Homework 5 for BST 631: Statistical Theory I Solutions, 09/21/2006


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1 Homewok 5 fo BST 63: Sisicl Theoy I Soluions, 9//6 Due Time: 5:PM Thusy, on 9/8/6. Polem ( oins). Book olem.8. Soluion: E = x f ( x) = ( x) f ( x) + ( x ) f ( x) = xf ( x) + xf ( x) + f ( x) f ( x) Accoing o Leiniz s ule n heoem.4.3, we cn ke iffeeniion ino he inegion, hen we hve E = f ( ) f ( ) + f ( ) + ( f ( x)) f ( ) ( f ( x)) + = f ( x) f ( x), Thus if such h min E = E, hen sisfies f ( x) f ( x) =. In he ohe hn, we hve f ( ) ( ) x+ f x= sisfies f ( x) = f ( x) = /, hus. The coninuiy of ssues h such exiss. Polem ( oins). Book olem.3. Soluion: () The suo se of Y is (,). Accoing o Theoem..8, we hve, < y < fy ( y) = y, y o y. () EY = E = 3 ( ) / 3 x + x = x = 6 x =. EY = E 4 = ( ) / 5. x + x = x = x = VY = EY ( EY ) = 4 / 45. Polem 3 (5 oins). Book olem.3.
2 Homewok 5 fo BST 63: Sisicl Theoy I Soluions, 9//6 Soluion: This funcion cn no e mgf. Becuse = =, u M () = f( x) =. Polem 4 (8 oins). Book olem.33. Soluion: () e λ ( λe ) M e e e e e x! x! λ x x x λ λ λe λ( e ) () = = = = i= i=, ( ) ( ) ( λ e λ e E = e ) = = λe e = = λ, λ( e ) λ( e ) λ( e ) E = ( e ) = = ( λe e ) = = ( λλ ( e + ) e e ) = = λ + λ, V = E ( E ) = λ. () M ( ) ( ) e (( ) e ) ( ) e x x x = = x= =, x= E ( ) e = ( ) = = = =, ( e ) ( ( e ) ) E ( ) e = = = ( ) = ( ) ( ) e ( ( ) e ). ( e ) ( ( e ) ) + ( e ) ( e ) ( )( ) = ) 3 = = ( ( e ) ) V = E ( E ) =. (c) ( x µ ) M ( ) = ex( )ex( x) πσ σ ( x ( µ + σ )) = ex( )ex( µ + σ / ) πσ σ / µ + σ = e, µ σ / µ σ / E = ( e + ) ( µ σ )e + = = + = = µ, E µ + σ / µ + σ / = ( e ) = = ( ( µ + σ )e ) = µ + σ / = = ( σ + ( µ + σ ) )e ) = σ + µ,
3 Homewok 5 fo BST 63: Sisicl Theoy I Soluions, 9//6 V = E ( E ) = σ. Polem 5 (5 oins). Book olem.38. Soluion: () This is negive inomil isiuion, we ley ge is mgf M () = ( ) e () If Y =, hen MY() = M ( ) = ( e ), while lim = lim =. ( ) e e ( ) e Thus, lim MY( ) = ( ). Polem 6 ( oins). Book olem.39. Soluion: I only give he nswes hee. x λ λx () e e =. () e λ λ =. λ (c) () x =. ( ) x /( ). = Polem 7 (5 oins) (Qulify Exm olem 4). As simle moel fo he movemen of isese eole fom one locion o nohe, consie wo uns (enoe Un n Un ), whee whie ll enoes nonisese eson n lck ll enoes isese eson. Un conins whie lls n lck ll; Un conins whie ll n lck lls. Consie he following EPERIMENT: "one ll is nomly wn fom Un n is u in Un ; hen, wo lls e nomly wn fom Un." () Fo his exeimen, le enoe he nume of whie lls conine in he smle of wo lls wn fom Un. Deive igoously he oiliy isiuion of he nom vile. 3
4 Homewok 5 fo BST 63: Sisicl Theoy I Soluions, 9//6 () Fo his exeimen, if oh lls selece fom Un e whie, wh is he oiliy h he ll selece fom Un ws lso whie? (3) Using hese sme wo uns, consie he following lenive EPERIMENT: "one ll is nomly wn fom Un n is u ino Un ; hen, lls e selece oneime wihou elcemen fom Un unil whie ll is oine." Fo his lenive exeimen, le Y enoe he nume of lls selece fom Un unil whie ll is oine (e.g., if he fis ll selece fom Un is lck n he secon one is whie, hen Y = ). Deive igoously he oiliy isiuion of he nom vile Y. Soluion: () cn kes vlues fom,, n. Le A i ={i lls e whie when we nomly selec wo lls fom Un }, hen P ( = i) = PA ( i )( i=,,). Le B = {he ll fom Un is lck} n B = {he ll fom Un is whie}. Then PB ( ) = /3 n PB ( ) = /3. Thus we hve PA ( ) = PA ( B) + PA ( B) = PA ( B) PB ( ) + PA ( B) PB ( ) While PA ( B ) is he oiliy o hve whie lls when we ke lls wihou 3 4 elcemen fom 4 lls conining 3 lck lls, hus PA ( B) = / = /. Similly, 4 we hve PA ( B) = / = /6. Then PA ( ) = 5/8. Similly, we hve PA ( ) = /8, PA ( ) = /8. Thus P ( = ) = 5/8, P ( = ) = /8, P ( = ) = /8. PB ( A) PA ( B) PB ( ) /6*/3 () This is coesoning o clcule PB ( A) = = = =. PA ( ) PA ( ) /8 (3) We cn follow he simil clculion fom () o o his, we hve PY= ( ) = /4*/3+ /4*/3= 5/, PY= ( ) = 3/4*/3*/3+ /4*/3*/3= /36, PY= ( 3) = 3/4*/3*/*/3+ /4*/3**/3= 7/36, PY= ( 4) = 3/4**/3*/**/3= /. Polem 8 (5 oins) (Qulify exm Jnuy 5 olem 5) In his olem we will use he Peo isiuion which is efine s follows: () Show h fo his isiuion f x β β x β x β ( β+ ) (, ) =, >, >, <. E if n only if < β. () Show h he momen geneing funcion fo he nom vile Y = ln( ) is: 4
5 Homewok 5 fo BST 63: Sisicl Theoy I Soluions, 9//6 ln( ) βe MY () =, < β. β Use his o show h Eln( ) = ln( ) + / β. Soluion: β ( β+ ) β ( β+ ) () E = x β x = β x ( β + ) <, his is equivlen o < β. (). Then M () = Ee = Ee = E Y Y ln β ( β+ ) β ( β + ) x β x β x β log = = β β βe = = = β ( β ) β β Fom (), we know h MY ( ) exiss if < β. E exiss if n only if. EY log log βe βe (( β )log + ) = = = ( ) = = log + / β. β ( β ) 5
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