THE ROYAL STATISTICAL SOCIETY 06 EAMINATIONS SOLUTIONS HIGHER CERTIFICATE MODULE 5 The Socety s provdg these solutos to assst cadtes preparg for the examatos 07. The solutos are teded as learg ads ad should ot be see as "model aswers". Users of the solutos should always be aware that may cases there are vald alteratve methods. Also, the may cases where dscusso s called for, there may be other vald pots that could be made. Whle every care has bee tae wth the preparato of these solutos, the Socety wll ot be resposble for ay errors or omssos. The Socety wll ot eter to ay correspodece respect of these solutos.
HC5 06 - solutos () P( ) P( ) P( ) P( 3)... (method) = ( ) ( ) ( )... (terms substtuted) = ( ) ( ( ) ( )...) (commo factor) Fal term s sum to fty of a GP wth frst term ad commo rato where 0. ( ) So P( ) ( ) as requred. (correct use of GP) ( ( )) (Full mars also for correct soluto by ducto) TOTAL 4 () ad ths s observed for 4 players. (correct term for 4) 0 P ( ) ( ) ad ths s observed for 48 players. (correct term for 48) P ( ) ( ) ad ths s observed for the other 8 players. P ( ) ( ) So the lelhood for the ta s gve by 4 48 8 (correct term for 8) L( ) ( ) ( ) (combg terms) 7 304 = ( ) as requred. l( ) log L( ) 7 log 304 log( ). (log lelhood) dl( ) 7 304 ad settg ths equal to zero we have d (correct dervatve), (set to 0) ˆ ˆ ˆ 7 7( ) 304 0.95. 376 (0.95) To chec that ths gves a maxmum we dfferetate aga: dl( ) 7 304 d ( ) whch s egatve. (egatve d derv ) TOTAL 9 () The stard error of ˆ ca be estmated by dl( ) d where the deomator s evaluated at ˆ. Ths gves 0.00048 0.003 7 304 0.95 0.8085 (method), (substtute), (0.003 or equv. varace) ( ) [ If cadtes wor out the expected formato, ths gves a varace of where ( ) 00. Substtutg ˆ to ths gves a varace of 0.000480. All 3 mars above, ad approprate follow-o mars below, should be awarded.] The the approxmate 95% cofdece terval s gve by ˆ.96 se. ( ˆ ).e. (.96) (method) 0.95.96 0.003 whch s (0.57, 0.33). (0.57), (0.33) TOTAL 7
HC5 06 - solutos. (a) () M '( t) R( t) log M ( t) R '( t). M () t (st derv correct) The usg the quotet rule, () We have that M '(0) E( ) ad So Also by defto, E ( ) R '(0) E( ). Also M ( t) M ''( t) M '( t) R''( t). ( d derv correct) M () t M 0 (0) ( ) (). TOTAL ''(0) E( ). (usg zero to fd expectatos) M E e E (M(0)=) E( ) ( E( )) R''(0) E( ) ( E( )) Var( ). (st derv show correct), ( d derv show correct) TOTAL 4 (b) () y t y e ( e ) MY ( t) E( e ) e e e. e y! y! = t ty ty e y0 y0 t ( e e ). (defto as expectato), (tegrato step), (recogto of sum) TOTAL 3 () t ( e ) 0 0 '( ) t Y ( ). M t e e E Y e e ( st derv correct), (correct substtuto) 3 3 3 3 E ( Y E( Y )) E ( Y ) E( Y ) 3 E( Y ) 3 E( Y ). (substtute for E(Y)), (expaso) 3 Usg E( Y ), E( Y ) gve the questo, we have 3 3 3 E ( Y E( Y )) 3 3 ( ) 3. 3 3 3 3 = 3 3 3 3 as requred. (substtuted ad smplfed) TOTAL 5 () Mgf of sum of depedet radom varables s the product of ther mgfs. (meto depedece), (state result) Here they are detcally dstrbuted, so we rase to the th power. (th power requred) ( e ) t M ( t) M ( t) e (power correctly appled) S Y Ths ca be detfed as the mgf of Posso ( ). (Posso detfed) Sce there s a uque oe-to-oe relatoshp betwee a dstrbuto ad ts mgf, S follows a Posso ( ) dstrbuto. (state uqueess) TOTAL 6
HC5 06 - solutos 3. () Probabltes sum to, so (sum to ) (4 8 4 6 4 3) 84. 84 (/84) TOTAL () Sample sze s oly 3, so there caot be whte dce ad blue dce. (correct reaso) 9 There are 84 ways of drawg the sample altogether. If ad Y 0 the there 3 3 are whte dce ad red de draw. The whte dce ca be draw 3 ways ad the 4 red de 4 ways, so the probablty s 3 4 as gve. (84 ways), (3 ways), (4 ways), (probablty calculato) TOTAL 5 4 8 35 () PY ( 0). So the codtoal dstrbuto of gve that Y = 0 s P( x, Y 0) gve by P( x Y 0).e. (35/84), (method) PY ( 0) x 0 3 4 8 P(=x Y=0) 35 35 35 35 (pmf correct) The E( Y 0) 8 4 3 45 9. (9/7) 35 35 7 8 48 9 75 5 5 9 4 E( Y 0) so Var( Y0). (5/7), (4/49) 35 35 7 7 7 49 TOTAL 6 (v) margal dstrbuto: 0 3 0 45 8 Y margal dstrbuto: 0 35 4 7 84 45 36 3 So E ( ). 84 ( margal correct), (E()=) 4 4 56 So EY ( ). (Y margal correct), (E(Y)=56/84) 4 6 3 4 E( Y ). (E(Y)=4/84) 4 56 4 So Cov(, Y ) E( Y ) E( ) E( Y ). (Cov = -/6) 84 6 Negatve covarace maes sese: the more whte dce the sample, the fewer spaces there are for blue dce. (vald commet) TOTAL 7
HC5 06 - solutos 4. () Sce we ow that E E E ( ), ( ) ( ) ( ). the sce Now E( ) E( ) ad so ( ) ( ), E E (alteratve form), (wrtte as expectatos) E ( ) ( ). (E( )), (multply by ) Var( ) E( ) E( ). (var of mea), (fal expresso) We requre E( ) ( ) ( ). So E ( ) ( ) as requred. (correctly combed) For a ubased estmator, we requre the expectato to equal. From the prevous le, ( ) E ( )., ( ) so ( ) s a ubased estmator for. (requre expectato = ), (fal estmator correct) TOTAL 9 () E( Sa ) E a( ) ae ( ) a( ). (costat out of sum), (correct fal expresso) So Bas( Sa ) a( ) ( a a ). (method), (bas correct) TOTAL 4 4 () Var( Sa ) Var a( ) a Var ( ) a ( ). (costat squared) 4 4 So MSE( S ) a ( ) ( a a ). (MSE correct) a For max/m, set dm equal to zero: (method) dm 4 4 4 a( ) ( a a )( ) 0. (derv correct) 4 Dvdg through by ( ) : a a a 0 a( ) a. ( /(+)) To show that ths gves a mmum value, loo at the secod dervatve: dm 4 4 4( ) ( ), whch s clearly postve, so the requred value of a s. ( d derv correct), (postve) TOTAL 7