Transcription of MTH135/STA104: Probability
1 MTH135/STA104: ProbabilityHomework# 8 Due:Tuesday, Nov 8, nea functionf(x; y) ontheplaneR2byf(x; y) =(1=x0< y < x <10otherx; ya)Show thatf(x; y) is a joint Probability density to check?Two things:f(x; y) 0 forall(x; y)2R2, andZZR2f(x; y)dx dy=Z10 Zx01xdy dx=Z101dx= 1 Iff(x; y) is thejoint , nd:b)Themarginaldensity functionsfx(x) =fy(y) =fx(x) =Rx01xdy= 1;0< x <1;fy(y) =R1y1xdx= logy;0< y < )TheexpectationsEX=EY=EX=R10x dx=12; easiestway forYis to usethejoint pdf,EY=ZZ0<y<x<1y1xdy dx=Z10x22xdx=14 two independent randomvariables,each withtheuni-formdistributionon(0;1).LetM= min(X; Y) be thesmallerof )Represent theeventM> xas a regionin theplane,and ndtheprobabilityP[M> x] as theareaof [M> x] = (1 x)2for0< x < )Useyourresultabove to ndthedensity functionforM.
2 PlotbothCDFandpdf forMontherange 1 x (x) = 1 P[M> x] = 1 (1 x)2for0< x <1, givingpdff(x) = 2(1 x) for0< x <1: 1012x01F(x).. 1012x012f(x).. ; U2; ; Unbenindependent uniformrandomvariablesdrawnfromtheset(0;1).Orderthemfromleastto greatestandcallthere-orderedvariablesU(1); U(2); ; U(n); forexample,U(1) minfU1; U2; ; Ung. Let0< x < y < )Findandjustifya simpleformulaforP[U(1)> x; U(n) y].Theindicatedevent occursif andonlyifalltheUj's fallwithintheinterval(x; y], which happenswithprobabilityP[U(1)> x; U(n) y] =P \nj=1[x < Uj y] = (y x)nb)Now nda simpleformulaforP[U(1) x; U(n) y].P[U(1) x; U(n) y] =P[U(1)>0; U(n) y] P[U(1)> x; U(n) y]=yn (y x)n;0< x < y <1:2c)Give thejoint pdf forU(1)andU(n). Becarefulabouttherangeofx; ywhereyourformulais correct|tryto give a correctoneforallx; (x; y) =n(n 1)(y x)n 2;0< x < y <1withf(x; y) = 0 forotherx; point (x; y) is saidto be drawn\uniformly"froma setA R2in theplaneif thejoint density functionf(x; y) hassomeconstant valuec >0 for(x; y)2A, andis zerooutsideA.
3 LetXandYbe thecoordinatesof a pointdrawnuniformlyfromthetriangleAwithc orners(0;0),(0;2),and(2;0).a)Whatis theconstant valueof thepdff(x; y) insideA? Why?Oneover theareaofA, or12, sincetheintegraloff(x; y) )LetR=pX2+Y2be thedistanceof (X; Y) [R 1] (Hint: draw a picture|nointegrationis needed)Thatquarter-circle,witharea =4, liescompletelywithinthetriangleA,so theprobability is simply12 ( =4) = = )Give thejoint CDFF(x; y) =P[X x;Y y] correctlyateverypointx; y2R2. (Hint: separateinto severalcases;draw pictures).F(x; y) =8>>>>>>>> <>>>>>>>>:0x 0 ory 0xy=20< x;0< y; x+y 2x+y 1 (x2+y2)=40< x 2;0< y 2; x+y >2x x2=40< x 2; y >2y y2=4x >2;0< y 21x >2; y > straight stick is broken at randomin two placeschosenindependentlyanduniformlyalo ngthelengthof thestick.
4 Whatis theprobability thatthepiecescanbe arrangedto forma triangle?DenotebyLthelengthof thestick, andbyxandythetwo triangleif noneis longerthanthesumof theothers;if0< x < y < Ltheconditionsarethatx <(L x) (orx < L=2),(L y)< y3(ory > L=2),and(y x)< x+ (L y) (ory < x+L=2).Thus thepartoftheallowableregionwithx < yis boundedby thelinesx=L=2,y=L=2,andy=x+L=2; hereis a diagramwiththeallowableregionshaded,forL = 10 halfof theshadedregionis a trianglewitharea12(L=2)2,so thetotalareaisL2=4 andtheindicatedprobability isP[Piecesforma triangle]= 1=4 joint density functionf(x; y) = 12xy(1 x)0< x <1;0< y <1andequalto )AreXandYindependent? Why?Yes,thejoint pdf is a productoffy(y) = 2y;0< y <1 andfx(x) =6x(1 x);0< x < ) (1 x)dx=Z 6x2 6x3 dx= 2x3 32x4 x=1x=0=12:c) dy= 2y3=3 y=1y=0=23:d) (EX)2=Z10x26x(1 x)dx 14= 6x44 6x55 x=1x=0 14=120:e) (EY)2=Z10y22y dy 49= 2y44 y=1y=0 49=118 ; X2; ; Xnbe independent, each withtheEx( ) distribution(soeach hasmean1= ).
5 LetV= minfXigandW= maxfXigbe theirminimumandmaximum,respectively. Findthejoint density 0 v w <1we haveP[v < Xi w] =e v e w, soP[V > v;W w] =Ph\ni=1[v < Xi w]i=nYi=1 e v e w = e v e w n;soF(v; w) =P[V v;W w]=P[V >0;W w] P[V > v;W w]=[1 e w]n [e v e w]nf(v; w) =@2@v @wF(v; w) =n(n 1) 2[e v e w]n 2e (v+w);0 v w <1;f(v; w) = 0 foru; thetimesof the rstandfourtharrival in a Poissonprocesswithrate , as in in of Homework7. For0< s < t <1, letXbe thenumber of events in theperiod (0; s] andletYbe thenumber of events in thetimeperiod (s; t]; noteXandYhaveindependent Poissondistributionswithmeans sand (t s), )Expresstheevents[s < T1] and[T4 t] in termsof therandomvariablesXandY. Youneedn' [s < T1] = [X= 0];the rstarrival is laterthansif andonlyif no arrivalshave comeby times.)
6 Theevents [T4 t] = [X+Y 4] areidentical,assertingthatthefourthevent occuredbeforetimetor,equivalently, thatthetotalnumber of eventsX+Yduringthetimeinterval (0; t] = (0; s][(s; t]is at )Expresstheevent [s < T1; T4 t] in termsof therandomvariablesXandY.[s < T1; T4 t] = [X= 0; Y 4].c)Expresstheevent [T1 s; T4 t] in termsof therandomvariablesXandY.[T1 s; T4 t] =[T1>0; T4 t]\[T1> s; T4 t]c=[X+Y 4]\[X= 0; Y 4]c=[X >0; X+Y 4]d)(extracredit)Findthejoint pdf numbers0 s t <1andletXbe thenumber of events intheinterval (0; s] andletYbe thenumber of events in theinterval (s; t];thesehave independent Poissondistributionswithmeans sand (t s),respectivelyandtheirsumZ=X+Y, thenumber of events in theinterval(0; t], hasa Poissondistributionwithmean t.)))
7 NowP[T1 s; T4 t] =P[T4 t] P[T1> s; T4 t]=P[Z 4] P[X= 0; Y 4]=1Xk=4e t( t)kk! e s1Xk=4e (t s) k(t s)kk!=e t1Xk=4 k[tk (t s)k]k!f(s; t) =@@t e t1Xk=4(t s)k 1(k 1)! k!= e t1Xk=4(t s)k 1(k 1)! k+e t1Xk=4(t s)k 2(k 2)! k=e t 4(t s)2=2;0< s < t <1;6thepdf is zerofors.
