THE ROYAL STATISTICAL SOCIETY 2016 EXAMINATIONS SOLUTIONS HIGHER CERTIFICATE MODULE 5

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1 THE ROAL STATISTICAL SOCIET 6 EAMINATIONS SOLUTIONS HIGHER CERTIFICATE MODULE The Society is providing these solutions to assist candidates preparing for the examinations in 7. The solutions are intended as learning aids and should not be seen as "model answers". Users of the solutions should always be aware that in many cases there are valid alternative methods. Also, in the many cases where discussion is called for, there may be other valid points that could be made. While every care has been taen with the preparation of these solutions, the Society will not be responsible for any errors or omissions. The Society will not enter into any correspondence in respect of these solutions.

2 3 x x y y y y.(i) ( ). x xy dxdy 3 dy dy Setting this equal to gives. (obtain /3), (set equal to one) TOTAL 3 x x y 3 y. (method correct integral), (.(+3y), (correct domain) (ii) y f ( y) f ( x, y) dx x xy dx 3 3y for 3y 3 7 (correct integral and limits) P( ) f ( y) dy y. (iii) So Then 3 f ( x, y) x xy 6( x xy) f ( x y) f ( y) 3 y ( 3 y) 6( x x) (7/) TOTAL for x. (correct formula used) (f(x y) correct), (correct domain) f ( x ) for x. (substitute y = correctly) x x E( ) xf ( x ) dx ( x x ) dx. 4 6 x x E E (correct method), (7/) E( ) x f ( x ) dx. (correct method), (7/) Then (iv) Var( ) ( ) ( ( )). x x 3 3 xy P( ) f ( x, y) dydx ( x xy) dydx x y dx = x 3 3 x 9 dx 4 4 (correct formula), (/). TOTAL. (correct integration limits) (integrating x^3/ at second stage) (correct final integral) (Note: integration order can be reversed, in which case limits for x are y and, then y from to. Full mars for this correct solution also). TOTAL 3 x

3 (i) P( ) P( ) P( ) P( 3)... (method) = ( ) ( ) ( )... (terms substituted) = ( ) ( ( ) ( )...) (common factor) Final term is sum to infinity of a GP with first term and common ratio where. ( ) So P( ) ( ) as required. (correct use of GP) ( ( )) (Full mars also for correct solution by induction) TOTAL 4 (ii) and this is observed for 4 players. (correct term for 4) P ( ) ( ) and this is observed for 48 players. (correct term for 48) P ( ) ( ) and this is observed for the other 8 players. P ( ) ( ) So the lielihood for the data is given by (correct term for 8) L( ) ( ) ( ) (combining terms) 7 34 = ( ) as required. l( ) log L( ) 7 log 34 log( ). (log lielihood) dl( ) 7 34 and setting this equal to zero we have d (correct derivative), (set to ) ˆ ˆ ˆ 7 7( ) (.9) To chec that this gives a maximum we differentiate again: dl( ) 7 34 d ( ) which is negative. (negative nd deriv ) TOTAL 9 (iii) The standard error of ˆ can be estimated by dl( ) d where the denominator is evaluated at ˆ. This gives (method), (substitute), (.3 or equiv. variance) ( ) [ If candidates wor out the expected information, this gives a variance of where n( ) n. Substituting ˆ into this gives a variance of.48. All 3 mars above, and appropriate follow-on mars below, should be awarded.] Then the approximate 9% confidence interval is given by ˆ.96 se. ( ˆ ) i.e. (.96) (method) which is (.7,.33). (.7), (.33) TOTAL 7

4 M '( t) 3. (a) (i) R( t) log M ( t) R '( t). M () t (st deriv correct) Then using the quotient rule, (ii) We have that M '() E( ) and So Also by definition, E ( ) R '() E( ). Also M ( t) M ''( t) M '( t) R''( t). ( nd deriv correct) M () t M () ( ) (). TOTAL ''() E( ). (using zero to find expectations) M E e E (M()=) E( ) ( E( )) R''() E( ) ( E( )) Var( ). (st deriv shown correct), ( nd deriv shown correct) TOTAL 4 (b) (i) y t y e ( e ) M ( t) E( e ) e e e. e y! y! = t t ty e y y t ( e e ). (definition as expectation), (integration step), (recognition of sum) TOTAL 3 (ii) t ( e ) '( ) t ( ). M t e e E e e ( st deriv correct), (correct substitution) E ( E( )) E ( ) E( ) 3 E( ) 3 E( ). (substitute for E()), (expansion) 3 Using E( ), E( ) given in the question, we have E ( E( )) 3 3 ( ) = as required. (substituted and simplified) TOTAL (iii) Mgf of sum of independent random variables is the product of their mgfs. (mention independence), (state result) Here they are identically distributed, so we raise to the nth power. (nth power required) n n ( e ) t M ( t) M ( t) e (power correctly applied) S This can be identified as the mgf of Poisson ( n ). (Poisson identified) Since there is a unique one-to-one relationship between a distribution and its mgf, S follows a Poisson ( n ) distribution. (state uniqueness) TOTAL 6

5 4. (i) Probabilities sum to, so (sum to ) ( ) (/84) TOTAL (ii) Sample size is only 3, so there cannot be white dice and blue dice. (correct reason) 9 There are 84 ways of drawing the sample altogether. If and then there 3 3 are white dice and red die drawn. The white dice can be drawn in 3 ways and the 4 red die in 4 ways, so the probability is 3 4 as given. (84 ways), (3 ways), (4 ways), (probability calculation) TOTAL (iii) P ( ). So the conditional distribution of given that = is P( x, ) given by P( x ) i.e. (3/84), (method) P ( ) x P(=x =) (pmf correct) Then E( ) (9/7) E( ) so Var( ). (/7), (4/49) TOTAL 6 (iv) marginal distribution: marginal distribution: So E ( ). 84 ( marginal correct), (E()=) So E ( ). ( marginal correct), (E()=6/84) E( ). (E()=4/84) So Cov(, ) E( ) E( ) E( ). (Cov = -/6) 84 6 Negative covariance maes sense: the more white dice in the sample, the fewer spaces there are for blue dice. (valid comment) TOTAL 7