Chpter 3 Single Rndom Vriles nd Proilit Distriutions (Prt )
Contents Wht is Rndom Vrile? Proilit Distriution Functions Cumultive Distriution Function Proilit Densit Function Common Rndom Vriles nd their Distriution Functions Trnsormtion o Single Rndom Vrile Averges o Rndom Vriles Chrcteristic Function Cheshev's Inequlit Computer Genertion o Rndom Vriles
Trnsormtion o Single Rndom Vrile Consider unctions o rndom vriles, i.e.: g( ) where g() is (single vlued) unction. (e.g.) e, W ln, U cos, V Recll: I is rndom vrile, is lso rndom vrile. Question: Given the proilit distriutions (F () or ()) o, ind corresponding proilit distriutions (F () or ()) o the newl deined r.v..
Trnsormtion o Single Rndom Vrile (Cont.) Figure 3.3: Emples o monotonic nd nonmonotonic trnsormtions o r.v.
Functions o Rndom Vrile (Cont.) Emple 3.3: Liner Function Let e r.v. Let r.v. =+ (0). Find cd F () given cd F (). 0 ), ( ] [ 0 ), ( ] [ ) ( ] [ ] [ ) ( F P F P F P P F
Trnsormtion o Single Rndom Vrile (Cont.). Cse #: g() is monotone incresing For n ritrr vlue 0 o, there unique corresponding vlue 0 o : 0 g( 0 ) Then, we hve: g( ) g( ) P( ) F ( ) P( 0) F ( 0) P 0 0 0 Chnging 0 nd 0 to ritrr vlues nd, we get F ( ) F ( ) where ( ) Dierentiting the ove F () w.r.t., we get the pd () s: df ( ) ( ) d df ( ) d d d g g ( ) d ( ) d g ( )
Trnsormtion o Single Rndom Vrile (Cont.). Cse #: g() is monotone decresing For n ritrr vlue 0 o, there unique corresponding vlue 0 o : 0 g( 0 ) But, in this cse we hve: F g( ) g( ) P( ) P( ) F ( ) ( 0) P 0 0 0 0 In generl, this cn e epressed s: F ( ) F ( ) where ( ) Dierentiting the ove F () w.r.t., we get the pd () s: df ( ) df ( ) ( ) d d d d g g ( ) d ( ) d g ( )
Trnsormtion o Single Rndom Vrile (Cont.) Remrk: Notice tht the slope (or derivtive) d/d > 0 or cse #, wheres d/d < 0 or cse #. Thereore, using the solute vlue o the derivtive, we cn comine the ove two cses, nd epress the proilit densit unction () s ollows: df ( ) ( ) d df ( ) d d d g ( ) ( ) d d g ( )
Trnsormtion o Single Rndom Vrile (Cont.) Emple 3.5 Suppose is n eponentil r.v. w/ prmeter, i.e.: ( ) e u( ), 0 Then ind the pd o newl deined r.v. vi the ollowing trnsormtion. Solution: The trnsormtion is monotone, nd solving the trnsormtion w.r.t., we get:
Trnsormtion o Single Rndom Vrile (Cont.) Emple 3.5 (Cont.) Solution (Cont.): The derivtive hs constnt vlue s: d d Thereore, the pd o cn e derived s: d ( ) ( ) d g ( ) e u( )
Functions o Rndom Vrile (Cont.) Emple 3.3: Liner Function (Cont.) Find pd () given pd (). ) ( ) ( 0 ), ( 0 ), ( ) ( 0 ), ( ] [ 0 ), ( ] [ ) ( ) ( ) ( F P F P F d df
Trnsormtion o Single Rndom Vrile (Cont.) Figure 3.4: The pd or Emple 3.5: () > 0; () < 0.
Trnsormtion o Single Rndom Vrile (Cont.) 3. Cse #3: g() is non-monotonic In this cse, there will more thn one solution o = or given vlue o =, i.e.: i gi ( ), i,,, m Thereore, we cn generlize the ormul o the newl derived r.v.'s pd s: ( ) m i di ( ) d i gi ( )
Functions o Rndom Vrile (Cont.) Nonliner Function Let e continuous r.v. nd let =g(). Equivlent events } { } { } { } { 3 3 3 d d d B d C k k k k d d d d d d d B P d P C ) ( / ) ( ) ( ) ( ) ( ) ( ] [ ) ( ] [
Trnsormtion o Single Rndom Vrile (Cont.) Emple 3.6 Let e Gussin r.v. w/ men m = 0. Find the pd o deined ollows: Solution: Note tht > 0, nd thereore () = 0 or < 0. For the cse when 0, solving the trnsormtion w.r.t., we otin: nd
Trnsormtion o Single Rndom Vrile (Cont.) Emple 3.6 (Cont.) Solution (Cont.): Thus Thereore, the pd o ecomes:,, i d d i 0, ) ( e e e
Trnsormtion o Single Rndom Vrile (Cont.) Figure 3.5: The pd o in Emple 3.6 or the cse o m = 0 nd =.
Averges o Rndom Vriles Epressing r.v.'s using its representtive vlues!!! : For the cses when complete description o the r.v. such tht the pd nd/or cd might not e necessr... Deinition 3.3 Epecttion o r.v.: The mthemticl epecttion o rndom vrile, using the r.v.'s pd, is deined ccording to the ollowing eqution: E( ) d ( ) where E() stnds or epecttion.
Averges o Rndom Vriles (Cont.) Note: The ove deinition pplies to oth continuous nd discrete rndom vriles, i.e., i is discrete, we hve: E( ) p n i n i p i i i n p i i i i d d where p i P( = i ), nd this might e the more milir orm o the mthemticl epecttion or discrete r.v.'s or ou.
Averges o Rndom Vriles (Cont.) Emple 3.8 The test scores o 00 students re summrized in Tle3.. Find the verge score using the mthemticl epecttion. Score # o students Reltive requenc 00 0.0 95 5 0.05 90 0 0.0 85 0 0.0 80 33 0.33 75 5 0.5 70 7 0.07 Tle3. Test scores or 00 students. 65 60 55 4 3 0.04 0.03 0.0
Averges o Rndom Vriles (Cont.) Emple 3.8 (Cont.) Solution: Using the reltive requenc pproch or proilit, nd the deinition o mthemticl epecttion, we hve: E( ) 000.0 80 750.5 550.0 950.05 700.07 900. 650.04 850. 600.03 (c) Compre the result with ordinr w o clculting verges, which ou m e more ccustomed to rom elementr school ds, elow: 00 955 900 65 4 603 55 00 800.33
Averges o Rndom Vriles (Cont.) Deinition 3.4 Epecttion o unctions o r.v.'s: In generl, or n unction g() o r.v., we deined the epecttion o this unction to e: g E g( ) ( ) d We tke this s deinition here, ut it cn ctull e proved: more dvnced course on proilit.
Averges o Rndom Vriles (Cont.). m-th moment: I g() = m where m is n integer, we cll it the m-th moment o r.v., i.e.: m - th moment E m m ( ) d (c) The irst moment (m = ) is clled the men nd denoted s, wheres the second moment (m = ) is clled its men squred vlue.
Averges o Rndom Vriles (Cont.) Emple 3.9 Find the men nd the men squred vlue o uniorm r.v. ~ U[, ]. Solution: The men is given : wheres the men squred vlue is s ollows: d E 3 3 3 3 3 3 d E
Averges o Rndom Vriles (Cont.). centrl moment: I g() = ( - ) n where n is n integer, we cll it the n-th centrl moment o r.v., i.e.: n n m E ( d n ) (c) The second centrl moment is especill clled the vrince, nd denoted the smol : E ( ) d
Averges o Rndom Vriles (Cont.) Note: The squre root o the vrince is clled the stndrd devition, nd it represents the verge mount o spred round the men: E E Note tht E[ - E()] is NOT dequte or representing the spred out men since positive nd negtive vlues o the dierence - E() will cncel out, thus smller mesure o devition m result. On the other hnd, E[ - E() ] would cure this prolem, ut hrd to hndle the solute vlue nlticll...
Averges o Rndom Vriles (Cont.) Emple 3.0 Consider Gussin r.v., whose pd is given : e ( ) m () Show tht the men is = m. () Find the centrl moments o.
Averges o Rndom Vriles (Cont.) Emple 3.0 (Cont.) Solution: () The men is given : m u e m e d (let u e u m du u d m e u m) u du
Averges o Rndom Vriles (Cont.) Emple 3.0 (Cont.) () B the deinition o the centrl moments, we hve: m n E n u n e u which is zero when n is odd. (wh?) For the cse when n eing even integers, the integrnd is smmetric out u = 0, nd emploing the tle o integrl, we otin: u k e k m k u du 3 k, k,, Note tht the specil cse o n = k = provides the vrince. n e du, d (let n,, u )
Averges o Rndom Vriles (Cont.) Properties o epecttion:. The epecttion o constnt is the constnt itsel: E[ ], constnt. The epecttion o constnt times unction o r.v. is the constnt times the epecttion o the unction o r.v.: E[ g( )] E[ g( )], constnt 3. The epecttion o the sum o two unctions o r.v. is the sum o ech epecttion: E g ( ) g ( )] E[ g ( )] E[ g ( )] [ Proo: ssignment Note: Comintion o the properties & 3 is clled the linerit propert o the epecttion: E[ g( ) g( )] E[ g( )] E[ g( )]
Averges o Rndom Vriles (Cont.) Emple 3. Show tht the vrince o r.v. cn e computed ccording to: Solution: Using the oregoing properties, E E E E E E E E E E E
Averges o Rndom Vriles (Cont.) Emple 3. Let two r.v.'s nd re linerl relted s: Find the men nd vrince o in terms o those o. Solution: Using the properties o epecttion, we hve the men s: E E E wheres the vrince is given : E[ ] E E
Chrcteristic Function Deinition 3.5 The chrcteristic unction: The chrcteristic unction o r.v. is specil cse o the mthemticl epecttion deined s ollows: jv M ( jv) E[ e ] ( ) e jv d Useulness o chrcteristic unction:. The m-th moment o r.v. cn e otined dierentiting the chrcteristic unction w.r.t. its rgument.. Sometimes the chrcteristic unction o r.v. is esier to otin thn the pd. 3. The chrcteristic unction nd the pd re Fourier trnsorm pirs.
Chrcteristic Function (Cont.) To show the irst sttement, we dierentite the chrcteristic unction w.r.t. v to otin: dm ( jv) d jv e d j ( ) e ( dv dv ) Now set v = 0, nd divide j to get: dm j ( jv) dv v0 ( ) d E( ) Repeting the sme procedure n times, the n-th moment o the r.v. cn generll e epressed s: E( n ) j n d n M ( jv) n dv v0 jv d
Chrcteristic Function (Cont.) Emple 3.4 Find the chrcteristic unction o Cuch r.v. w/ its pd given s: ( ) Solution: Appling the deinition o the chrcteristic unction, we otin: jv M ( jv) e d cosv j sin vd cosv d which, use o tle o indeinite integrl, cn e epressed s: M ( jv) e v Note tht M (jv) is not dierentile t v = 0, nd thus we cnnot use it to evlute the moments. In ct, its moments do not eists in this cse.
Chrcteristic Function (Cont.) Emple 3.5 Find the chrcteristic unction o the doule sided eponentil r.v. (clled the Lplcin r.v.), whose pd is given : ( ) e, 0 Solution: B the deinition o the chrcteristic unction, we get: jv M ( jv) e e d e cosv j sin vd cosve d use o tle o indeinite integrl, cn e epressed s: ( ) cos M jv ve d 0 v
Cheshev s Inequlit Recll: The stndrd devition o r.v. gives mesure o spred out its men. The Cheshev s inequlit provides ound on the proilit tht r.v. devited more thn k stndrd devitions rom its men. It is ver loose ound, ut its merit is the ct tht ver little need to e known out the r.v. to otin the ound.
Cheshev s Inequlit (Cont.) Cheshev's Inequlit: For n rndom vrile, the proilit o eing devited rom its men more thn k stndrd devition must stis the ollowing inequlit: P k k or P k k Note tht two events - k nd - < k re mutull eclusive to ech other.
Cheshev s Inequlit (Cont.) Cheshev's Inequlit (Cont.): Proo: Let = - nd = k. Then, the LHS o the irst inequlit ecomes: P P P which ollows rom the ct is the union o two mutull eclusive events nd -. ( c) E
Cheshev s Inequlit (Cont.) Cheshev's Inequlit (Cont.): Proo (Cont.): Now, consider the second moment o, which is: E( ) ( ) d ( ) d ( ) d ( ) d ( ) d P P, 0 Solving, we otin: P P P E
Cheshev s Inequlit (Cont.) Cheshev's Inequlit (Cont.): Proo (Cont.): Replcing = - with E[ ] =, nd = k, we hve the Cheshev's inequlit s: P k k Q.E.D
Cheshev s Inequlit (Cont.) Emple 3.6 () Find ound on the proilit tht r.v. is within three stndrd devitions o its men. () Find the ect proilit o this event, i the r.v. is Gussin, nd compre with the ound. Solution: () From the Cheshev's inequlit, we hve: P 3 0. 889 3
Cheshev s Inequlit (Cont.) Emple 3.6 () The pro. o the given event or Gussin r.v. is: (c) Note tht the Cheshev's inequlit does NOT provide tight ound in this cse. 0.9973 0.0035 (3) 3 3 0 3 3 3 3 Q du e du e d e P u u
Computer Genertion o Rndom Vriles Recll:. Genertion o uniorm pseudorndom numers ~ U[0,]: = rnd(, 000);. Genertion o Gussin pseudorndom numers ~N(0,): = rndn(p, q) This genertes n rr o Gussin pseudorndom numers with p rows nd q columns. 3. Genertion o Gussin pseudorndom numers Z~N(m, ): Z = + m B w o trnsormtion
Computer Genertion o Rndom Vriles (Cont.) Genertion o rndom numers with n ritrr distriution: Let U e r.v. uniorml distriuted in [0, ], nd deine new r.v. V s: V g(u) where g() is ssumed to e monotonic. Then, the pd o the newl deined r.v. V is given : V ( v) U ( u) du dv u g ( v) du dv dg ( v), dv 0, where the lst eqution ollows ecuse U (u) is unit in [0, ] nd zero elsewhere. 0 u otherwise
Computer Methods or Generting Rndom Vriles Suppose tht U is uniorml distriuted in the intervl [0, ]. Let F () e the cd o the r.v. we re interesting in generting. Deine the r.v. Z s Z F ( U ). cd o Z P[ Z ] P[ F ( U ) ] P[ U F ( )]. P[ Z ] F ( ) P[ U h] h Trnsormtion method or generting ) GenerteU uniorml distriuted in [0,]. ) Let Z F ( U ).
Computer Genertion o Rndom Vriles (Cont.) Genertion o rndom numers with n ritrr distriution (Cont.): Re-writing the ove result or the cse o 0 u : dg ( v) dg ( v), 0 ( ) dv dv V v dg ( v) dg ( v), 0 dv dv Integrting nd solving or g - (v), we otin: v V ( ) d FV ( v), dg ( v) dv 0 g ( v) v ( ) d F ( v), dg ( v) dv 0 V V where F V (v) represents the desired cd o r.v. V.
Computer Genertion o Rndom Vriles (Cont.) Emple 3.7 Using uniorm r.v. U uniorml distriuted in [0, ], ind the required trnsormtion V = g(u) so tht it will generte n eponentil pd given : V ( v) e v u( v) Solution: The cd o the desired eponentil r.v. is: F V v 0, v 0 v) V ( v) dv v e, v 0 (
Computer Genertion o Rndom Vriles (Cont.) Emple 3.7 (Cont.) Solution: From which, we otin: u g ( v) e v, v Note tht this inverse trnsormtion lws hs positive slope. Solving or v, epressing it into the reltionship etween two r.v.'s U nd V: V 0.5ln U 0.5ln U which mens tht the required trnsormtion is V = g(u) = - 0.5 ln(u). Here we use tht ct: i U is uniorm on [0, ], so is - U. 0