THE ROYAL STATISTICAL SOCIETY 4 EXAINATIONS SOLUTIONS GRADUATE DIPLOA PAPER I STATISTICAL THEORY & ETHODS The Societ provides these solutions to assist candidates preparing for the examinations in future ears and for the information of an other persons using the examinations The solutions should NOT be seen as "model answers" Rather, the have been written out in considerable detail and are intended as learning aids Users of the solutions should alwas be aware that in man cases there are valid alternative methods Also, in the man cases where discussion is called for, there ma be other valid points that could be made While ever care has been taken with the preparation of these solutions, the Societ will not be responsible for an errors or omissions The Societ will not enter into an correspondence in respect of these solutions RSS 4
Graduate Diploma, Statistical Theor & ethods, Paper I, 4 Question X and Y are independent, each with probabilit function given b P(W = w) = ( θ) w θ, for w =,, and where < θ < (i) Let U = X + Y U ma take the values, 3, B the independence of X and Y, u P ( U = u) = P( X = x and Y = u x) for u =, 3, x= Hence u { θ θ} {( θ) θ} θ ( θ) θ ( θ) u x u x u u, PU= u = = = u x= x= for u =, 3, [this is a negative binomial distribution] (ii) If U (= X +Y) = u, X must take values in the range,,, u Then ( = and = ) PU ( = u) P X x Y u x P X = xu = u = x {( ) } ( ) ( u ) θ ( θ) u x { } θ θ θ θ = = u u (for x =,,, u ) This is a discrete uniform distribution on the integers,,, u (iii) Suppose And takes X attempts and Bob takes Y attempts Then X and Y are independent geometric random variables with θ = 4 Thus from part (i), P(X + Y = 6) = (6 )(4) (6) 4 = 37 From part (ii), P(X < Y X + Y = 6) = P(X = or X + Y = 6) = 6 + 6 = 5
Graduate Diploma, Statistical Theor & ethods, Paper I, 4 Question (i) (a) We have P(U = u) = /m, for u =,,, m So EU [ ] upu ( m) Also, m m+ m + = = + + + = = m m ( + )( + ) ( + )( + ) m m m m m E U = u P( u) = ( + + + m ) = = m m 6 6 and therefore Var(U) = E[U ] {E[U]} m + m + m+ m+ m m m ( 4m 3m 3) + = = + = = 6 m + (b) V = U + k, so E [ V] = + k and Var ( V ) m = (ii) Given that Y =, the first rolls must have been 6s and the final roll not a 6 Thus X is a discrete uniform random variable with, using the notation of part (i)(b), 5+ 5 k = 6( ) and m = 5 Hence E[ X] = + 6 ( ) = 6 3 and Var ( X ) = = These are the values conditional on Y = 6 5/6 5 Now, Y is a geometric random variable with EY [ ] = = and Var ( Y ) 6 = = 5 5 6 6 5 36 E[ X] = E E( X Y) = E[ 6Y 3] = 6E[ Y] 3 = 3 = 5 5 We also need [ ] E Var X Y = E = and ( E X Y ) ( Y ) ( Y) Var = Var 6 3 = 36 Var = 6 / 5, from which (using the result quoted in the question) 6 66 Var ( X ) = + = 5 5
Graduate Diploma, Statistical Theor & ethods, Paper I, 4 Question 3 (i) θ u e θ E U = u du = u e du Γ Γ α α θu α m m α+ m θu [put t = θu] ( α) ( α) α m ( α ) θ Γ( a) ( α m) α α θ α + m t θ Γ + = t e dt = + α+ m Γ θ Hence EU [ ] = α Also θ ( + ) = and so Var(U) = E[U ] {E[U]} = E U α α θ α θ (ii) e x f X x = 8xe d = 8x = 4xe = x (for x > ) This is gamma with α = and θ = = = = fy xe dx e x e (for > ) 8 8 4 x This is gamma with α = 3 and θ = Therefore from part (i), E[X] = and Var(X) = /, and E[Y] = 3/ and Var(Y) = 3/4 [ ] 8 ( 8 = x= ) 4 E XY x xe dxd e x dx d = = = e d [put t = ] 4 8 t t dt 4! e 3 6 = = Γ 5 = = 8 3 3 ( XY) E( X) EY Cov, = = =, and the correlation between X and Y is ( XY) ( X) Var ( Y) 3 4 Cov, = = Var 3
Graduate Diploma, Statistical Theor & ethods, Paper I, 4 Question 4 X and Y are uniform(, ) U = X Y, V = X + Y So X = ½(U + V) and Y = ½(V U) (i) The limits of U are (, ) and of V are (, ) The diagram shows the square region where f (u, v) v u The Jacobian of the transformation is J x u u x v v = = = Therefore, for u, v in the region shown, and using the independence of X and Y, f u, v = J f x, = f x f = = (ii) For u, For u, For v, For v, + u u f u = dv= + u u u f u = dv= u f v = du = v v v v f v = du = v v (f (u) and f (v) are zero elsewhere) These distributions are identical except for a shift of one unit in the x-direction (iii) We now have W = a + tx and Z = b + ty (t > ) So W + Z = (a + b) + t(x + Y) = (a + b) + tv where V is as in parts (i) and (ii) So, using the result for V in part (ii), t t / / P W + Z < a+ b+ = P tv < = P V < = vdv= v = 8
Graduate Diploma, Statistical Theor & ethods, Paper I, 4 Question 5 (i) The method is to complete the square in the exponent in the integral and then recognise that this re-creates the pdf of a Normal distribution so that the value of the integral is simpl tx tx x µ X () t = E e = e exp dx Thus the exponent is σ π σ { } x µ tx = ( x µ x + µ σ tx) = x ( µ + σ t) µσ t σ t σ σ σ 4 x µ σ t µ t+ σ t + Hence X () t = e exp dx = exp µ t σ t + σ π σ tax ( b) bt ( atx ) bt (ii) ax b () t E e + + = e E e = = e X at Taking a = /σ and b = µ/σ we have Z, and so µ σ µ t σ t Z () t = e + = e σ σ t/ t / exp Putting µ = and σ = in the result of part (i), we see that this is the moment generating function of the standard Normal distribution Therefore (assuming without proof the uniqueness propert between distributions and their moment generating functions) Z follows the standard Normal distribution (iii) We expand Z (t) as a power series in t and use the result that the rth moment of Z about the origin, E[Z r ], is the coefficient of t r /r! in this expansion [Alternativel, use the result that it is given b the rth derivative evaluated at t = ] () ( ) ( ) 3 t t 4 6 t t t Z t = + t + + + = + + + +! 3! 8 48 We need E[Z ] and E[Z 4 ], which we see are and 3 respectivel So Var(Z ) = E[Z 4 ] {E[Z ]} = 3 =
Graduate Diploma, Statistical Theor & ethods, Paper I, 4 Question 6 (i) F( x) = P( X x) = P( X x bo) P( bo) + P( X x girl) P( girl) = { } F x + F x d = = + dx Hence the pdf of X is f ( x) F( x) { f( x) f( x) } ( ) ( ) E X = xf x dx= xf x dx+ xf x dx = µ + µ (ii) [ ] ( ) = = + E X x f x dx x f x dx x f x dx [now use σ = E[X ] {E[X]} ] {( σ ) } µ ( σ µ σ µ µ ) = + + + = + + X = σ + µ + µ µ + µ = σ + µ + µ µ µ 4 4 and thus ( ) Var ( ) = σ + ( µ ) µ 4 (iii) As the sample is large, we ma use the central limit theorem to sa that the distribution of X is approximatel Normal, whatever the distributions f and f r The mean of X is equal to the mean of X, ie r µ + µ Var n n σ 4 µ µ ( X ) r = Var X = + (iv) Let the (independent) sample means for bos and girls be X and X Their means are µ and Xst = X+ X has mean µ, and their variances are each σ /n So σ σ σ ( µ + µ ) and variance + = Thus both X r and 4 n n n the latter has smaller variance; it is more precise X st are unbiased but
Graduate Diploma, Statistical Theor & ethods, Paper I, 4 Question 7 (i) (a) The distribution of X is x 3 P(x) 64/5 48/5 /5 /5 cdf F(x) 5 896 99 36 < 5 and so corresponds to x = 6789 is between 5 and 896 and so corresponds to x = 355 < 5 and so corresponds to x = 898 < 5 and so corresponds to x = So the sample is,,, (b) x 4 4 x 4 3 t / / / t x 4 / F x = t e dt = e = e Thus the inverse cdf method gives u = e x, where u is the random number, so x= log( u) /4 u u log( u) [ log( u)] /4 36 6388 8963 973 6789 3 7 77 355 6448 8776 9679 898 7 6844 996 These are the required random numbers (ii) Inter-arrival times, X, are exponential with parameter ½ Given u (as above), the inverse cdf method leads to u = e x/, ie x = log( u) [as is tabulated in column 3 above] First arrival time is at 8963 Service starts immediatel and lasts (uniform distribution on (5, 5)) 5 + 6789 = 789, and thus ends at 8963 + 789 = 375 Second arrival time is at 8963 + 8776 = 7739 Service cannot start until 375 and then lasts 5 + 898 = 7898, and thus ends at 375 + 7898 = 4865 Expressing the simulated results to the nearest minute after 9 am, the first customer arrives at 9 and leaves at 3, the second arrives at 8 and leaves at 49
(i) Graduate Diploma, Statistical Theor & ethods, Paper I, 4 Question 8 Consider the number of balls in urn A, an integer from to Define the p = P j balls in urn A at step n i balls in urn A at step n transition probabilit ij Then we have p =, p = for j ; j i i for i =,,,, pii =, pii+ =, pij = for j i or i+ ; p =, p = for j j (ii) Let the stationar probabilities be π, π,, π We have π = p π Hence j ij i i= j+ j+ π = π; for j =,,,, π j = π j + π j+ ; π = π It is required to show that π j = j ( j =,,, ) satisf these equations Immediatel, = = and π π π / π = = / Thus π = π and π = π, as required To confirm that the general equation is satisfied, consider j+ j+ j+! j+! + = + j j+ j! j+! j+! j! ( j j )!!!! = + = + = = j! j! j! j! j! j! j! j! j ultipling b (/), we see that the general equation ( j =,,, ) is satisfied (iii) Let X be the number of balls in urn A The given probabilities form a binomial distribution with parameters (6, ½) So E(X) = 6 ½ = 3 and Var(X) = 6 ½ ½ = 5 P(X = 34) can be approximated b P(335 < N(3, 5) < 345) Standardising to N(, ) gives P(937 < N(, ) < 69) = 8774 869 = 65