Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 Solutos from Prevous Homework
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 000, Problem Page, Problem () Let X ~ U (0,) ad let Y a(l( X)). Fd the pdf for the radom varable Y. () Let the radom varables X f x e for 0 x. x/ ~ X ( ) (.) Fd the momet geeratg fucto for each radom varable. (.) Fd the dstrbuto fucto for Y X X. x x (3) Let X ~ fx ( x, p) p ( p) ( x0,,, ) ad let x p~ g( p, ) p ( p) / B(, ), 0 p, 0, 0, where Soluto: ( a) ( ) B(, ). Fd EX. ( ) y () The pdf of Y s fy( y) exp( ) I( al(), ) ( y). a a () M X () t. t Let Y X X, Z X, the y z z fyz, ( y, z) exp( )exp( )(0 z y ). y y y Thus, fy( y) fy, Z( y, z) dz [exp( ) exp( )]. 0 For, y fy ( y) yexp( ), whch s also equal to (3) EX ( ) EEX ( ( P)) EP ( ). y y lm 0 [exp( ) exp( )].
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 000, Problem Page, Problem The game of craps s played by lettg the thrower toss dce utl he ws or loses. The thrower ws o the frst toss f he gets a total of 7 or ; he loses f he gets a total of, 3, or. If he gets ay other total o hs frst toss, that total s called hs pot. He the tosses the two dce repeatedly utl he obtas a total of 7 or hs pot. He ws f he gets hs pot ad loses f he gets a total of 7. What s the thrower's probablty of wg? Soluto: Let X deote the umber of tosses whe he ws, the PX ( ) P{get a total of 7 or }=8/36. For k, we have: PX ( k) P{frst toss gets 4,secod to k- tosses does ot gets 4 or 7, ktoss gets 4} P{frst toss gets 5,secod to k- tosses does ot gets 5 or 7, k toss gets 5} P{frst toss gets 6,secod to k- tosses does ot gets 6 or 7, k toss gets 6} P{frst toss gets 8,secod to k- tosses does ot gets 8 or 7, k toss gets 8} P{frst toss gets 9,secod to k- tosses does ot gets 9 or 7, k toss gets 9} P{frst toss gets 0,secod to k- tosses does ot gets 0 or 7, k toss gets 0} 3 3 6 k 4 4 6 k 5 5 6 k ( ) ( ) ( ) ( ) ( ) ( ) 36 36 36 36 36 36 36 36 36 5 5 6 k 4 4 6 k 3 3 6 k ( ) ( ) ( ) ( ) ( ) ( ) 36 36 36 36 36 36 36 36 36 3 3 6 k 4 4 6 k 5 5 6 k ( ) ( ) ( ) ( ) ( ) ( ) 36 36 36 36 36 36 36 36 36 Actually, we use the followg probablty table for oe toss: Total 3 4 5 6 7 8 9 0 p /36 /36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 /36 /36 The fal probablty s 8 4 50 PX ( k) 0.499. k 36 36 45 396 3
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 00, Problem 4 Page 4, problem 4 As a smple model for the movemet of dseased people from oe locato to aother, cosder two urs (deoted Ur ad Ur ), where a whte ball deotes a o-dseased perso ad a black ball deotes a dseased perso. Ur cotas whte balls ad black ball; Ur cotas whte ball ad black balls. Cosder the followg EXPERIMENT: "oe ball s radomly draw from Ur ad s put Ur ; the, two balls are radomly draw from Ur." () For ths expermet, let X deote the umber of whte balls cotaed the sample of two balls draw from Ur. Derve rgorously the probablty dstrbuto of the radom varable X. () For ths expermet, f both balls selected from Ur are whte, what s the probablty that the ball selected from Ur was also whte? (3) Usg these same two urs, cosder the followg alteratve EXPERIMENT: "oe ball s radomly draw from Ur ad s put to Ur ; the, balls are selected oe-at-a-tme wthout replacemet from Ur utl a whte ball s obtaed." For ths alteratve expermet, let Y deote the umber of balls selected from Ur utl a whte ball s obtaed (e.g., f the frst ball selected from Ur s black ad the secod oe s whte, the Y ). Derve rgorously the probablty dstrbuto of the radom varable Y. Soluto: () X ca takes values from 0,,ad. Let A ={ balls are whte whe we radomly select two balls from Ur }, the PX ( ) PA ( )( 0,,). Let B0 {the ball from Ur s black} ad B {the ball from Ur s whte}. The 0 PB ( ) /3 ad PB ( ) /3. Thus we have PA ( ) PA ( B) PA ( B) PA ( B) PB ( ) PA ( B) PB ( ) 0 0 0 0 0 0 0 0 Whle PA ( 0 B 0) s the probablty to have 0 whte balls whe we take balls wthout 3 4 replacemet from 4 balls cotag 3 black balls, thus PA ( 0 B0) / /. Smlarly, 4 we have PA ( 0 B0) / /6. The PA ( 0) 5/8. Smlarly, we have PA ( ) /8, PA ( ) /8. Thus 4
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 PX ( 0) 5 /8, PX ( ) /8, PX ( ) /8. PB ( A) PA ( B) PB ( ) /6*/3 () Ths s correspodg to calculate PB ( A). PA ( ) PA ( ) /8 (3) We ca follow the smlar calculato from () to do ths, we have 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 /. 5
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 00 September, Problem Page 7, Problem Assume that the geders of cosecutve chldre bor to a partcular couple are depedet of oe aother, ad further assume that the probablty of ay partcular chld beg a grl s equal to, 0< <. Cosder the followg three "famly plag" strateges: Strategy #: The couple decdes to have chldre utl the frst grl s bor. Strategy #: The couple decdes to have chldre utl they have both a grl ad a boy. Strategy #3: The couple decdes to have chldre utl they have ether two grls or two boys. () Fd explct expressos for the probablty dstrbuto ad expected value for the total umber of chldre bor to ths couple uder strategy. () Fd explct expressos for the probablty dstrbuto ad expected value for the total umber of chldre bor to ths couple uder strategy. Show that ths probablty dstrbuto s a vald probablty dstrbuto fucto. (3) Fd explct expressos for the probablty dstrbuto ad expected value for the total umber of chldre bor to ths couple uder strategy 3. (4) What ca you say regardg whch of these three strateges produces the smallest expected umber of chldre? What ca you say regardg whch of these three strateges produces the largest expected umber of chldre? Soluto: X total umber of chldre. x () PX ( x) ( ), x,,, thus X follows a geometrc dstrbuto wth the parameter, ad EX /. () x x PX ( x) ( ) ( ), x,3,. Usg the results from the geometrc dstrbuto, we ca show ths s a vald pdf: ( ( ) x ( ) x ) ( ( ) x ( ) x ) ( ) x, x ad easly get EX ( ) ( ( )). ( ) (3) PX ( ) P(Grl,Grl)+ P (boy,boy)= ( ), PX ( 3) P (Grl,Boy,Boy)+ P (Grl,Boy,Grl)+ P (Boy,Grl,Grl)+ P (Boy,Grl, Boy), PX ( 3) ( ) ( ). EX. 6
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 (4) We frst deote the expectatos from () to (3) as f ( ), f ( ), ad f ( ) 3. To dscuss the questo, we do a par-wse comparso: f( ) f( ) ( ) ( ), thus f( ) f( ) for all 0. ( ) 3 ( ) 3( ) f( ) ( ) f ( ) ( ). Cosder the fucto gx ( ) x 3x, t s easy to prove that gx ( ) s creasg for x [0,0.5], whle g(0.5) 0. Thus gx ( ) 0 for x [0,0.5]. Because whe 0, we have 0 ( ) 0.5. Ths mples f3( ) f( ) 0 for all 0. 3 f( ) f3( ) ( ). Cosder the fucto 3 g( ). Because g( ) s a cotuous fucto respect to, lm 0 g( ), d g( ) 6 4 0 lm g( ), ad for all 0. These mply that: () g( ) d s a deceasg fucto; () a uque 0 such that g( 0) 0; (3) whe 0 0, g( ) 0; (4) 0, g( ) 0 ; (5) t s easy to get 0 0.403. Therefore, the strategy has the largest expected umber of chldre. The strategy 3 has the smallest expected umber of chldre whe 0.403, otherwse the strategy has the smallest expected umber of chldre. 7
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 00 September, Problem 3 Page 8, Problem 3 () Suppose that X s the cocetrato ( parts per mllo) of a certa arbore pollutat, ad suppose that the radom varable Y l X has a dstrbuto that ca be adequately modeled by the desty fucto f y e y / Y ( ) ( ),,0. (.) Fd a explct expresso for F ( y ), the cumulatve dstrbuto fucto (CDF) Y assocated wth the desty fucto fy ( y ). If ad, use ths CDF to fd the exact umercal value of PX ( 4 X ). (.) For the desty fucto f ( y) gve above, derve a explct expresso for a geeratg Y fucto ( t ) that ca be used to geerate the absolute cetral momets Y r v E( Y EY ) for r a o-egatve teger, ad the use ( t ) r Var( Y ), the varace of Y. Y drectly to fd () Let U be depedetly ad detcally dstrbuted as Uform(0,) for,. Let X be defed as X U U. Show that the probablty desty fucto for the radom varable x,0 x X s the followg fucto: f( x) x, x. Soluto: () EY y exp( ), y ; FY ( y) y exp( ), y. ad cosder PX ( 4 X) PX ( 4) / PX ( ) PY ( l(4)) / PY ( l()) ( F (l(4))) / ( F (l())) 0.84. Y Y ( t ) E (exp( Y EY t ) E (exp( Y t )) Y t. d Var Y dt () Let Y U, the the jot pdf of ( X, Y ) s t ( ) t 0. f( x, y) (0 y, y x y ). 8
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 From ths, t s easy to get the margal pdf of X. 9
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 003 August, Problem 4 Page, Problem 4 Suppose X has a ch-square dstrbuto v/ x/ fx ( x) x e, x0, v0, v/ ( v /) ad Y has a beta dstrbuto gve by ( ) fy ( y) y ( y), 0, 0,0 y. ( ) ( ) () Obta formula for r EX ad r EY, where r s a postve teger. () If r [ v/ ]*, where [ s ]* deotes the largest teger s, show that (3) Fd r EY r EX ( ) [( v)( v4) ( v r)]. What codto must r satsfy?. (4) Assume that X ad Y are depedet, ad let Z X / Y. Fd EZ ( ). Soluto: () r r v/ x/ ( rv)/ x/ EX x x e dx x e dx 0 v/ 0 v/ ( v/) ( v/) ( rv)/ ( rv/) r ( rv/) v/. ( v/) ( v/) EY r r r y r y dy ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 ( ) ( ) ( r) ( ) ( r) () From (), we have r r ( v/ r) EX ( ). r ( v/) ( v/) ( v/ r) ( v) ( v r) (3) From (), we have r ( ) ( r) EY, but requre r. ( ) ( r) ( v /) ( ) ( ) (4) EX ( / Y) EXEY ( ) ( ) v. ( v /) ( ) ( ). 0
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 004 Jauary, Problem Page 3, Problem ( ) c Cosder the pdf f( x) x, x, c 0. c () Show that the dstrbuto of X s a member of the expoetal famly. () Let Y l( X).Fd the dstrbuto of Y. (3) Fd the frst ad secod momets of Y. (4) Derve the momet geeratg fucto of Y. Soluto: () f( x) exp{ (/ c)log( x)}( x ). c () Let gx ( ) log( x), we have gx ( ) s creasg, pdf of Y s y y (/ c) y y/ c fy( y) fx( g ( y)) e ( e ) e e ( y 0). c c (3) yc / z ; EY ye dy cze dz c 0 c 0 EY y e dy c z e dz c ze dz c 0 c 0 0 (4) yc / z z. M t e e dy e dz t c c ct yt y/ c z( ct ) Y () ( / ) 0 0. g ( y) exp( y), ad d e dy y e y, thus the
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 004 Jue, Problem 3 Page 7, Problem 3 Aswer each of the followg questos: () Let the radom varable X has probablty desty fucto. x( ) f( x ) e h( x), x Derve the momet geeratg fucto M X ( t) ad from t fd EX ad VarX. () Let U ~ Uform (0,) ad let TU ( ) be a mappg from (0,) to (-,) as follows: Fd each of the probabltes: (.) PT ( [, /]). [0,/ 4) [, / ) T [/ 4,/ ) [ /,3 / 8) T [ /, 5 / 8) [3 / 8, 3 / 4) T [5 / 8,3 / 4) [3 / 4,7 / 8) T [3 / 4,7 / 8] [7 / 8,] T (7 / 8,] (, / ] T (.) PT ( (3/8,7/8)). (.3) PT ( [ /,3/8) [7/8,]). (3) A far de s cast utl a 6 appears. What s the probablty that t wll take more tha 5 trals? Soluto: () Because f( x ) s vald pdf, thus : We have: x ( ) x ( ) f ( x ) d e h( x) dx e h( x) dxe. ( ) ( ) ( ) ( ) () Xt xt x t t X ( ) ; M t Ee e f x dx e e hxdx e d d ( t) ( ) ( t) ( ) EX M X( t) t0 e t0 '( t) e t0 '( ) ; dt dt
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 d d ( t) ( ) ( t) ( ) EX M () X t t0 '( te ) t0 '( te ) t0 dt dt ''( te ) ( '( t)) e ''( ) ( '( )) ; ( t) ( ) ( t) ( ) t0 t0 VarX EX EX ( ) "( ). (.) PT ( [, / ]) PU ( (0,/ 4) (7 / 8,)) / 4 / 8 3 / 8. (.) PT ( (3 / 8,7 / 8)) PU ( (/,5 / 8) (5 / 8,3 / 4)) / 8 / 8 / 4. (.3) PT ( [ /,3/8) [7/8,]) PU ( (/4,/) (3/4,7/8)) /4/8 3/8. (3) Let X deote the umber of trals, the X has geometrc dstrbuto wth the parameter p /6, thus the probablty s: ( 5) ( 5) ( ) x PX PX p p 0.408. 5 x0 3
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 005 Jauary, Problem 5 Page, Problem 5 I ths problem we wll use the Pareto dstrbuto whch s defed as follows: f x x x ( ) (, ), 0, 0,. () Show that for ths dstrbuto r EX f ad oly f r. () Show that the momet geeratg fucto for the radom varable Y l( X) s: Soluto: t l( ) e MY () t, t. t Use ths to show that El( X) l( ) /. () r r ( ) ( r) EX x x dx x dx. The ( r), ths s equvalet to r. () r EX exsts f ad oly f M () t Ee Ee EX Y Yt t l X t t ( ) ( t) x x dx x dx t tlog e t ( t) t t From (), we kow that MY ( t ) exsts f t.. EY tlog tlog d e e (( t) log ) t0 t0 ( ) log /. dt t ( t) 4
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 005 July, Problem 3 Page 5, Problem 3 The tme to death ( days) after a autologous boe marrow trasplat follows a log ormal dstrbuto wth 3.77 ad.084. Ht: For the log-ormal dstrbuto, we ca wrte ((logt ) / ) f() t ad St () PT ( t) [(l t )/ ], where () ad () represet t the pdf ad cdf, respectvely, of a stadard ormal radom varable. () Fd the mea ad meda tmes to death. () Fd the probablty that a dvdual survves 0, 50 ad 00 days followg a trasplat. (3) I survval aalyss, the hazard fucto s used to measure the rsk that a dvdual expereces the evet of terest. The hgher the hazard fucto at a gve tme pot, the greater the rsk. Takg advatage of the fact that the hazard rate may be wrtte as ht () f()/ t St (), plot the hazard rate of the tme to death at t = 0., t =0.35, t =, t = 0, t = 50, ad t = 00 days. What does the shape of ths fucto tell you about the hazard of death over tme? (4) Derve the geeral formula for the varace of a radom varable T that has a log ormal dstrbuto wth parameters ad. (Ht: You may fd t useful to use the mgf of a ormal radom varable.) Soluto: () Accordg to the results of the log-ormal dstrbuto, we have meda of Y s e 3.97. To get the meda of Y, y0 0, we have EY / e 67.97. The X l( y ). 0 / PY ( y0) P( X l( y0)) P( ) X l( y0) Because ~ (0,), thus we have 0, the y0 e. () See the followg table: Survval Tme ( t ) (l t 3.77) (l t 3.77) ( ) St () PT ( t) sqrt(.084) sqrt(.084) 5
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 t 0-0.6057 0.74 0.7764 50 5.096 0.6947 0.3053 00 0.9893 0.8387 0.63 (3) See the followg table ad fgure 3. It ca be see that the hazard rate creases sharply the drops. (l t 3.77) (l t 3.77) ( ) sqrt(.084) sqrt(.084) f () t (l t 3.77) ( ) St () PT ( t) ht () f()/ t St () sqrt(.084) 0. -3.7958 0.0003 0.00 0.000 0.9999 0.00 0.35 -.980 0.0055 0.009 0.007 0.9983 0.009 -.706 0.0908 0.034 0.047 0.9573 0.038 0-0.6057 0.33 0.030 0.74 0.776 0.036 50 0.509 0.3504 0.0049 0.6947 0.3053 0.059 00 0.9893 0.446 0.007 0.8388 0.63 0.005 (4) The calculato based o the momet geeratg fucto has bee llustrated the class: ( ) VarT e e. 6
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 0 0 40 60 80 00-0 -5 0 5 0 Fgure. 7
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 0 4 6 8 0 0 4 6 8 0 Fgure. 8
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 hazard rate 0.005 0.00 0.05 0.00 0.05 0.030 0 0 40 60 80 00 tme t Fgure 3. The hazard rate. 9
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 006 July, Problem Page 30, Problem Aswer the followg questos. () A commttee s to be formed whch s to cota 4 republcas ad 3 democrats out of respectve groups of 0 republcas ad 8 democrats. a. How may dstct commttees ca be formed? b. Suppose that oe of the republcas s amed George. What s the probablty that George s o the commttee? () How may 3-dgt eve postve tegers are there? Note that whe we say a 3-dgt umber (XYZ), we mea that X s ot zero. (3) A expermet s performed whch a co s flpped ad the a de s tossed. Cosder Soluto: the outcome of the expermet to be a bvarate observato ( X, Y ) where X s the result of the co toss ad Y s the umber o the de. a. What s the sample space of the expermet? b. Fd the probablty of a head ad a eve umber. c. Fd the probablty of a tal ad a or 3. d. Fd the probablty of a head ad a umber greater tha or equal to 3. 08 () (.) 760. 4 3 9 8 () (.) P /760 0.4. 33 () 9*0*5 450. (3) (3.) The sample space cotas elemets wth equal probablty (3) (3.) P(X=head, Y=eve)=/4. (3) (3.3) P(X=tal, Y= or 3)=/6. (3) (3.4) P(X=head, Y=3, or 4, or 5, or 6)=/3. 0
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 006 July, Problem Page 30, problem Cosder the dstrbuto fucto, where 0, 0, 0. x x f( x;,, ) exp,0 x ( ) () Is ths a member of the expoetal famly of dstrbutos? Justfy your aswer. () Fd the r th momet. Ht: Make a chage of varable y ( x/ ). (3) Gve that the radom varable N has a Posso dstrbuto wth parameter X ad that X s a radom varable wth the dstrbuto of f( x;,, ). Fd the mea ad varace of N. Soluto: x () Ths s ot a member of expoetal famly of dstrbutos because exp wrtte as exp( w(,, ) c( x)). ca ot be x () If we do the trasformato y ( ) momet s:, the x / y ad dx x x x EX x dxy ( ) r r ( ) ( ) exp( ( ) ) ( ( ) ) 0 0 exp( ) ( ) r r/ / / y y y y dy r ( r / ) exp( ). ( ) ( ) r r/ y y dy 0 / y. Thus, the r th (3) ( / ) EN ( ) EEN [ ( X)] EX ( ). ( ) Var( N) E[ Var( N X )] Var[ E( N X )] E( X ) Var( X ) EX ( ) EX ( ) [ EX ( )] ( ) ( / )( ( / )) ( / ).
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 007 Jauary, Problem Page 33, Problem To play a lottery, a cotestat purchases a dollar tcket ad chooses 5 dstct umbers from the tegers through 49. The 6 th umber s the chose from a separate lst of the tegers through 4. To w the lottery, you must match each of the frst fve umbers (the order of ths sequece does ot matter) ad match the 6th umber. A player ca select ay sequece of admssble umbers, so several players could potetally match the wg set of umbers ad share the prze. () What s the chace of wg (or sharg) the prze, f you purchase oe tcket ad choose the 6 umbers at radom? () Suppose you purchase 00 tckets, ad choose the 6 umbers o each tcket at radom, ad depedetly from oe tcket to the ext. Gve a expresso for the probablty that you ca w (or share) the prze. (3) A smaller prze s gve f you ether pck exactly 4 of the frst 5 umbers or pck the frst 5 umber correctly but ot the 6th umber. What s the chace of wg ths smaller prze, f you purchase oe tcket ad select the 6 umbers at radom? Soluto: () Defe the radom varable X f you w the prze ad X 0 otherwse, the P(You w the prze) P( X ) th P(You match the frst 5 umbers ad match the 6 umber) th P(You match the frst 5 umbers) P(match the 6 umber). 49 4 5 () Defe A as the evet that you match the frst 5 umbers ad match the 6 th umber for the th tcket (,,,00). The you ca w (or share) s: p P( A ) P( X ) (,,00). The probablty that 49 4 5 P (you ca w (or share) the prze) th = P (You match the frst 5 umbers ad 6 umber for at least oe of tckets)
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 =- P (You ca ot w (or share) the prze) 00 00 00 = P( A) ( c ) ( c P A P A ) 00 = = PA ( ) ( ) 49 4 5 (3) P(You w the smaller prze) P(You match exactly 4 of the frst 5 umbers) P th (You match the frst 5 umbers ad ot match the 6 umber) P(You match exactly 4 of the frst 5 umbers) th P(You match the frst 5 umbers) P(You do ot match the 6 umber) 544 4 4. 49 49 4 5 5 You ca get the frst part by ths way. Ths s the samplg wthout replacemet ad we ca use the u-ordered sample space to calculate the probablty. The umber of elemets the sample space s 49, because we choose 5 from 49 wthout replacemet. For a set of 5 umbers, you 5 5 44 have way to match exactly 4 of them ad you have way to choose aother umber 4 that s ot ths set. 00. 3
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 007 Jauary, Problem Page 33, Problem Assume that X has a Normal(0, ) dstrbuto ad that X s depedet of X ad also has a Normal(0, ) dstrbuto. Let Y a X b X ad let Y ax bx. Usg the formato about the dstrbuto of X ad X, () What are the expected values ad varaces of Y ad Y? () Calculate the Cov( Y, Y ) as a fucto of a, a, b, b. (3) What are the margal dstrbutos of Y ad Y? (4) Show how you would determe a, a, b, b, so that Y ad Y are bvarate Normally Soluto: dstrbuted wth: EY ( ) EY ( ) 0, Var( Y) Var( Y), ad Cor( Y, Y). () EY ae X be X ( ) ( ) ( ) 0, EY ( ) ae ( X) be ( X) 0. VarY ( ) var( ax bx ) var( ax) var( bx ) a b. VarY ( ) var( ax bx) var( ax) var( bx) a b. () E( YY ) E[ aa Xab XXabXX bb X] aa bb, ad EY ( ) EY ( ) 0, thus Cov( Y, Y) aa bb. (3) Y ~ (0, a b ) ad Y ~ (0, a b ). (4) a, a, b, b must stratfy the followg codtos: aa bb. To show that Y ad Y has a bvarate ormal, we frt get a b, a b, ad The Jacoba s: J X. ab ab by ab by ab X ay ay ad ab ab. The jot pdf of ( X, X ) s f XX x x ( x, x) exp( ), thus the jot pdf of ( Y, Y ) s 4
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 ( byby) ( ayay) f ( y, y ) exp( ) J YY ( ab ab ) ( a b ) y ( aa bb ) y y ( a b ) y J exp( ) ( ) ab ab y /( a b ) ( aa bb ) y y /(( a b )( a b )) y /( a J exp( ( ab ab) / (( a b )( a b)) b ) ). Deote a b, a b, ad ( aa bb ) /(( a b )( a b )), the we have ( aa bb ) / (( a b )( a b )) ( ab a b) / (( a b )( a b )), Ad J y y y y f ( y, y ) exp( ( YY ( ). 5
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 007 Jue, Problem Page 36, Problem Let X be the umber of heads that occurs 3 flps of a co for whch p(0 p ) s the probablty of obtag a head o each flp. Gve X x where x {0,,,3}, cosder a secod expermet cosstg of flppg the same co utl a addtoal x heads occur. Let Y be the umber of tals that occur ths secod expermet before we observe the addtoal x heads. () What s the codtoal probablty mass fucto of Y X? () Verfy that the codtoal pmf of Y X belogs to the expoetal famly. (3) Fd the jot pmf of X ad Y. (4) Evaluate E( Y ). Soluto: 3 x 3x () X has bomal(3, p ), thus PX ( x) p( p) ( x0,,,3). The codtoal pmf of x Y gve X x s: f( y x) P( Y y X x) P(Amog x y tossg, there are y tals ad the last tossg s head) x y p x y ( p ) ( y 0,, ). y () c( p) x y x y f y x p p p y p y y x y x ( ) ( ) exp( log( )) t y x p, ( ) belogs to the expoetal famly. (3) The jot pmf of X ad Y s: x y. Defe hy ( ) y y,ad wp ( ) log( p), we ca see that the codtoal pmf of Y X x y 3 f x y f y x f x p p p p x y y x x y x 3x XY, (, ) ( ( ) X( ) ( ) ( ) ( 0,,,3; 0,, ). (4) EY ( ) EEY [ ( X)], where Y X has a egatve bomal dstrbuto wth the parameter X ad p, thus p EY ( X) ( X ) ad p, 6
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 p p p EY ( ) E[( X) ( E( X)) ( 3 p). p p p 7
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 006 Jauary, Problem Page 7, Problem Let the cotuous radom varable X wth support [-,] have desty fucto, x x0 fx ( x). x 0 x () Let the radom varable T X. Fd PT ( /4). () Fd the momet geeratg fucto for X. (3) Let U be dstrbuted Uform ( /,/ ). Let U ad U be a..d sample from ths Soluto: () dstrbuto. Show that the desty fucto of the radom varable X U U s the fucto f X ( x) from part (). PT PX PX PX P( X /) P( X /) ( / 4) ( / 4) ( / ) ( / ) / ( xdx ) ( x) dx/8/8/4. / () The momet geeratg fucto for X s: (3) For x 0, we have For x 0, we have 0 Xt xt xt 0 M () t Ee e ( x) dx e ( x) dx X exp( t) exp( t) t t t t t t (exp( t ) exp( t ) ). t F ( x) P( U U x) du du X uu x, / u, u/ x/ xu x/ dudu ( x u /) du / / / x/ x/ x/ / (/ ) / (/ ) / xu u u (/) x x/. 8
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 F ( x) P( U U x) du du X uu x, / u, u/ x/ / / xu du / / du du x / / du x(/) x /. Therefore, X U U has the pdf fucto f X ( x) from part (). 9
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 Solutos from Gradg 30
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 008 July, Problem Page 4, Problem As a customer, you receve a promoto letter from a compay. The letter cotas several umbers, such as AB-03, ZH-009, etc. You ca w the prze f oe of your umbers matches oe of thrtee dfferet wg umbers ther lst. There are oe frst prze of $00,000, two secod przes of $50,000 each, ad 0 thrd przes of $5,000 each. However, to be elgble to w, you eed to mal the promoto letter back to the compay at a cost of $0.4 postage. No purchase s requred. From the structure of the umbers that you receved, t s obvous that umbers set out cosst of two letters followed by four dgts. Assumg that the umbers you receved were geerated at radom. () If the letter oly cotas oe umber, what s the probablty that you ca w a prze? How much moey do you expect to w? () If the letter cotas four umbers, what s the probablty that you ca w at least oe prze? (3) If the letter cotas umbers, please fd the smallest that you thk t s worth $0.4 to mal the letter ad justfy your aswer. Soluto: () (5 pots) Let S be the sample space ad A be the evet of wg. The the umber of elemets the sample space s: 0 4 6 ad the umber of elemets A s 3. So the probablty of A s: PA ( ) A / S 3 / (0 6 ).9 0 4 6. () (5 pots) Let X be the amout of moey you wll w. The the support set of X s {00000,50000,5000}. Ad 4 PX ( 00000) / (0 6 ), 4 PX ( 50000) / (0 6 ), ad 4 PX ( 5000) 0 / (0 6 ), so the expected value of X s 4 EX ( ) 50000 / (0 6 ) 0.037. () (5 pots) Let A (,,3,4) be the evet of wg for the th umber. From (), we kow that PA 6 ( ).96 0 (,,3, 4). The the probablty that you ca w at least oe prze s:. Sce 4 4 c 4 c P( A) P(( A) ) P( A ) too. We have: A are depedet, so P( A) PA ( ) ( PA ( )) 7.69 0. 4 4 c 4 6 c A are depedet 3
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 (3) (5 pots) Let X (,, ) be the amout of moey you wll w for the th umber. The Y X s the total moey you wll w. We wat to fd a smallest such that EY ( ) 0.4. Sce EX ( ) 0.037 ad EY ( ) E( X ) *0.037, we ca get. 3
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 008 July, Problem 3 Page 43, Problem 3 Soluto: () (7 pots) Suppose that X, X,, Xm ad Y, Y,, Y are depedet radom samples, we have: EX ( ) E( X)) EX ( ) m m m m m m ; m Var( X ) Var( X ) Var( X ) Var( X ) m m m m Smlarly, we have EY ( ) ad m m. Var( Y ) m, the: EX ( Y) EX ( ) EY ( ) ad Var( X Y ) Var( X ) Var( Y ). m () (8 pots) The mgf of the ormal dstrbuto (, ) s: I addto, ( x) ( x) tx tx tx Mt () Ee ( ) e e dx e dx ( x( t)) ( t) t t ( x( t)) e dxe e dx t t exp( ). m m t X t X m m m () ( / ) X X m ( t/ m) t/ m ( / m) t t M t Ee Ee M t m exp( ) exp( ). Thus, the momet geeratg fucto of X Y s: M () t M () t M () t M () t M ( t) XY X Y X Y ( / mt ) t ( / )( t) ( t) exp( )exp( ) ( / m / ) t ( ) t exp( ). Therefore, X Y has a ormal dstrbuto of (, / m / ). 33
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 (3) (5 pots) From (), we kow that X Y has a ormal dstrbuto of (,4.5/ ), we wat to fd such that X Y ( ) P X Y P 4.5 / 4.5 / ( ( ) ) ( ) 0.95. X Y ( ) Sce Y has a stadard ormal dstrbuto, thus we eed / 4.5 /.96, 4.5 / that s.96 4.5 7.8. Thus 8. 34
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 009 Jauary, Problem Page 45, problem Two far dce are rolled. Let A be the evet that the sum of pots o the faces show s odd ad B be the evet that at least oe three (umber 3) s obtaed. Fd the probabltes of the followg evets: () A B. () A B. (3) A B. (4) ( AB) A. Soluto: A {(,),(,),(,4),(4,),(,6),(6,),(,3),(3,),(,5), (5, ),(3, 4),(4,3),(3,6),(6,3),(4,5),(5, 4),(5,6),(6,5)} B {(3,3),(,3),(3,),(,3),(3,),(3,4),(4,3),(3,5),(5,3),(3,6),(6,3)} 35
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 009 Jauary, Problem Page 45, problem Problem. Let X,, X be a radom sample from the expoetal dstrbuto wth the commo pdf: f ( x) exp( x) for 0 x. Let X () X () X ( ) be ther order statstcs. Please aswer the followg questos: () For j, what s the jot pdf of X () ad X ( j)? (Ht: Please provde the fal aswer. So you oly eed to specfy the fucto ad ts support set. () For j, let S X() ad T X( j) X( ), please derve the jot pdf of S ad T. (3) For j, derve the pdf of T X( j) X( ) usg the jot pdf of S ad T (). Soluto: () The jot pdf s: for 0 x y.! f( x, y) ( )!( j )!( j)! exp( )exp( )[ exp( )] [exp( ) exp( )] j x y x x y [exp( y)] j () Do the trasformato S X() ad T X( j) X( ), the X () S ad X ( j) S T. The Jacob s ad ths s a oe to oe trasformato from ( X(), X ( j) ) to ( ST, ). The jot pdf of S ad T s:! f(,) s t ( )!( j )!( j)! j j exp( s)exp( s t)[ exp( s)] [exp( s) exp( s t)] [exp( s t)]! ( )!( j )!( j)! [exp( s)] [ exp( s)] [exp( t)] [ exp( t)] for 0 s,0 t. (3) The margal pdf of T s: j j 36
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 f () t T 0 0 f(,) s t ds! [exp( t)] [exp( t)] ( )!( j)!( j)! [exp( )] [ ( )] s exp s ds y y dy y s 0 j j! [exp( t)] [exp( t)] ( )!( j)!( j)! [ ] ( exp( )) y j j! j j ( )!( )! [exp( t)] [exp( t)]. ( )!( j )!( j)!! ( )! ( j)!( j)! j j [exp( t)] [ exp( t)]. 37
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 00 July Problem Page 53, Problem A system uses two fuses wth lfetme deoted by Y ad Y, where Y ad Y are depedet ad have expoetal dstrbutos wth meas ad, respectvely, where 0 s a parameter. () Cosder System whch oly eeds oe fuse to work. The other fuse s the used as a backup. Therefore, the lfetme of System, deoted by Z, s the sum of the lfetmes of the two fuses,.e., Z Y Y. Fd the probablty desty fucto of Z. () Cosder System whch fals f ether fuse fals. Therefore, the lfetme of System, deoted by Z, s a fucto of ( Y, Y ) as defed by Z m( Y, Y ). Fd the probablty desty fucto of Z. (3) Calculate the expected lfetme of the systems () ad (). Based o the expected lfetmes that you calculated, whch system has a loger expected lfetme? Does ths make practcal sese? Justfy your aswer. Soluto: The pdf of Y ad Y are respectvely. The cdf of Y ad Y are fy ( y ) exp( y)( y 0) ad fy ( y ) exp( y)( y 0), PY ( y ) FY ( y ) exp( y) ( y 0) ad PY ( y ) FY ( y ) exp( y)( y 0), respectvely. B.. Because Y ad Y are depedet, so the jot pdf of Y ad Y s For y 0, f ( y, y ) f ( y ) f ( y ) Y, Y Y Y exp( y) exp( y)( y 0, y 0). 38
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 PY ( Y y) f ( y, y ) dydy y yy, y, 0, y0 Y Y y y y { [ exp( y 0 0 )] dy} fy ( y ) dy So the pdf of Z Y Y, y [ exp( ( y y 0 )] exp( y) dy y y exp( y ) dy exp( y) exp( y ) dy exp( y) exp( y)[ exp( y)] exp( y) exp( y) 0 0 f ( ) Z z for z 0, s (5 pots) B.. For z 0, d d fz ( z ) F ( ) ( Z z P Y Y z) dz dz d [ exp( z) exp( z)] ( pots) dz exp( z) exp( z) So the pdf of Z, PZ ( z) PZ ( z) P(m( YY, ) z) PY ( zpy ) ( z) [ PY ( z)][ PY ( z)] 3 exp( z)exp( z) exp( z) f ( ) Z z for z 0, s (5 pots) d d 3 3 3 fz ( z ) P( Z z) [ exp( z)] exp( z). ( pots) dz dz EZ E( Y Y ) EY EY 3. ( pots) B.3. Note the Z has a expoetal dstrbuto wth parameter, so EZ 3 Therefore we have EZ EZ. ( pots). 3 I practcal, because both Y ad Y have the support of (0, ) ( pot`). So Z YY Y m( Y, Y) Z( pot), therefore we have EZ EZ. 39
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 00 July, Problem 3 Page 53, Problem 3 The umber of customers to shop a grocery store a day, X, ca be descrbed as the follow x probablty mass fucto: f( x ) exp( ) for x 0,,, where 0 s a ukow x! parameter. We are terested estmatg the probablty of o customer a day, ( ) PX ( 0). Let X,, X dstrbuto: be a radom sample from a populato wth the above () Fd the Cramer-Rao lower boud for a ubased estmator of ( ). () Fd the uformly mmum varace of ubased estmator (UMVUE) of ( ) ad deote t by ( ˆ ). (3) Compare the varace of ( ˆ ) ad the Cramer-Rao lower boud from () for ay gve by drectly calculatg the varace of ( ˆ ). 0 Soluto: ( ) PX ( 0) exp( ) exp( ) ( pot). 0! x () Sce f( x ) exp( ) belogs to the expoetal famly ad X x!,, X are a radom sample from f( x ), so the CRLB of a ubased estmator of ( ) s d [ ( )] CRLB d ( pot). E[ log f ( X )] d d ( ) exp( ) exp( ) d d ( pot). x Because log f( x ) log[ exp( )] xlog( ) log( x!), so we have x! x log f( x ) ad x log f( x ). So X E[ log f( X )] E EX ( pots). 40
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 [ exp( )] So CRLB= ( ) exp( ) ( pot). x () Sce f( x ) exp( ) belogs to the expoetal famly ad the parameter space 0 x! cotas a ope set, so T X s a complete, suffcet statstc ( pots) ad the pmf of t ( ) T s ft ( t ) exp( ) ( pot). t! Defe hx (,, X) f X 0 ad hx (,, X) 0 otherwse. The Eh( X,, X) P( h( X,, X) ) P( X 0) ( ) so hx (,, X ) s a ubased estmator of ( ) ( pots). E( h( X,, X) T t) P( h( X,, X) T t) PhX ( (,, X), Tt) PT ( t) PX ( 0, X t) PX ( 0, X t) P( X t) P( X t) t [( ) ] exp( ( ) ) PX ( exp( ) 0) P( X ) t t! t P( X ) [ ] exp( ) t t! ( ) t T So the UMVUE of ( ) s ( ) ( ) X ( pot). (3 pots) T X (3) Sce ( ) ( ) X s the UMVUE of ( ), so E( ) exp( ) ( pot). 4
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 t X X ( ) t E[( ) ] E[( ) ] ( ) exp( ) t0 t! ( ) t ( ) exp( ) t0 t! ( ) t ( ) (3 pots) exp( ) ( ) exp( ) t0 ( ) t! exp( ) exp( ) exp( )exp( ) ( ) exp( ) So the varace of Sce ( ˆ ) s ( pot), exp( )exp( ) [exp( )] exp( )[exp( ) ] exp( ) ( ) /! ( ) /!, so the varace of UMVUE of ( ) 0 s greater tha the CRLB of ubased estmator of ( )( pot). 4
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 00 July, Problem 4 Page 54, Problem 4 Oe observato, X, from the followg probablty desty fucto: x 0, where 0 s a ukow parameter. x f( x ) exp( ) for () Show that X has the mootoe lkelhood rato (MLR). () Fd the maxmum lkelhood estmator (MLE) of based o X. (3) Fd the restrcted rage MLE of based o X for 0, where 0 s a kow postve costat. (4) Costruct a UMP level (0 ) test for H0 : 0 vs. H: 0 based o X ad fd the rejecto rego. Soluto: () For ay 0 ad x 0 ( pot), we f f Sce 0, so X X x exp( ) ( x ) exp[( ) x] ( pots) ( x ) x exp( ) ad X has the MLR ( pots). () The lkelhood fucto s fx ( x ) 0, so s a creasg fucto of x, so f ( x ) x x log L( x) log( ), ad log L( x). Let log L( x) 0 X x L( x) f( x ) exp( )( x 0) ( pot), so, we have a uque soluto ˆ x ( pots). I addto, x log L( x) ( ) 0 x 3 x x ( pots), the MLE of s ˆ X. (3) The lkelhood fucto s x L( x) f( x ) exp( )( x 0). If 0 x, the we kow that L( x) acheves ts maxmum at x from () ( pot). 43
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 x If 0 x, the we kow that log L( x) ( x) 0 for all 0 x, so L( x) s a creasg fucto of ad acheves ts maxmum at 0 ( pots). Therefore the restrcted rage MLE of based o X for 0 s ˆ X, 0 X ; ( pot) 0, 0 X. (4) X s a suffcet statstc ad t has the MLR accordg to () ( pot), so accordg to the Karl-Rub theorem ( pot), the UMP level test for H0 : 0 vs. H: 0 based o X has the followg rejecto rego: R { X c} ( pot), where c satsfes P ( X c) ( 0 pot). Because x x c P ( X c) exp( ) dx exp( ) exp( ) 0, so we have c c 0 0 0 0 c log( ). The rejecto rego s : 0 R { X 0 log( )} ( pots). 44
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 0 Jauary, Problem 3 Page 56, Problem 3 Suppose X has a gamma (, ) desty wth 3 : Codtoal o X f x x x x ( ) (, ) exp( ), 0, 3, 0. x, the dstrbuto of Y s x exp( yx) for y 0. () Fd E( Y ) ad Var ( Y ) accordg to: E( Y) E[ E( Y X)] ad Var( Y ) E[ Var[ Y X ]] Var[ E( Y X )]. () Wrte dow a expresso for the jot desty of X ad Y. (3) Determe the margal desty of Y. (4) Determe the codtoal desty of X gve Y y. Soluto (0 pots): The codtoal pdf of Y gve X Part ( pots): x s f ( y x) xexp( yx) for y 0. E( Y X x) yfyx x( y x) dy yx exp( xy) dy 0 0 x YX x ( pots) (they ca also drectly use the codtoal pdf of Y s a expoetal dstrbuto wth parameter x, so EY ( X x) ). x So EY ( X) We have: ( pot). X EY ( ) EEY [ ( X)) E f( x, ) dx X 0 x ( ) x exp( xdx ) x exp( xdx ) 0 x ( ) 0 ( ) ( ) ( ) Smlarly, we have ( pots) VarY ( X x) EY ( X x) [ EY ( X x)] YX x 0 0 ad E( Y X x) y f ( y x) dy y x exp( xy) dy x (3 pots) (they ca also drectly use the x 45
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 codtoal pdf of Y s a expoetal dstrbuto wth parameter x, so Var( Y X x) ), so x we have Var( Y X ) ad EVarY [ ( X)] E[ ] ( pot). X X We have: Var( E( Y X )) Var( ) E( ) ( E ) E( ) ( ) X X X X E( ) f( x, ) dx 0 X x ( ) x exp( xdx ) x exp( xdx ) 0 x ( ) 0 ( ) ( ). ( ) ( )( ) So the varace of Y s: ( pots) Var( Y ) E( ) E( ) ( ) X X ( pot) ( ) ( )( ) ( ) ( ) Part ( pots): The jot pdf of X ad Y s fxy, ( x, y) f( x, ) fyx x( y x) ( pots) ( ) x exp( xx ) exp( xy) x exp( ( yx ) ) ( ) ( ) for x 0 ad y 0. Part 3 (3 pots): The margal dstrbuto of Y s for y 0. 0 X, Y ( ) x 0 f ( y) f ( x, y) dx Y exp( ( yxdx ) ) (3 pots) ( ) ( ) ( ) ( y) ( y) 46
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 Part 4 (3 pots): The codtoal dstrbuto of X gve Y y s: fxy y( x y) fx, Y( x, y)/ fy( y) ( ) x exp( ( y) x) / [ ] (3 pots) ( ) ( y) ( y) ( ) x exp( ( y) x) ( ) for x 0, whch s a gamma (, y). 47
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 0 Jauary, Problem 4 Page 57, Problem 4 Suppose Y ~ depedet Posso ( ),,,, where 0 s ukow parameter ad are fxed postve costats. () Fd a oe dmesoal suffcet statstc for. () Fd the maxmum lkelhood estmator of, ˆ. (3) Show ˆ s a ubased estmator of ad calculate the mea squared error (MSE) of ˆ. (4) A alteratve estmator of s * ( Y / ). Show * s a ubased estmator of ad calculate the mea squared error (MSE) of *. (5) Dscuss whch estmator of ˆ ad * s better for. Soluto (0 pots): Part (4 pots): The jot pdf s: ( ) f( y,, y ) f ( ) exp( ) Y y y! exp( ) y y y! y ( pot) Let T T Y, ( y ( (,, y gt y ),, y) ) exp( y ), ad hy (,, y), y! the f ( y,, y ) g( T( y,, y) )* h( y,, y). ( pots). Accordg to the factorzato theorem ( pot), T Y s a oe dmesoal suffcet statstc for. Part (5 pots): The lkelhood fucto s: So we have: y y L( y,, y) exp( ) ( pot). y! y log L( y,, y) ( )log( ) log( ) y y! 48
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 log L ( y,, y ) y ( pot). log (,, ) 0, we have ˆ y /. ( pot) Let L y y y log L( y,, y ) ˆ y ˆ 0 ( pots), so log L( y,, y ) acheve ts maxmum at ˆ y /, so ˆ Y / s the MLE of. Part 3(4 pots): Because Y ~ depedet Posso ( ), so EY ( ) ad Var( Y ). E( ˆ ) E( Y / ) E( Y)/ ( pots) So ˆ s a ubased estmator of. EY ( )/ / Sce ˆ s a ubased estmator of, the MSE of ˆ s the varace of ˆ. Ad sce Y are depedet, we have MSE( ˆ ) Var( ˆ ) Var( Y / ) Var( Y )/( ) ( pots) [ ( )]/( ) [ ]/( ) / Part 4 (4 pots): Var Y So * s a ubased estmator of. E( *) E( Y / ) ( / ) E Y ( pots) EY ( )/ / Sce * s a ubased estmator of, the MSE of * depedet, we have: Part 5 (3 pots): s the varace of * MSE( *) Var( ˆ *) Var( Y / ) Var( Y / ) ( / ) ar( )/ Var Y V Y ( pots) / /. Ad sce Y are 49
Problem Solutos for BST 695: Specal Topcs Statstcal Theory, Ku Zhag, 0 Compare the MSE of ˆ ad * for : ˆ MSE( ) MSE( *) ( ) 4 ( ) *( ) 4( ) 4( ) ( ) 0 So ˆ s a better estmator tha * because t has a smaller MSE. (3 pots) To prove the geeral case, we eed the equalty betwee the arthmetc mea ad the harmoc mea. Specfcally, we have the followg equalty to the root of mea square, arthmetc mea, geometrc mea, ad harmoc mea:. 50