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SPRING2020EconomicMathematicsIIPROBLEMSET#61.(#7.1,p.222)LetXbearandomvariablewithprobabilityFindtheprobabilitydistributionoftherandomvariableY=2X 1.Solution:fY(y)={13;y=1;3;50;otherwise2.(#7.2,p.222)LetXbeabinomialrandomvariablewithprobabilitydistributionFindtheprobabilitydistributionoftherandomvariableY=X2.Solution:fY(y)=8:(3py)(25)py(35)3 py;y=0;1;4;90;otherwise3.(#7.6,p.223)GiventherandomvariableXwithprobabilitydistributionFindtheprobabilitydistributionofY=8X3.1Solution:fY(y)=ddyP(Yy)=ddyP(8X3y)==ddyP(X12y13)=fX(12y13)ddy12y13=y1316y 23=16y 13for0y8;andzerootherwise.4.(#7.8,p.223)Adealer’sprofit,inunitsof$5000,onanewautomobileisgivenbyY=X2,whereXisarandomvariablehavingthedensityfunction(a)FindtheprobabilitydensityfunctionoftherandomvariableY.(b)UsingthedensityfunctionofY,findtheprobabilitythattheprofitonthenextnewautomobilesoldbythisdealershipwillbelessthan$500.Solution:(a)fY(y)=ddyP(Yy)=ddyP(X2y)=ddyP( pyXpy)ddyP(Xpy)=fX(py)ddypy=2(1 py)12py=1py 1for0y1;andzerootherwise.(b)∫1100(1py 1)dy=p40 1105.SupposearandomvariableXhasapdffX(x)=12 e jxj; 1x1;where 0:Thisiscalledthedoubleexponential(orLaplace)distribution.FindthepdfforY=jXj:2Solution:fY(y)=ddyP(Yy)=ddyP(jXjy)=ddyP( yXy)=ddyP(Xy) ddyP(X y)=fX(y)ddy(y) fX( y)ddy( y)=12 e y1 12 e y( 1)= e y;fory0andzerootherwise.6.SupposefX(x)=12for0x2:FindthePDFofY=X(2 X):Solution:fY(y)=ddyP(Yy)=ddyP(2X X2y)=ddyP(2X X2 y0)=ddyP(X1 √1 y)+ddyP(X1+√1 y)=ddyP(X1 √1 y)+ddy(1 P(X1+√1 y))=fX(1 √1 yddy(1 √1 y) fX(1+√1 y)ddy(1+√1 y)=12( 12p1 y) 12(+12p1 y)=12p1 y;for0y1andzerootherwise.7.(#4.2,p.117)TheprobabilitydistributionofthediscreterandomvariableXisFindthemeanofX.3Solution:E(X)=∑Xxf(x)=0f(0)+1f(1)+2f(2)+3f(3)=(31)(14)1(34)2+2(32)(14)2(34)1+3(33)(14)3(34)0=2764+1864+364=4864=348.(#4.13,p.117)ThedensityfunctionofthecontinuousrandomvariableX,thetotalnumberofhours,inunitsof100hours,thatafamilyrunsavacuumcleaneroveraperiodofoneyear,isgiveninExercise3.7onpage92asFindtheaveragenumberofhoursperyearthatfamiliesruntheirvacuumcleaners.Solution:E(X)=∫10xf(x)dx+∫21xf(x)dx=∫10x2dx+∫21x(2 x)dx=13x3 10+x2 13x3 21=1100hours.9.(#4.34,p.127)LetXbearandomvariablewiththefollowingprobabilitydistri-bution:FindthestandarddeviationofX.Solution:E(X)=∑Xxf(x)= 20:3+30:2+50:5=2:5E(X2)=∑Xx2f(x)=4( 2)20:3+320:2+520:5=15:5)2X=E(X2) [E(X)]2=15:5 6:25=9:25)ThestandarddeivationX=√2X=p9:253:041410.(#4.43,p.127)Thelengthoftime,inminutes,foranairplanetoobtainclearancefortakeoffatacertainairportisarandomvariableY=3X 2,whereXhasthedensityfunctionFindthemeanandvarianceoftherandomvariableY.Solution:WeknowE(Y)=E(3X 2)=3E(X) 2andVar(Y)=Var(3X 2)=9Var(X):Let’scalculateE(X)andVar(X)first.E(X)=∫1 1xf(x)dx=∫10x14e x=4dx=x( e x=4) 10+∫10e x=4dx=4E(X2)=∫1 1x2f(x)dx=∫10x214e x=4dx=(x2( e x=4) 10+∫102xe x=4dx=0+8∫10x14e x=4dx=8E(X)=32)2X=E(X2) [E(X)]2=32 42=16Therefore,E(Y)=E(3X 2)=3E(X) 2=12 2=10;and2Y=92X=169=14411.(#4.50,p.127)Foralaboratoryassignment,iftheequipmentisworking,thedensityfunctionoftheobservedoutcomeXis5FindthevarianceandstandarddeviationofX.Solution:E(X)=∫1 1xf(x)dx=∫10x2(1 x)dx=13E(X2)=∫10x22(1 x)dx=∫10(2x2 2x3)dx=(23x3 12x4) 10=1=6:E(X2)=∫0x22(1 x)dx=∫0(2x2 2x0)dx=)2X=E(X2) [E(X)]2=1=6 1=9=1=18)X=√1=1812.(#4.59,p.138)IfarandomvariableXisdefinedsuchthatfindand2.Solution:E(X2 2X+1)=E(X2) 2E(X)+1=10andE(X2 4X+4)=E(X2) 4E(X)+4=6E(X2)=16;andE(X)=72=E(X)=72;and2=var(X)=E(X2) 2=16 494=1546