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1、Chapter5:MultivariateProbabilityDistributions5.1a.ThesamplespaceSgivesthepossiblevaluesforY1andY2:SAAABACBABBBCCACBCC(y1,y2)(2,0)(1,1)(1,0)(1,1)(0,2)(1,0)(1,0)(0,1)(0,0)Sinceeachsamplepointisequallylikelywithprobably1/9,thejointdistributionforY1andY2isgivenbyy101201/
2、92/91/9y212/92/9021/900b.F(1,0)=p(0,0)+p(1,0)=1/9+2/9=3/9=1/3.5.2a.Thesamplespaceforthetossofthreebalancedcoinsw/probabilitiesarebelow:OutcomeHHHHHTHTHHTTTHHTHTTTHTTT(y1,y2)(3,1)(3,1)(2,1)(1,1)(2,2)(1,2)(1,3)(0,–1)probability1/81/81/81/81/81/81/81/8y10123–11/8000y210
3、1/82/81/8201/81/80301/800b.F(2,1)=p(0,–1)+p(1,1)+p(2,1)=1/2.5.3NotethatusingmaterialfromChapter3,thejointprobabilityfunctionisgivenby⎛4⎞⎛3⎞⎛2⎞⎜⎜⎟⎟⎜⎜⎟⎟⎜⎜⎟⎟p(y⎝y1⎠⎝y2⎠⎝3−y1−y2⎠1,y2)=P(Y1=y1,Y2=y2)=,where0≤y1,0≤y2,andy1+y2≤3.⎛9⎞⎜⎜⎟⎟⎝3⎠Intableformat,thisisy10123003/846/8
4、41/84y214/8424/8412/840212/8418/840034/84000Chapter5:MultivariateProbabilityDistributions94Instructor’sSolutionsManual5.4a.Alloftheprobabilitiesareatleast0andsumto1.b.F(1,2)=P(Y1≤1,Y2≤2)=1.Everychildintheexperimenteithersurvivedordidn’tandusedeither0,1,or2seatbelts.1
5、/21/35.5a.P(Y≤1/2,Y≤1/3)=3ydydy=.1065.12∫∫112001y1/2b.P(Y≤Y/2)=3ydydy=.5.21∫∫11200.51.5.515.6a.P(Y−Y>.5)=P(Y>.5+Y)=1dydy=[]ydy=(.5−y)dy=.125.1212∫∫∫121y2+.52∫220y2+.500111b.P(YY<.5)=1−P(YY>.5)=1−P(Y>.5/Y)=1−1dydy=1−(1−.5/y)dy121212∫∫12∫22.5.5/y2.5=1–[.5+.5ln(.5)]=.84
6、66.1∞1∞⎡⎤⎡⎤5.7a.P(Y<1,Y>5)=e−(y1+y2)dydy=e−y1dye−y2dy⎥=[]1−e−1e−5=.00426.12∫∫12⎢∫1⎥⎢∫205⎣0⎦⎣5⎦33−y2b.P(Y+Y<3)=P(Y<3−Y)=e−(y1+y2)dydy=1−4e−3=.8009.1212∫∫1200115.8a.Sincethedensitymustintegrateto1,evaluatekyydydy=k/4=1,sok=4.∫∫121200yy2122b.F(y,y)=P(Y≤y,Y≤y)=4ttdtdt=yy
7、,0≤y1≤1,0≤y2≤1.121122∫∫1212120022c.P(Y1≤1/2,Y2≤3/4)=(1/2)(3/4)=9/64.1y25.9a.Sincethedensitymustintegrateto1,evaluatek(1−y)dydy=k/6=1,sok=6.∫∫21200b.NotethatsinceY1≤Y2,theprobabilitymustbefoundintwoparts(drawingapictureisuseful):113/41P(Y1≤3/4,Y2≥1/2)=6(1−y)dydy+6(1−y
8、)dydy=24/64+7/64=31/64.∫∫212∫∫2211/21/21/2y15.10a.Geometrically,sinceY1andY2aredistributeduniformlyoverthetriangularregion,usingthe