TRANSFORMATIONS OF RANDOM VARIABLES

TRANSFORMATIONS OF RANDOM VARIABLES 1. INTRODUCTION 1.1. Definition. We are often interested in the probability distributions or densities of functions of oneormorerandomvariables.supposewehaveasetofrandomvariables,x 1,X 2,X 3,...X n,with a known joint probability and/or density function. We may want to know the distribution of some functionoftheserandomvariablesyφ(x 1,X 2,X 3,...X n ).Realizedvaluesofywillberelatedto realized values of the X s as follows y Φ (x 1, x 2, x 3,..., x n ) (1) A simple example might be a single random variable x with transformation y Φ (x) log (x) (2) 1.2. Techniques for finding the distribution of a transformation of random variables. 1.2.1.Distributionfunctiontechnique. Wefindtheregioninx 1,x 2,x 3,...x n spacesuchthat Φ(x 1, x 2,...x n ) φ.wecanthenfindtheprobabilitythat Φ(x 1,x 2,...x n ) φ,i.e.,p[ Φ(x 1,x 2,...x n ) φ]byintegratingthedensityfunctionf(x 1,x 2,...x n )overthisregion.ofcourse,f Φ (φ)isjust P[ Φ φ].oncewehavef Φ (φ),wecanfindthedensitybyintegration. 1.2.2. Method of transformations(inverse mappings). Suppose we know the density function of x. Also supposethatthefunction y Φ (x)isdifferentiableandmonotonicforvalueswithinitsrangefor whichthedensityf(x).thismeansthatwecansolvetheequation y Φ (x)forxasafunction ofy.wecanthenusethisinversemappingtofindthedensityfunctionofy.wecandoasimilar thingwhenthereismorethanonevariablexandthenthereismorethanonemapping Φ. 1.2.3. Method of moment generating functions. There is a theorem(casella[2, p. 65]) stating that if two random variables have identical moment generating functions, then they possess the same probability distribution. The procedure is to find the moment generating function for Φ and then compareittoanyandallknownonestoseeifthereisamatch.thisismostcommonlydonetosee if a distribution approaches the normal distribution as the sample size goes to infinity. The theorem is presented here for completeness. Theorem1. LetF X (x)andf Y (y)betwocumulativedistributionfunctionsallofwhosemomentsexist. Then a:ifxandyhaveboundedsupport,thenf X (u)f Y (u)foralluifandonlyifex r EY r forall integersr,1,2,.... b:ifthemomentgeneratingfunctionsexistandm X (t)m Y (t)foralltinsomeneighborhoodof, thenf X (u)f Y (u)forallu. For further discussion, see Billingsley[1, ch. 21-22]. Date: February 18, 28. 1

2 TRANSFORMATIONS OF RANDOM VARIABLES 2. DISTRIBUTION FUNCTION TECHNIQUE 2.1. Procedure for using the Distribution Function Technique. As stated earlier, we find the regioninthex 1,x 2,x 3,...x n spacesuchthat Φ(x 1,x 2,...x n ) φ.wecanthenfindtheprobability that Φ(x 1,x 2,...x n ) φ,i.e.,p[ Φ(x 1,x 2,...x n ) φ]byintegratingthedensityfunctionf(x 1,x 2,...x n )overthisregion.ofcourse,f Φ (φ)isjustp[ Φ φ].oncewehavef Φ (φ),wecanfindthe density by differentiation. 2.2.Example1.LettheprobabilitydensityfunctionofXbegivenby { 6 x(1 x) < x < 1 f(x) otherwise NowfindtheprobabilitydensityofYX 3. (3) LetG(y)denotethevalueofthedistributionfunctionofYatyandwrite G(y ) P (Y y ) P (X 3 y ) P (X ) y 1/3 y 1/3 6 x (1 x)dx (4) y 1/3 ( 6 x 6 x 2 ) dx ( 3 x 2 2 x 3 ) y1/3 3 y 2/3 2y Now differentiate G(y) to obtain the density function g(y) g(y) dg(y) d ( 3 y 2/3 2 y 2 y 1/3 2 ) (5) 2 (y 1/3 1 ), < y < 1 2.3.Example2.Lettheprobabilitydensityfunctionofx 1 andofx 2 begivenby f(x 1, x 2 ) { 2 e x 1 2 x 2 x 1 >, x 2 > otherwise NowfindtheprobabilitydensityofYX 1 +X 2 orx 1 Y-X 2.GiventhatYisalinearfunction ofx 1 andx 2,wecaneasilyfindF(y)asfollows. (6)

TRANSFORMATIONS OF RANDOM VARIABLES 3 LetF Y (y)denotethevalueofthedistributionfunctionofyatyandwrite F Y (y) P (Y y) y y y y y y x2 Nowintegratewithrespecttox 2 asfollows 2 e x1 2 x2 dx 1 dx 2 2 e x1 2 x2 y x2 dx 2 [( 2 e y + x 2 2 x 2 ) ( 2 e 2x2 ) ] dx 2 2 e y x2 + 2 e 2x2 dx 2 2 e 2 x2 2 e y x2 dx 2 (7) F Y (y) P (Y y) y 2 e 2x2 2 e y x2 dx 2 e 2x2 + 2 e y x2 y e 2y + 2 e y y [ e + 2 e y] (8) e 2y 2 e y + 1 NowdifferentiateF Y (y)toobtainthedensityfunctionf(y) f Y (y) df (y) d ( e 2y 2 e y + 1 ) 2 e 2y + 2 e y (9) 2 e 2 y ( 1 + e y ) 2.4.Example3.LettheprobabilitydensityfunctionofXbegivenby f X (x) 1 σ 2π e 1 2 ( x µ σ ) 2, < x < [ exp 1 2 π σ 2 ] (x µ )2 2 σ 2, < x < NowletYΦ(X)e X.WecanthenfindthedistributionofYbyintegratingthedensityfunction ofxovertheappropriateareathatisdefinedasafunctionofy.letf Y (y)denotethevalueofthe distributionfunctionofyatyandwrite (1)

4 TRANSFORMATIONS OF RANDOM VARIABLES F Y (y) P (Y y) P ( e X y ) P (X lny), y > (11) ln y [ ] 1 (x µ )2 exp 2 π σ 2 2 σ 2 dx, y > NowdifferentiateF Y (y)toobtainthedensityfunctionf(y).inthiscasewewillneedtherulesfor differentiating under the integral sign. They are given by theorem 2 which we state below without proof. Theorem2.Supposethatfand f x arecontinuousintherectangle R {(x, t) : a x b, c t d} andsupposethatu (x)andu 1 (x)arecontinuouslydifferentiablefora x bwiththerangeofu (x)and u 1 (x)in(c,d).if ψisgivenby then ψ (x) u1 (x) u (x) f(x, t)dt (12) dψ dx x u1(x) u (x) f(x, t)dt f (x, u 1 (x)) du 1(x) dx f (x, u (x)) du (x) dx + u1(x) u (x) f(x, t) dt x Ifoneoftheboundsofintegrationdoesnotdependonx,thentheterminvolvingitsderivativewillbezero. Foraproofoftheorem2see(Protter[3,p.425]).Applyingthistoequation11whereytakesthe roleofx, lnytakestheroleofu 1 (x),andxtakestheroleoftinthetheoremweobtain F Y (y) ln y F Y (y) f Y (y) + ln y [ 1 exp 2πσ 2 ( 1 2πσ 2 exp [ d ( 1 [ y 2πσ exp 2 The last line of equation 14 follows because ( [ 1 exp 2πσ 2 d ( 1 2πσ 2 exp [ ] (x µ)2 2σ 2 dx, y > ] ( ) ) (lny µ)2 1 2σ 2 y ]) (x µ)2 2σ 2 dx ]) (lny µ)2 2σ 2 ] ) (x µ)2 2σ 2 (13) (14)

TRANSFORMATIONS OF RANDOM VARIABLES 5 3. METHOD OF TRANSFORMATIONS(SINGLE VARIABLE) 3.1. Discrete examples of the method of transformations. 3.1.1. One-to-one function. Find a formula for the probability distribution of the total number of headsobtainedinfourtossesofacoinwheretheprobabilityofaheadis.6. Thesamplespace,probabilitiesandthevalueoftherandomvariablearegivenintable1. TABLE 1. Outcomes, Probabilities and Number of Heads from Tossing a Coin Four Times. Value of random variable X Element of sample space Probability (x) HHHH 81/625 4 HHHT 54/625 3 HHTH 54/625 3 HTHH 54/625 3 THHH 54/625 3 HHTT 36/625 2 HTHT 36/625 2 HTTH 36/625 2 THHT 36/625 2 THTH 36/625 2 TTHH 36/625 2 HTTT 24/625 1 THTT 24/625 1 TTHT 24/625 1 TTTH 24/625 1 TTTT 16/625 From the table we can determine the probabilities as P(X ) 16 96 216 216 81, P(X 1), P(X 2), P (X 3), P (X 4) 625 625 625 625 625 We can also compute these probabilities using counting rules. The probability of one head and then three tails is ( ) ( ) ( ) ( ) 3 2 2 2 5 5 5 5 or ( ) 1 ( 3 2 5 5 Theprobabilityof3headsandthenonetailis ( ) ( 3 3 5 5 or ) 3 24 625 ) ( 3 5 ) ( ) 2 5

6 TRANSFORMATIONS OF RANDOM VARIABLES ( ) 3 ( ) 1 3 2 54 5 5 625. Ofcoursethereareotherwaystoobtain1headandthreetailsbesidesoneheadandthenthree tails.inparticularthereare ( ( 4 1) 4waystoobtainonehead.Andthereare 4 ) 1waytoobtain zeroheads.similarly,therearesixwaystoobtaintwoheads,fourwaystoobtainthreeheadsand onewaytoobtainfourheads.wecanthenwritetheprobabilitymassfunctionas f(x) ( ) ( ) x ( ) 4 x 4 3 2 for x, 1, 2, 3, 4 (15) x 5 5 This, of course, is the binomial distribution. The probabilities of the various possible random variables are contained in table 2. TABLE2. ProbabilityofNumberofHeadsfromTossingaCoinFourTimes Number of Heads x Probability f(x) 16/625 1 96/625 2 216/625 3 216/625 4 81/625 NowconsideratransformationofXintheformY2X 2 +X.Therearefivepossibleoutcomes fory,i.e.,,3,1,21,36.giventhatthefunctionisone-to-one,wecanmakeupatabledescribing the probability distribution for Y. TABLE3. ProbabilityofaFunctionoftheNumberofHeadsfromTossingaCoin Four Times. Y2*(#heads) 2 +#ofheads Number of Heads x Probability f(x) y g(y) 16/625 16/625 1 96/625 3 96/625 2 216/625 1 216/625 3 216/625 21 216/625 4 81/625 36 81/625 3.1.2.Casewherethetransformationisnotone-to-one.NowletthetransformationofXbegivenbyZ (6-2X) 2.ThepossiblevaluesforZare,4,16,36.WhenX2andwhenX4,Y4.Wecan findtheprobabilityofzbyaddingtheprobabilitiesforcaseswhenxgivesmorethanonevalueas shownintable4.

TRANSFORMATIONS OF RANDOM VARIABLES 7 TABLE4. ProbabilityofaFunctionoftheNumberofHeadsfromTossingaCoin Four Times(not one-to-one). y Y(6-(#heads)) 2 Number of Heads x g(y) 216 3 625 216 4 2, 4 625 + 81 625 297 625 96 16 1 625 16 36 625 3.2. Intuitive Idea of the Method of Transformations. The idea of a transformation is to consider thefunctionthatmapstherandomvariablexintotherandomvariabley.theideaisthatifwecan determinethevaluesofxthatleadtoanyparticularvalueofy,wecanobtaintheprobabilityofy bysummingtheprobabilitiesofthosevaluesofxthatmappedintoy.inthecontinuouscase,to findthedistributionfunction,wewanttointegratethedensityofxovertheportionofitsspace thatismappedintotheportionofyinwhichweareinterested.supposeforexamplethatbothx andyaredefinedonthereallinewith X 1and Y 1.IfwewanttoknowG(5),weneed tointegratethedensityofxoverallvaluesofxleadingtoavalueofylessthanfive,whereg(y)is theprobabilitythatyislessthanfive. 3.3. General formula when the random variable is discrete. Consider a transformation defined byyφ(x).thefunction ΦdefinesamappingfromthesamplespaceofthevariableX,toasample space for the random variable Y. IfXisdiscretewithfrequencyfunctionp X,then Φ(X)isdiscreteandhasfrequencyfunction p Φ(X) (t) p X (x) x: Φ(x)t p X (x) xǫφ 1 (t) (16) Theprocessissimpleinthiscase. Oneidentifiesg 1 (t)foreachtinthesamplespaceofthe randomvariabley,andthensumstheprobabilitieswhichiswhatwedidinsection3.1. 3.4. General change of variable or transformation formula. Theorem3. Letf X (x)bethevalueoftheprobabilitydensityofthecontinuousrandomvariablexatx.if the function y Φ(x) is differentiable and either increasing or decreasing(monotonic) for all values within therangeofxforwhichf X (x),thenforthesevaluesofx,theequationyφ(x)canbeuniquelysolved forxtogivexφ 1 (y)w(y)wherew( )Φ 1 ( ).Thenforthecorrespondingvaluesofy,theprobability densityofyφ(x)isgivenby

8 TRANSFORMATIONS OF RANDOM VARIABLES [ f X Φ 1 (y) ] dφ 1 (y) f g(y) f Y (y) X [ w (y)] d w (y) dφ(x) dx f X [ w (y)] w (y) otherwise Proof. Consider the digram in figure 1. (17) FIGURE1. yφ(x)isanincreasingfunction. Ascanbeseenfrominfigure1,eachpointontheyaxismapsintoapointonthexaxis,that is,xmusttakeonavaluebetween Φ 1 (a)and Φ 1 (b)whenytakesonavaluebetweenaandb. Therefore P(a < Y < b) P ( Φ 1 (a) < X < Φ 1 (b) ) Φ 1 (b) Φ 1 (a) f X (x)dx (18)

TRANSFORMATIONS OF RANDOM VARIABLES 9 Whatwewouldliketodoisreplacexinthesecondlinewithy,and Φ 1 (a)and Φ 1 (b)witha andb. Todosoweneedtomakeachangeofvariable.Considerhowwemakeausubstitution whenweperformintegrationorusethechainrulefordifferentiation.forexampleifuh(x)then duh (x)dx.soifxφ 1 (y),then Thenwecanwrite dx d Φ 1 (y ) ( f X (x) dx f X Φ 1 (y) ) d Φ 1 (y) (19) For the case of a definite integral the following lemma applies. Lemma1.Ifthefunctionuh(x)hasacontinuousderivativeontheclosedinterval[a,b]andfiscontinuous ontherangeofh,then b a f (h(x)) h (x)dx. h(b) h (a) f (u)du (2) Usingthislemmaortheintuitionfromequation19wecanthenrewriteequation18asfollows P (a < Y < b ) P ( Φ 1 (a) < X < Φ 1 (b) ) Φ 1 ( b ) f Φ 1 (a) X (x) dx b a f ( X Φ 1 (y ) ) d Φ 1 (y ) d y Theprobabilitydensityfunction,f Y (y),ofacontinuousrandomvariableyisthefunctionf( ) that satisfies P (a < Y < b) F (b) F (a) b a (21) f Y (t)dt (22) Thisthenimpliesthattheintegrandinequation21isthedensityofY,i.e.,g(y),soweobtain g (y) f Y (y) f X [ Φ 1 (y) ] d Φ 1 (y) d y (23) aslongas d Φ 1 (y) exists.thisprovesthelemmaif Φisanincreasingfunction.Nowconsiderthecasewhere Φisa decreasing function as in figure 2. Ascanbeseenfromfigure2,eachpointontheyaxismapsintoapointonthexaxis,thatis, Xmusttakeonavaluebetween Φ 1 (a)and Φ 1 (b)whenytakesonavaluebetweenaandb. Therefore P (a < Y < b) P ( Φ 1 (b) < X < Φ 1 (a) ) (24) Φ 1 ( a) Φ 1 ( b ) f X (x)dx

1 TRANSFORMATIONS OF RANDOM VARIABLES FIGURE2. yφ(x)isadecreasingfunction. MakingachangeofvariableforxΦ 1 (y)asbeforewecanwrite P (a < Y < b) P ( Φ 1 (b) < X < Φ 1 (a) ) (25) Because Φ 1 (a) Φ 1 (b) a b f X (x)dx ( f X Φ 1 (y) ) d Φ 1 (y) b a f X ( Φ 1 (y) ) d Φ 1 (y) d Φ 1 (y) whenthefunctionyφ(x)isincreasingand dx 1 dx

TRANSFORMATIONS OF RANDOM VARIABLES 11 d Φ 1 (y) ispositivewhenyφ(x)isdecreasing,wecancombinethetwocasesbywriting g (y) f Y (y) f X [ Φ 1 (y) ] d Φ 1 (y ) d y (26) 3.5. Examples. 3.5.1. Example 1. Let X have the probability density function given by { 1 2 f X (x) x, x 2, elsewhere FindthedensityfunctionofYΦ(X)6X-3. (27) Noticethatf X (x)ispositiveforallxsuchthat x 1.Thefunction ΦisincreasingforallX. Wecanthenfindtheinversefunction Φ 1 asfollows y 6 x 3 6 x y + 3 x y + 3 6 Φ 1 (y) Wecanthenfindthederivativeof Φ 1 withrespecttoyas dφ 1 d ( ) y + 3 6 (28) (29) 1 6 Thedensityofyisthen g(y) f Y (y) f X [ Φ 1 (y) ] ( ) ( ) 1 3 + y 1 2 6 6, dφ 1 (y) 3 + y 6 Forallothervaluesofy,g(y).Simplifyingthedensityandtheboundsweobtain 2 (3) g(y) f Y y) { 3 + y 72, 3 y 9 elsewhere (31)

12 TRANSFORMATIONS OF RANDOM VARIABLES 3.5.2.Example2.LetXhavetheprobabilitydensityfunctiongivenbyf X (x). Thenconsiderthe transformationyφ(x)σx+µ, σ.thefunction ΦisincreasingforallX.Wecanthenfind theinversefunction Φ 1 asfollows y σ x + µ σ x y µ x y µ Φ 1 (y) σ Wecanthenfindthederivativeof Φ 1 withrespecttoyas Thedensityofyisthen dφ 1 d 1 σ ( y µ [ f Y (y) f X Φ 1 (y) ] d Φ 1 (y ) [ ] y µ f X 1 σ σ 3.5.3. Example 3. Let X have the probability density function given by σ ) (32) (33) (34) f X (x) FindthedensityfunctionofYX 1/2. { e x, x, elsewhere (35) Noticethatf X (x)ispositiveforallxsuchthat x.thefunction ΦisincreasingforallX. Wecanthenfindtheinversefunction Φ 1 asfollows y x 1 2 y 2 x (36) x Φ 1 (y) y 2 Wecanthenfindthederivativeof Φ 1 withrespecttoyas Thedensityofyisthen d Φ 1 d y2 2y f Y (y) f X [ Φ 1 (y) ] e y2 2y dφ 1 (y) (37) (38)

TRANSFORMATIONS OF RANDOM VARIABLES 13 Agraphofthetwodensityfunctionsisshowninfigure3. 1 FIGURE 3. The Two Density Functions. Value of Density Function.8.6.4.2 f x g y 1 2 3 4

14 TRANSFORMATIONS OF RANDOM VARIABLES 4. METHOD OF TRANSFORMATIONS(MULTIPLE VARIABLES) 4.1.Generaldefinitionofatransformation.Let ΦbeanyfunctionfromR k tor m,k,m 1,such that Φ 1 (A){x ǫr k : Φ(x)ǫA} ǫß k foreveryaǫß m whereß m isthesmallest σ-fieldhavingall theopenrectanglesinr m asmembers.ifwewriteyφ(x),thefunction Φdefinesamappingfrom thesamplespaceofthevariablex(ξ)toasamplespace(y)oftherandomvariable Ψ.Specifically and Φ (x) : Ξ Ψ (39) Φ 1 (A) { x ǫ Ψ : Φ (x) ǫ A } (4) 4.2. Transformations involving multiple functions of multiple random variables. Theorem4. Let f X1 X 2 (x 1 x 2 )bethevalueofthejointprobabilitydensityofthecontinuousrandom variablesx 1 andx 2 at(x 1,x 2 ).Ifthefunctionsgivenbyy 1 u 1 (x 1,x 2 )an 2 u 2 (x 1,x 2 )arepartially differentiablewithrespecttox 1 andx 2 andrepresentaone-to-onetransformationforallvalueswithinthe rangeofx 1 andx 2 forwhich f X1 X 2 (x 1 x 2 ),then,forthesevaluesofx 1 andx 2,theequations y 1 u 1 (x 1,x 2 )an 2 u 2 (x 1,x 2 )canbeuniquelysolvedforx 1 andx 2 togive x 1 w 1 (y 1, y 2 )and x 2 w 2 (y 1, y 2 )andforcorrespondingvaluesofy 1 an 2,thejointprobabilitydensityofY 1 u 1 (X 1,X 2 ) andy 2 u 2 (X 1,X 2 )isgivenby f Y1 Y 2 (y 1 y 2 ) f X1 X 2 [ w 1 (y 1, y 2 ), w 2 (y 1 y 2 ) ] J (41) wherejisthejacobianofthetransformationandisdefinedasthedeterminant J x 1 x 1 y 1 y 2 x 2 x 2 y 1 y 2 (42) Atallotherpoints f Y1Y 2 (y 1 y 2 ). 4.3.Example.Lettheprobabilitydensityfunctionof X 1 and X 2 begivenby f X1X 2 (x 1, x 2 ) { e (x 1 + x 2) x 1, x 2 elsewhere (43) ConsidertworandomvariablesY 1 andy 2 bedefinedinthefollowingmanner. Y 1 X 1 + X 2 Y 2 X 1 (44) X 1 + X 2 TofindthejointdensityofY 1 andy 2 wefirstneedtosolvethesystemofequationsinequation 44forX 1 andx 2.

TRANSFORMATIONS OF RANDOM VARIABLES 15 Y 1 X 1 + X 2 X 1 Y 2 X 1 + X 2 X 1 Y 1 X 2 Y 1 X 2 Y 2 Y 1 X 2 + X 2 Y 2 Y 1 X 2 Y 1 (45) TheJacobianisgivenby Y 1 Y 2 Y 1 X 2 X 2 Y 1 Y 1 Y 2 Y 1 (1 Y 2 ) X 1 Y 1 ( Y 1 Y 1 Y 2 ) Y 1 Y 2 x 1 x 1 y J 1 y 2 x 2 x 2 y 1 y 2 y 2 y 1 1 y 2 y 1 y 2 y 1 y 1 (1 y 2 ) y 2 y 1 y 1 + y 1 y 2 (46) y 1 Thistransformationisone-to-oneandmapsthedomainofX(Ξ)givenbyx 1 > andx 2 > in thex 1 x 2 -planeintothedomainofy(ψ)inthey 1 y 2 -planegivenbyy 1 > and < y 2 < 1.Ifwe apply theorem 4 we obtain f Y1Y 2 (y 1 y 2 ) f X1 X 2 [ w 1 (y 1, y 2 ), w 2 ( y 1 y 2 ) ] J e ( y1 y2 + y1 y1 y2 ) y 1 e y1 y 1 y 1 e y1 Consideringallpossiblevaluesofvaluesofy 1 an 2 weobtain { y1 e y1 for y 1, < y 2 < 1 f Y1 Y 2 (y 1, y 2 ) elsewhere WecanthenfindthemarginaldensityofY 2 byintegratingovery 1 asfollows (47) (48) f Y2 (y 1, y 2 ) f Y1 Y 2 (y 1, y 2 ) 1 y 1 e y1 1 (49)

16 TRANSFORMATIONS OF RANDOM VARIABLES Wemakeauvsubstitutiontointegratewhereu,v,du,anddvaredefinedas This then implies u y 1 v e y1 du 1 dv e y1 1 (5) f Y2 (y 1, y 2 ) forally 2 suchthat < y 2 < 1. y 1 e y1 1 y 1 e y1 ( ) ( e y1 ) ( e ) y1 ( e e ) + 1 1 e y1 1 (51) Agraphofthejointdensitiesandthemarginaldensityfollows.Thejointdensityof(X 1,X 2 )is showninfigure4. x 1 4 3 1 2 FIGURE4. JointDensityof X 1 and X 2..2.15 f 12 x 1,x 2.1.5 1 2 3 4 x 2 Thisjointdensityof(Y 1,Y 2 )iscontainedinfigure5.

TRANSFORMATIONS OF RANDOM VARIABLES 17 y 1 8 6 2 4 FIGURE5. JointDensityof Y 1 and Y 2..3.2 f y 1,y 2.1.25.5.75 1 y 2 ThismarginaldensityofY 2 isshowngraphicallyinfigure6. y 1 8 6 2 4 FIGURE6. MarginalDensityof Y 2. 2 1.5 1 f 2 y 2.5.25.5.75 1 y 2

18 TRANSFORMATIONS OF RANDOM VARIABLES REFERENCES [1] Billingsley, P. Probability and Measure. 3rd edition. New York: Wiley, 1995 [2] Casella, G. and R.L. Berger. Statistical Inference. Pacific Grove, CA: Duxbury, 22 [3] Protter, Murray H. and Charles B. Morrey, Jr. Intermediate Calculus. New York: Springer-Verlag, 1985