Chapter 2 Multivariate Distributions - MyWeb

1 1/115 Chapter 2 Multivariate Distributions of Two Random VariablesBoxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 20182/115 Bivariate random vectorDefinitionArandom variableis a function from a sample vectoris a function random vector is also called a bivariate :X= (X1,X2) assigns to each elementcof thesample spaceCexactly one ordered pair of numbersX1(c) =x1andX2(c) = and weight of consumption and hours on an Wang, The University of IowaChapter 2 STAT 4100 Fall 2018 Discrete Random Variables3/1153/115 Joint probability mass functionDefinitionAjoint probability mass functionpX1,X2(x1,x2) =p(X1=x1,X2=x2)(orp(x1,x2))with space(x1,x2) Shas the properties that(a)0 p(x1,x2) 1,(b) (x1,x2) Sp(x1,x2) = 1,(c)P[(X1,X2) A] = (x1,x2) Ap(x1,x2).Boxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 20184/115 ExampleA restaurant serves three fixed-price dinners costing$7,$9, and$10. For a randomly selected couple dinning at this restaurant, letX1=the cost of the man s dinner andX2=the cost of the woman s joint pmf ofX1andX2is given in the following is the probability ofP(X1 9,X2 9)?

2 Man s dinner cost more?Boxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 20185/115 Marginal probability mass functionDefinitionSuppose thatX1andX2have the joint pmfp(x1,x2). Then thepmf forXi, denoted bypi( ),i= 1,2is themarginal (x1) = x2p(x1,x2)andp2(x2) = x1p(x1,x2).Example Find the marginal pmf of the previous Wang, The University of IowaChapter 2 STAT 4100 Fall 20186/115 ExampleLetX1=Smaller die face,X2=Larger die face, when rolling a pairof two dice. The following table shows a partition of the samplespace 0000022/36 1/36 0000x232/36 2/36 1/36 00042/36 2/36 2/36 1/36 0052/36 2/36 2/36 2/36 1/36 062/36 2/36 2/36 2/36 2/36 1/36 Find the marginal pmf Wang, The University of IowaChapter 2 STAT 4100 Fall 20187/115 Expectation discrete random variablesDefinitionLetY=u(X1,X2). Then,Yis a random variable andE[u(X1,X2)] = x1 x2u(x1,x2)p(x1,x2)under the condition that x1 x2|u(x1,x2)|p(x1,x2)|< ExampleFindE(max{X1,X2})for the restaurant Wang, The University of IowaChapter 2 STAT 4100 Fall 2018 Continuous Random Variables8/1158/115 Joint density functionAjoint density functionfX1,X2(x1,x2)(orf(x1,x2)) with space(x1,x2) Shas the properties that(a)f(x1,x2)>0,(b) (x1,x2) Sf(x1,x2)dx1dx2= 1,(c)P[(X1,X2) A] = (x1,x2) Af(x1,x2) Wang, The University of IowaChapter 2 STAT 4100 Fall 20189/115 ExampleLetX1andX2be continuous random variables with joint densityfunctionf(x1,x2) ={4x1x2for0< x1,x2< (1/4< X1<3/4; 1/2< X2<1).}

3 2 FindP(X1< X2).3 FindP(X1+X2<1).Solution: 11/2 3/41/44x1x2dx1dx2= 3/8 = 10 x204x1x2dx1dx2= 1/2 = 10 1 x204x1x2dx1dx2= 1/6 = Wang, The University of IowaChapter 2 STAT 4100 Fall 201810/115 Marginal probability density functionSuppose thatX1andX2have the joint pdff(x1,x2). Then the pdfforXi, denoted byfi( ),i= 1,2is themarginal :f1(x1) = x2f(x1,x2)dx2andf2(x2) = x1f(x1,x2) the marginal pdf from the previous :f1(x) =f2(x) = Wang, The University of IowaChapter 2 STAT 4100 Fall 201811/115 ExampleLetX1andX2be continuous random variables with joint densityfunctionf(x1,x2) ={cx1x2for0< x1< x2< (X1+X2<1).3 Find marginal probability density function Wang, The University of IowaChapter 2 STAT 4100 Fall 201812/115 Solution:We havec= 8because 10 1x1x1x2dx1dx2= 1/8 = 1/20 1 x1x18x1x2dx1dx2= 1/6 = the marginal pdf, we havefX1(x1) = 1x18x1x2dx2= 4x1 4x31,fX2(x2) = x208x1x2dx1= Wang, The University of IowaChapter 2 STAT 4100 Fall 201813/115 LetX1andX2be continuous random variables with joint pdff(x1,x2) ={cx1x2for0< x1< x2< isP{[X1< X2] [X2>4(X1 1/2)2]}?}}

4 Solution:We see1/4is the solution ofx= 4(x 12)2on0< x <1. Therange ofX2is(1/4,1). WhenX2=x2is given, we next get therange ofX1. ByX2= 4(X1 1/2)2, we haveX1=12 determine the lower bound ofX1is12 X24because theintersection ofX1=X2andX2= 4(X1 1/2)2is less than1/2whenX1 (0,1). We also haveX1<1, so the probability is 114 x112 x248x1x2dx1dx2= Wang, The University of IowaChapter 2 STAT 4100 Fall 201814/115 Expectation continuous random variablesLetY=u(X1,X2). Then,Yis a random variable andE[u(X1,X2)] = x1 x2u(x1,x2)f(x1,x2)dx2dx1under the condition that x1 x2|u(x1,x2)|f(x1,x2)dx2dx1< Boxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 201815/115 ExampleLetX1andX2be continuous random variables with joint densityfunctionf(x1,x2) = (36/5)x1x2(1 x1x2)for0< x1,x2< (X1X2).Solution: 10 10365(x21x22(1 x1x2))dx1dx2= Wang, The University of IowaChapter 2 STAT 4100 Fall 201816/115 TheoremLet(X1,X2)be a random vector.

5 LetY1=g1(X1,X2)andY2=g2(X1,X2)be random variables whose expectations for all real numbersk1andk2,E(k1Y1+k2Y2) =k1E(Y1) +k2E(Y2).We also note thatEg(X2) = g(x2)f(x1,x2)dx1dx2= g(x2)fX2(x2) Wang, The University of IowaChapter 2 STAT 4100 Fall 201817/115 Example & (X1,X2)be a random vector with pdff(x1,x2) ={8x1x20< x1< x2< 7X1X22+ 5X2andY2=X1/X2. DetermineE(Y1)andE(Y2).Boxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 2018 Discrete & Continuous cumulative distribution functionDefinitionThejoint cumulative distribution functionof(X1,X2)isFX1,X2(x1,x2) =P[{X1 x1} {X2 x2}]for all(x1,x2) with pmf and pdf:1 Discrete random variables:FX1,X2(x1,x2) = X1 x1 X2 x2p(x1,x2).2 Continuous random variables:FX1,X2(x1,x2) = x10 x20fX1,X2(x1,x2) Wang, The University of IowaChapter 2 STAT 4100 Fall 201819/115 Joint cumulative distribution function (cont d)DefinitionThejoint cumulative distribution functionof(X1,X2)isFX1,X2(x1,x2) =P[{X1 x1} {X2 x2}]for all(x1,x2) :1F(x1,x2)is nondecreasing ( ,x2) =F(x1, ) = ( , ) = a rectangle(a1,b1] (a2,b2], we haveP{(X1,X2) (a1,b1] (a2,b2]}=F(b1,b2) F(a1,b2) F(b1,a2) +F(a1,a2).))))}

6 Boxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 201820/115 Example the discrete random vector(X1,X2). Its pmf is given inthe following table:X1\X2012301/8 1/800102/8 2/802001/8 1/8 Find the value of the joint cdfF(x1,x2)at(1,2).Solution:3 Wang, The University of IowaChapter 2 STAT 4100 Fall 201821/115 Example1. Find the joint cdf offX1,X2(x1,x2) ={2e x1 x20< x1,x2< :FX1,X2(x1,x2) = x10 x202e t1 t2dt1dt2= 2(1 e x1)(1 e x2).2. Find the joint cdf offX1,X2(x1,x2) ={2e x1 x20< x1< x2< :FX1,X2(x1,x2) = min(x1,x2)0 x2t12e t1 Wang, The University of IowaChapter 2 STAT 4100 Fall 201822/115 Moment generating function (mgf)DefinitionLetX= (X1,X2)>be a random vector. IfM(t1,t2) =E(et1X1+t2X2)exists for|t1|< h1and|t2|< h2, whereh1andh2are positive,then we callM(t1,t2)themoment generating function(mgf) ofX= (X1,X2)>.We may writeM(t1,t2) =E(et1X1+t2X2)=E(et>X)wheret>is a row vector(t1,t2)andXis a column vector(X1,X2)>.}}

7 Boxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 201823/115 Example the continuous-type random variablesXandYhave the jointpdff(x,y) ={e y0< x < y < the joint :MX,Y(t1,t2) = 0 xexp(t1x+t2y y)dydx=1(1 t1 t2)(1 t2),provided thatt1+t2<1andt2< Wang, The University of IowaChapter 2 STAT 4100 Fall 201824/115 Marginal mgfRecall thatMX1,X2(t1,t2) =E(et1X1+t2X2).The marginal mgf is given byMX1(t1) =E(et1X1)=MX1,X2(t1,0),MX2(t2) =E(et2X2)=MX1,X2(0,t2).Boxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 201825/115 Example (cont d)Let the continuous-type random variablesXandYhave the jointpdff(x,y) ={e y0< x < y < the marginal :MX,Y(t1,t2) = 0 xexp(t1x+t2y y)dydx=1(1 t1 t2)(1 t2),provided thatt1+t2<1andt2< (t1) =MX,Y(t1,0) =11 t1, t1<1,MY(t2) =MX,Y(0,t2) =1(1 t2)2, t2< Wang, The University of IowaChapter 2 STAT 4100 Fall 201826/115 Example (cont d)Let the continuous-type random variablesXandYhave the jointpdff(x,y) ={e y0< x < y < the marginal :MX(t1) =MX,Y(t1,0) =11 t1, t1<1,MY(t2) =MX,Y(0,t2) =1(1 t2)2, t2< thatf1(x) = xe ydy=e x,0< x < ,f2(x) = y0e ydx=ye y,0< y <.}}}

8 Boxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 201827/115 Fact: It can be shown thatE(XY) =dMX,Y(t1,t2)dt1dt2 t1=0,t2= : Method 1: In the previous example,E(XY) = 0 y0xye ydxdy= 2:MX,Y(t1,t2) =1(1 t1 t2)(1 t2),dMX,Y(t1,t2)dt1dt2= t1+ 3t2 3(t2 1)2( t1 t2+ 1)3,where we seedMX,Y(t1,t2)dt1dt2 t1=0,t2=0= 3as Wang, The University of IowaChapter 2 STAT 4100 Fall 201828/115 Chapter 2 Multivariate Transformation: Bivariate Random VariablesBoxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 201829/115 Transformation of discrete random vectorsIAssume there is a one to one mapping betweenX= (X1,X2)>andY= (Y1,Y2)>:Y1=u1(X1,X2),X1=w1(Y1,Y2),Y2=u2 (X1,X2),X2=w2(Y1,Y2).ITransformation ofdiscreterandom variable:pY1,Y2(y1,y2) =pX1,X2(w1(y1,y2),w2(y1,y2)).Boxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 201830/115 Example independent random variables such thatpX(x) = x1x!e 1, x= 0,1,2.

9 AndpY(y) = y2y!e 2, y= 0,1,2,..IFind the pmf ofU=X+ the mgf Wang, The University of IowaChapter 2 STAT 4100 Fall 201831/115 Transformation of continuous random variablesLetJdenote theJacobianof the transformation. This is thedeterminant of the2 2matrix( x1 y1 x1 y2 x2 y1 x2 y2)The determinant isJ(y1,y2) = x1 y1 x2 y2 x1 y2 x2 formula:The joint pdf of thecontinuousrandomvectorY= (Y1,Y2)>isfY1,Y2(y1,y2) =fX1,X2)(w1(y1,y2),w2(y1,y2)) |J(y1,y2)|.Notice the bars around the functionJ, denoting absolute Wang, The University of IowaChapter 2 STAT 4100 Fall 201832/115 ExampleA device containing two key components fails when, and onlywhen, both components fail. The lifetimes,X1andX2, of thesecomponents have a joint pdff(x1,x2) =e x1 x2, wherex1,x2>0and zero otherwise. The costY1, of operating thedevice until failure isY1= 2X1+ the joint pdf ofY1,Y2whereY2= the marginal pdf forY1(Ans:e y1/2 e y1, fory1>0)Boxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 201833/115 Example (X1,X2)has joint pdffX1,X2(x1,x2) ={10x1x220< x < y < Find the joint and marginal pdf s Wang, The University of IowaChapter 2 STAT 4100 Fall 201834/115 Solution sketch1.}

10 One to one transformation:y1=x1/x2, y2=x2,0< x1< x2<1x1=y1y2, x2=y2,0< y1<1,0< y2< Give the joint pdf:fY1,Y2(y1,y2) = 10y1y2y22|y2|,whereyis defined above Give the marginal pdf ofY1:fY1(y1) = 10fY1,Y2(y1,y2)dy2= 2y1,0< y1< Wang, The University of IowaChapter 2 STAT 4100 Fall 201835/115 Example (X1,X2)has joint pdffX1,X2(x1,x2) ={14exp( x1+x22)0< x1< ,0< x2< 1/2(X1 X2)andY2=X2. Find the joint and marginalpdf s Wang, The University of IowaChapter 2 STAT 4100 Fall 201836/115 Solution sketch1. One to one transformation:y1=12(x1 x2), y2=x2,0< x1< ,0< x2< .x1= 2y1+y2, x2=y2, 2y1< y2,0< y2< .2. Give the joint pdf:fY1,Y2(y1,y2) =e y1 y2/4 |2|,whereyis defined above Give the marginal pdf ofY1:fY1(y1) ={ 2y1fY1,Y2(y1,y2)dy2=ey1/2, < y1<0, 0fY1,Y2(y1,y2)dy2=e y1/2,0 y1< ,which givesfY1(y1) =e |y1|, < y < .Boxiang Wang, The University of IowaChapter 2 STAT 4100 Fall 201837/115 Example (X1,X2)has joint pdffX1,X2(x1,x2) ={14exp( x1+x22)0< x1< ,0< x2< 1/2(X1 X2).}}}