Chapter 2 Transformations and Expectations

Transcription

1 Chapter Trasformatios a Epectatios Chapter Distributios of Fuctios of a Raom Variable Problem: Let be a raom variable with cf F ( ) If we efie ay fuctio of, say g( ) g( ) is also a raom variable whose istributio epes o F a the fuctio g, the Discrete Case: Let be a iscrete raom variable with pmf f ( ) I this case is a coutable set Defie g( ) for some fuctio g so that { y: y g( ), } is also a coutable set The the pmf of is give by f ( y) P( y) P( ) f ( ) g ( y) g ( y) Eample (Biomial trasformatio): Eample: Suppose that has the followig pmf: Fi the pmf of ep( ) f ep( ), 0; ep( )/(*!),,, ; ( ) ep( )/(*( )!),,, ; 0,otherwise

2 Solutio: The support set if is A {,ep(),ep(), } For ay y A We have: ep( ), y ; f ( y) P( y) P(ep( ) y) P( log( y)) ep( )/(log( y))!, yep(),ep(), ; 0,otherwise Cotiuous Case: The cf of g( ) is: F ( y) P( y) P( g( ) y) P({ : g( ) y}) f ( ) Sometimes, it ca be ifficult to fi { : g ( ) y} { : g( ) y} Eample (Uiform trasformatio): Suppose has a uiform istributio o the iterval (0, ), that is f /( ) 0 ( ) 0 Otherwise Cosier si ( ) The ( ) ( ) ( ( ) ) ({ :si ( ) }) F y P y P g y P y Defiitio: The support set or the support of a istributio or a raom variable is efie as { : f ( ) 0}

3 Defiitio: A fuctio g is icreasig (or ecreasig) if u v implies gu ( ) gv ( ) (or gu ( ) gv ( )) If g is either icreasig or ecreasig, the g is sai to be a mootoe fuctio Remark: Whe we trasfer { : g( ) y}, we oly ee to cosier the support set of If g is a mootoe, the g is sai to be oe-to-oe a oto from 3 If g is a mootoe, the g ( y) has a sigle value a g is a mootoe fuctio Theorem 3: Let have cf F ( ), let g( ), a let a be { : f ( ) 0} a { y: y g( ) for some }, the a If g is a icreasig fuctio o, the F y F g y for y ( ) ( ( )) b If g is a ecreasig fuctio o a is a cotiuous raom variable, the y F y F g y for ( ) ( ( )) Eample 4 (Uiform-epoetial relatioship-i): Obtai the cf of log( ) if has a uiform istributio at (0,): f ( ) if 0, a f ( ) 0 for other Solutio: The support set of is (0,) a is a cotiuous raom variable Therefore, (0, ) 3 g ( ) logis a ecreasig fuctio a y g ( y) e 3

4 4 F y F g y e y y ( ) ( ( )) ( (0, )) Solutio: We ca have the followig solutio too: F ( y) P( y) P( log( ) y) P( ep( y)) ep( y) for y 0 Questio: How o we get the pf of g( ) from the pf of? Theorem 5: Let have pf f ( ) a let g( ), where g is a mootoe fuctio Let a be { : f ( ) 0} a { y: y g( ) for some } Suppose that f ( ) is cotiuous o a that g ( y) has a cotiuous erivative o The the pf of is Proof: Applicatio of Theorem 3 f f ( g ( y)) g ( y) y ( y) y 0 Otherwise Eample 6: (Iverte gamma pf): Let f ( ) be the gamma pf f e ( )! / ( ), 0<< where is a positive costat a is a positive iteger Let / Obtai the pf f ( y ) of 4

5 Eample: Let have the probability esity fuctio give by f ( y) yfor 0 y a f ( y) 0 otherwise Fi the esity fuctio of U 4 3 Solutio: The support set of U is [,3] The fuctio g( y) 4y 3 is a cotiuous a ecreasig fuctio We have g ( u) (3 u)/4 therefore the pf of U is: u fu( u) f( g ( u)) g ( u) *(3 u)/4*( /4) for u 3 u 8 Solutio: We ca first get the cf a the get pf of U For u 3: (3 u) PU ( u) P( 43 u) P ( (3 u)/4) yy (3 u)/4 6 Thus, 3 u fu( u) FU( u) for u 3 u 8 3 Questio: What happes if g is ot a mootoe trasformatio? Eample 7 (Square trasformatio): Let be a cotiuous raom variable with cf F ( ) a pf f ( ) Fi the cf a pf of Theorem 8: Let have a pf f ( ) a g( ), a efie f ( ), let = g(), a efie the sample space { : f ( ) 0} Suppose there eists a partitio of, A0, A,, Ak, such that P ( A0 ) 0 a f ( ) is cotiuous o each A i Further, suppose there eist fuctios g ( ),, ( ) gk, efie o A,, Ak, respectively, satisfyig 5

6 i g ( ) gi( ), for Ai ii gi( ) is mootoe o A i iii the set { y: y g ( ) for some A} is the same for each i,,, k, a iv i i g ( y) has a cotiuous erivative o, for each i =,,k, the f i k f ( ( )) ( ) i gi y gi y y ( y) y 0 otherwise Note: If ay of the coitios of Theorem 8 is ot satisfie, it will be very ifficult to fi the istributio of g( ) Eample 9 (Normal-chi square relatioship): Let have a staar ormal istributio, ie, / f ( ) e, Show that has pf f ( y) e y y /, 0 y I this case, the pf of is the pf of a chi square raom variable with egree of freeom 6

7 Theorem 0 (Probability itegral trasform): Let have cotiuous cf F ( ) a efie the raom variable as F ( ) The is uiformly istribute o (0, ), ie, P ( y) y, 0 y Proof: Defie F ( y) if{ : F ( ) y} for 0 y, the P ( y) PF ( ( ) y) PF F F y F F ( y) is icreasig (why?); ( [ ( )] ( )) ( is icreasig) P F y ( ( )) F F y F ( ( )) (efiitio of ) y (cotiuity of F ) Note: We require that is a cotiuous raom variable If is ot cotiuous, may properties of i the proof o ot hol aymore F ( y) use Eample: Suppose that a cf i eample 5, what is Solutio: 0 if 0; /8 if 0 ; F ( ) 4/8 if ; 7/8 if 3; if 3 F ( y) F 3 if 7/8< y ; if 4/8<y 7/8; ( y) if /8<y 4/8; 0 if 0<y /8; if y=0 7

8 Basics of Supremum a Ifimum Defiitio: For ay set A (, ), the supremum of A, eote by sup A, is efie as: () A, sup A; () 0, 0 A, such that sup A 0 Defiitio: For ay set A (, ), the ifimum of A, eote by if A, is efie as: () A, if A; () 0, 0 A, such that 0 if A Eample: A {,,,,, } 3, the sup A a if A 0 Eample: A {,,,, }, the sup A a 3 if A Theorem: For ay set A (, ), if M, such that M for A, the sup A eists Theorem: For ay set A (, ), if M, such that M for A, the if A eists Defiitio: If sup A (if A ) oes ot eists, the sup A (if A ) 8

9 Note: Theorem 0 has a very importat applicatio If oe is itereste i geeratig a raom variable from a populatio with cf F ( ), oe oly ees to geerate a uiform raom umber u, betwee 0 a, a solve for i the equatio F ( ) u Eample: Let be a cotiuous raom variable with cf F ( ) e epoetial istributio The if U F ( ), the u F ( ) e log( u) I this case, we say that has a Sectio Epecte Values Defiitio : The epecte value or mea of a raom variable g( ), eote by Eg( ), is Eg( ) gf ( ) ( ) if is cotious; gf ( ) ( ) gp ( ) ( ) if is iscrete provie that the itegral or sum eists If E g( ), we say that Eg( ) oes ot eit Eample (Epoetial mea): Suppose has a epoetial ( ) istributio, that is, it has pf give by Fi the mea of f ( ) e /, 0, 0 9

10 Theorem (Itegratio with parts): If the fuctio u( ) a v( ) have erivatives, a v ( ) u ( ) a u ( ) v ( ) eist, the we have u ( ) v ( ) uv ( ) ( ) v ( ) u ( ) Eample: Suppose has the pf of Solutio: f ( ) E for (5 ) Eample: Suppose has the pf of ( gamma(, )), please fi E f ( ) ( ) ep( / ) for 0 where 0 a 0 Eample 4 (Cauchy mea): A Cauchy rv has pf f ( ) Fi E Solutio:, 0

11 Thus M log( ) M log( M ) 0 0 M log( M ) 0 M E limm lim Questio: Does the itegral eist? 6 Eample: Suppose that ~ f ( ),,, Does E eists? 6 6 Solutio: a (iverget) Theorem 5: Let be a raom variable a let a, b, a c be costats The for ay fuctios g ( ) a g ( ) whose epectatios eist, we have E( ag ( ) bg ( ) c) aeg ( ) beg ( ) c a b If g ( ) 0for all, the Eg ( ) 0 c If g( ) g( ) for all, the Eg( ) Eg( ) If a g ( ) b for all, the a Eg ( ) b

12 Eample 6 (Miimizig istace): Fi the value b that miimizes Solutio : E( ) b E( b) E( E E b) E E E E E b E E b ( ) ( )( ) ( ), otice that E( E)( E b) ( E b)( E E( E)) 0, the we have E( b) E( E) E( E b), therefore, mi E ( b b ) E ( E ) Solutio : whe b E E( b) E( be b ) E b be E ( b E ) ( E ) Eample 7 (Uiform-epoetial relatioship II): Let have a uiform (0,) istributio, ie, Defie g( ) log Fi Eg( ) f ( ) for 0 Sectio 3 Momets a Momet Geeratig Fuctios ' ' Defiitio 3: For each iteger, the th momet of (or F ( )), is E momet of, is E ( ) ', where E Note: The first momet of is the mea, ie, ' E The th cetral

13 Defiitio 3: The variace of a raom variable is its seco cetral momet, Var E( E ) The positive square root of Var is the staar eviatio of Questio: What is the first cetral momet of? ( E( E) 0) Some Notes: ) variace a staar eviatio are measures of sprea; ) always greater tha or equal to 0; 3) Whe oes the variace=0? 4) A alterate formula for the variace is Var E ( E ) Eample 33: (Epoetial variace) Let have a epoetial istributio We fou that the E Fi the Var Eample: If ~ f ( ) / a for a0,0, fi E a Var Theorem 34: If is a raom variable with fiite variace, the for ay costats a a b, the Var( a b) a Var Eample: If E a ( )/ Z, where E a Var, fi EZ a VarZ 3

14 Eample 35: (Biomial variace) Let ~Biomial(, p) We fou that E p Fi Var Defiitio 36: Let be a raom variable with cf F The momet geeratig fuctio (mgf) of (or F ), eote by M () t, is t M () t Ee, provie that the epectatio eists for t i some eighborhoo of 0, ie, there is a h 0 such that, for all t t ( h, h), Ee eists If the epectatio oes ot eist i a eighborhoo of 0, we say that the momet geeratig fuctio oes ot eist t Cotiuous Case: M () t e f ( ) Discrete Case: M () t t e P( ) Note: We ca use mgf of to obtai its momets Theorem 37: If has mgf M () t, the ( E M ) (0), where ( ) M (0) M( t) t0 t Eample 38 (Gamma mgf): Let have a gamma istributio with parameters a, the Fi the mgf of f ( ) e ( ) /, 0, 0, 0 4

15 Eample 39 (Biomial mgf): Fi the mgf of a Biomial( p, ) raom variable Note: Characterizig the set of momets is ot eough to etermie a istributio uiquely because there may be two istict raom variables havig the same momets but ifferet istributios (See Eample 30) Theorem 3: Let F ( ) a F ( y) be two cfs all of whose momets eist a If a have boue support, the F( u) F( u) for all u if a oly if E r 0,, r r E for all itegers b If the mgfs eist a M () t M () t for all for t i some eighborhoo of 0, the F ( u) F ( u) for all u Theorem 3: (Covergece of mgfs) Suppose { i, i,, } is a sequece of raom variables, each with mgf M i () t Furthermore, suppose that lim i M ( t) M ( ) i t for all t i a eighborhoo of 0, a M () t is a mgf The there is uique cf F whose momets are etermie by M () t a for all where is cotiuous, we have lim i F ( ) F( ) That is, covergece, for t h of mgfs to a mgf implies covergece of cfs i Eample 33: (Poisso approimatio to Biomial) Show that if ~ Biomail(, p ) a ~ Poisso( p), the that the cf of coverges to the cf of Theorem 35: For ay costats a a b, the mgf of the raom variable a bt M a b() t e M ( at) b is give by 5

16 Eample: Suppose that the istributio of? ( ) ~ f ( ) ep( )( (, ), fi mgfs of a a b What is Sectio 4 Differetiatig Uer a Itegral Sig Recall: the proof of a theorem that uses the mgf to calculate the momet: t t t M ( t) e f( ) e f( ) e f( ) t t t Theorem 4: (Leibitz s Rule) If f (, ), a( ), a b( ) are ifferetiable with respect to, the b( ) b( ) f (, ) f( b( ), ) b( ) f( a( ), ) a( ) f(, ) a( ) a( ) Notes: If a( ) a b( ) are costat, the Leibitz s Rule simplifies to b b f (, ) f(, ) a a 6

17 If b( ) a f (, ) g( ) (implies that f (, ) is oly a fuctio of ) The Leibitz s Rule simplifies to f(, ) g( ) g( a( )) a( ) a( ) a( ) 3 If b( ), coul we have a( ) a( )? f (, ) f( a( ), ) a( ) f(, ) I the followig theorems, we ee to some coitios to establish the followig formulas: f (, ) f(, ) 3 h(, ) h(, ) 0 0 b h (, ) b h (, ) a 0 0 a Eample 47; If has the pmf P ( ) ( ), 0,,,0 (geometric istributio), fi E Solutio : Let 0 E ( ) ( ) T ( ), the ( ) T ( ), we have ( ) ( ) T ( ) ( ) ( ) ( ) We ca get E lim ( ) / T 7

18 Solutio : Let h (, ) ( ) ( 0,, ), the h (, ) ( ) ( ) 0 ( ) ( ) 0 0 / E /( ) a [( ) ( ) ] ( ) ( ) Check the coitios i the Theorem 48: () () h (, ) ( ) ( ) is cotiuous for every ( ) [( ) ( ) ] 0 ( ) ( ) S ( ) ( )( ) S lim S 0 To prove S coverges uiformly to 0 o a close set [ c, ] (0,), we ee prove for a give 0, N, such that S for all [, c] whe N We ca choose N such that ( N )( c) N T Eample (Iterchagig the ifferetiatio a itegratio): If is a raom variable a has the mgf, the t t t M ( t) e f( ) e f( ) e f( ) t t t 8