1

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e EXERCISE P.1 (b) X s a random varable because attendance s not known pror to the outdoor concert. Before the concert, attendance s uncertan because the weather s uncertan. Expected attendance s gven by E( X) x f( x) 5000.10000.60000.1100 x (c) Expected proft s gven by EY ( ) E(5X000) 5 E( X) 00051100000 3500 (d) The varance of proft s gven by var( Y) var(5 X 000) 5 var( X) 50,000 6,000,000

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 3 EXERCISE P. The completed table s f ( xy, ) 0 1 f ( x ) 10 0.18 0.00 0.18 X 0 0.00 0.30 0.30 Y 10 0.07 0.5 0.5 f ( y ) 0.5 0.75 (b) E( X) x f( x) 100.1800.3100.53. x You should take the bet because the expected value of your wnnngs s postve. (c) The probablty dstrbuton of your wnnngs f you know she dd not study s It s gven n the followng table f ( x,1) f x y1 for x 10, 0,10 f (1) Y f x f f x y 1 X (,1) (1) Y 10 0.00 0.75 0.0 0 0.30 0.75 0. 10 0.5 0.75 0.6 (d) Gven that she dd not study, your expected wnnngs are E X Y 1 xf x y 1 100.000.100.66 x

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e EXERCISE P.3 Assume that total sales X are measured n mllons of dollars. Then, X N.5,0.3, and P X 3.5 3PZ 0.3 P Z 1.6667 1P Z 1.6667 10.95 0.078

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 5 EXERCISE P. Extendng the table to nclude the margnal dstrbutons for poltcal afflaton (PA) and CITY yelds Poltcal Afflaton (PA) R I D f ( CITY ) CITY Southern 0. 0.0 0.1 0. Northern 0.18 0.1 0.30 0.6 f ( PA ) 0. 0.16 0. (, ) 0.18 f R Northern 0.3 f ( Northern) 0.6 PR CITY Northern CITY (b) Poltcal afflaton and regon of resdence are not ndependent because, for example, f( R, Northern) 0.18 f ( R) f ( Northern) 0.0.6 0.5 PA CITY (c) E( PA) Rf ( R) I f ( I) D f ( D) PA PA PA 00.0.1650.. (d) E( X) EPAPA E( PA) EPA where E PA R f ( R) I f ( I) D f ( D) PA PA PA 0 0. 0.16 5 0. Thus, 11.1 EX EPA EPA ( ) ( ).11.1 7.1

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 6 EXERCISE P.5 (b) The probablty that the NFC wns the 1 th flp, gven they have won the prevous 11 flps s 0.5. Each flp s ndependent; so the probablty of wnnng any flp s 0.5 rrespectve of the outcomes of prevous flps. Because the outcomes of prevous flps are ndependent and ndependent of the outcomes of future flps, the probablty that the NFC wll wn the next two consecutve flps s 0.5 multpled by 0.5. That s, 0.5 0.5. Go Sants!

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 7 EXERCISE P.6 ESALES ( ) E(071030 PRICE) 071030 EPRICE ( ) 07103075 860 var( SALES ) var(0710 30 PRICE) 30 var( PRICE) 30 5,6,500 (b) (c) PSALES 6300 6300 860 PZ 6500 P Z 1.0065 P Z 1.0065 0.85

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 8 EXERCISE P.7 After ncludng the margnal probablty dstrbutons for both C and B, the table becomes C B 0 1 f () c 0 0.05 0.05 0.05 0.15 1 0.05 0.0 0.15 0.0 0.05 0.5 0.15 0.5 f () b 0.15 0.50 0.35 The margnal probablty dstrbuton for C s gven n the last column of the above table. (b) EC ( ) cf( c) 00.1510.00.51.3 (c) C c f c E C c var( ) ( ) ( ) 0 0.15 1 0.0 0.5 (1.3) 0.51 c (d) For the two companes advertsng strateges to be ndependent, the condton f (,) cb f () c f () b must hold for all c and b. We fnd that C B f(0,0) 0.05 f (0) f (0) 0.150.15 0.05 C Thus, the two companes advertsng strateges are not ndependent. B (e) Values for A are gven by the equaton A 5000 1000B. Its probablty dstrbuton s obtaned by matchng values obtaned from ths equaton wth correspondng probabltes for B. A f ( a ) 5000 0.15 6000 0.50 7000 0.35 (f) Snce the relatonshp between A and B s an exact lnear one, they are perfectly correlated. The correlaton between them s 1.

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 9 EXERCISE P.8 X f ( x ) 1 16 16 3 16 16 5 16 6 16 1 6 1 1 1 6 6 3 (b) PX PX or X 5 (c) 1 1 1 1 1 1 EX ( ) xf( x) 1 3 5 6 3.5 6 6 6 6 6 6 x The result EX ( ) 3.5 means that f a de s rolled a very large number of tmes, the average of all the values shown wll be 3.5; t wll approach 3.5 as the number of rolls ncreases. (d) EX 1 1 1 1 1 1 x f( x) 1 3 5 6 15.16667 6 6 6 6 6 6 (e) X EX E X x var( ) ( ) 15.16667 3.5.91667 (f) The results for ths part wll depend on the rolls obtaned by the student. Let X n denote the average value after n rolls. The values obtaned by one of us and ther averages are: 0 values of X,1,5,3,,1,5,5,,,,,,,,,3,,6,3 X5 3.000 X10 3.00 X 0 3.00 These values are relatvely close to the mean of 3.5 and are expected to become closer as the number of rolls ncreases.

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 10 EXERCISE P.9 fx 0...6.8 f(x)=/3 - /9x 0 1 3 x The area under the curve s equal to one. Recallng that the formula for the area of a trangle s half the base multpled by the heght, t s gven by 1 3 1. 3 (b) When x 1, f( x) 5 9. The probablty s gven by the area under the trangle between 0 and 1/. Ths can be calculated as 1 P1/ X 3. The latter probablty s Therefore, 1 1 5 5 5 P1/ X 3 bh 0.69 9 36 5 11 P(0 X 1/ ) 1P1/ X 31 0.30555 36 36 (c) To compute ths probablty we can subtract the area under the trangle between 3/ to 3 from the area under the trangle from 1/ to 3. Dong so yelds 1 3 1 3 3 3 1 3 1 1 1 3 f f 1 11 11 1 9 1 P X P X P X 18 11 9 1 16 5 0.7778 18

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 11 EXERCISE P.10 X Y 1 1 ( ) ( ) ( ) ( ) EZ E EX EY (b) Assumng X and Y are ndependent, X Y 1 1 var( Z) var var( X) var( Y) ( ) (c) Assumng that cov(, ) 0.5 XY, 1 X Y var( Z) var var( X) var( Y) cov( X, Y) 1 0.5 3

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 1 EXERCISE P.11 Let X denote the length of lfe of a personal computer selected at random. The fracton of computers that fal wthn a gven tme nterval s equal to the probablty that X les n that nterval. 1 1 3. 1.897 0.089 1.6 PX P Z PZ 3. 0.73 0.3176 1.6 (b) PX P Z PZ 3. 1.1068 0.8658 1.6 (c) PX P Z PZ.5 3. 3. 1.6 1.6 (d) P.5 X P Z P0.7115 Z 0.73 (e) We want 0 X 0 s such that 0.73 PZ 0.7115 P Z 0.68 0.38 0. X such that Solvng for X 0 yelds P X X 0 0.05. Now, PZ ( 1.65) 0.05, and thus a sutable X 0 3. 1.65 1.6 X 0 3. 1.65 1.6 1.319 (whch s approxmately 16 months)

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 13 EXERCISE P.1 The probablty functon of X s shown below..0 f(x).35.30.5.0.15.10.05.00 0 1 3 5 6 7 x (b) The probablty that, on a gven Monday, ether, or 3, or students wll be absent s x f( x) f() f(3) f() 0.60.30.0.8 (c) The probablty that, on a gven Monday, more than 3 students are absent s 7 x f( x) f() f(5) f(6) f(7) 0.0.080.00.010.35 (d) 7 EX ( ) xf. ( x) 00.010.030.630.30. x0 3.16 50.0860.070.01 Based on nformaton over many Mondays, the average number of students absent on Mondays s 3.16. var( X ) E X [ E( X)] (e) E X 7 x f( x) 0 0.0 1 0.03 0.6 3 0.3 x0 var( X ) 11.58 (3.16) 1.59 1.67 0. 5 0.08 6 0.0 7 0.01 = 11.58 (f) EY ( ) E(7X3) 7 E( X) 373.163 5.1 var( Y) var(7 X 3) 7 var( X) 91.59 78.156

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 1 EXERCISE P.13 Let X be the annual return from the mutual fund. Then, ~ 0.05,0.0 X N. (b) 0 0.05 PX ( 0) PZ PZ ( 1.5) 0.1056 0.0 0.15 0.05 PX ( 0.15) PZ PZ (.5) 0.006 0.0 (c) Let Y be the return from the alternatve portfolo. Then, ~ 0.07,0.07 0 0.07 PY ( 0) PZ PZ ( 1) 0.1587 0.07 Y N. 0.15 0.07 PX ( 0.15) PZ PZ ( 1.19) 0.165 0.07 The calculatons show that the probablty of a negatve return has ncreased from 10.56% to 15.87%, whle the probablty of a return greater than 15% has ncreased from 0.6% to 1.65%. Whether fund managers should or should not change ther portfolos depends on ther rsk preferences.

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 15 EXERCISE P.1 Expressng the returns n terms of percentages, we have RA, 8 and B 8, 1 E( P) E0.5R 0.75R 0.5ER 0.75ER A B A B 0.5 0.758 7 (b) var P var 0.5R 0.75R Now, Hence, 1 P A B R. R R R R 0.5 var A 0.75 var B 0.5 0.75 cov A, B var RA RB R var R cov, A cov( RA, RB) AB 81 96 B var( P) 0.5 8 0.75 1 0.50.7596 11 11 11 P (c) When 0.5 var RA RB R var R cov, A cov R, R 0.5 0.581 8 A B A B B (d) When 0 var( P) 0.5 8 0.75 1 0.50.758 103 103 10.15 P, cov R, R 0, and the varance and standard devaton of the portfolo are A B var( P) 0.5 8 0.75 1 85 85 9. P

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 16 EXERCISE P.15 1 x x x 79 1 1 1 77 1.5 (b) x x x1 x x3 x 1 (c) x xx1xx xx3xx x 1 (7 1.5) ( 1.5) ( 1.5) ( 7 1.5) 5.5 0.5.5 8.5 0 (d) x x x1x x x x3x x x 1 (7 1.5) ( 1.5) ( 1.5) ( 7 1.5) 5.5 0.5.5 ( 8.5) 109 1 1 531 5.5 (e) y y y1 y y3 y 1 x xy y 1 (7 1.5) (55.5) ( 1.5) ( 5.5) ( 1.5) (35.5) ( 71.5) (15.5) 5.5 ( 0.5) 0.5 ( 3.5).5 (.5) ( 8.5) 6.5.75 1.75 6.5 55.5 66 (f) xy xy x y x y x y x y xy 1 1 1 3 3 x x1 x x3 x x x 1 753 ( 7) 11.55.5 9 16 9.5 66 0.6055 109

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 17 EXERCISE P.16 (b) (c) (d) (e) (f) x1 x x3 x x 3 3 x x x 1 x1y1xyx3y3xy xy 1 x y x y x y x y x y 1 3 3 5 6 1 3 3 3 x y x y x y ( x y ) ( x y ) ( x y ) ( x y ) 1 1 3 3 3 1

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 18 EXERCISE P.17 (b) (c) (d) (e) 1 3 1 ( abx ) ( abx ) ( abx ) ( abx ) ( abx ) 1 3 ab( x x x x ) 1 3 1 3 191 3 ( x x) (0 0) (1 1) ( ) (3 3) x0 x x0 = + 5 + 10 + 17 = 3 f( x) f() f(3) f() f () f(5) f(6) f ( xy, ) f(0, y) f(1, y) f(, y) (f) ( x y) ( x1) ( x) (x6) x y1 x x ( 6) (36) ( 6) 10 1 1 36

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 19 EXERCISE P.18 x x ( x x x x ) (1353) 3 1 1 3 (b) x xx1xx xx3 xx x 1 13 33 53 33 0 (c) x x x1 x x x x3 x x x 1 13 33 53 33 8 (d) 1 x x x x x x x 19593 8 1 3 (e) x x x xx x x xn x nx 1 n n n n n n n 1 1 1 1 1 1 n n x nx nx x nx 1 1

Probablty Prmer, Exercse Solutons, Prncples of Econometrcs, e 0 EXERCISE P.19 n n n n n x x y y x y x y xy xy 1 1 1 1 1 n x y yn n n 1 x xn 1 y nxy 1 n 1 n 1 n x y nxy nxy 1 n 1 xy nxy