November 2002 Course 4 solutions

Transcription

1 November Course 4 solutios Questio # Aswer: B φ ρ = = 5. φ φ ρ = φ + =. φ Solvig simultaeously gives: φ = 8. φ = 6. Questio # Aswer: C g = [(.45)] = [5.4] = 5; h= =.4. ˆ π =.6 x +.4 x =.6(36) +.4(4) = (5) (6) Questio # 3 Aswer: D N is distributed Poisso(λ) µ = E( λ) = αθ = (.) =.. v= E λ = a= Var λ = αθ = =. 5 k = = ; Z = = / 6 7 Thus, the estimate for Year 3 is 5 (.5) + (.) = ( ).; ( ) (.).44. Note that a Bayesia approach produces the same aswer.

2 Questio # 4 Aswer: C At the time of the secod failure, ˆ 3 Ht () =. + = 3 = At the time of the fourth failure, ˆ Ht () = = Questio # 5 Aswer: E R ˆ x 4 = =.65 = i β yi Questio # 6 Aswer: B The likelihood is: x j rr ( + ) L( r+ x ) j β r+ x j j= x j!( + β) j= x r x j j L = β ( + β). The loglikelihood is: l = xjl β ( r+ xj)l( + β) j= xj r+ xj l = = j= β + β = x ( + β) ( r+ x ) β = x rβ j j j j= j= = x rβ; βˆ = x / r.

3 Questio # 7 Aswer: C The Bühlma credibility estimate is Zx+ ( Z ) µ where x is the first observatio. The Bühlma estimate is the least squares approximatio to the Bayesia estimate. Therefore, Z ad µ must be selected to miimize + µ + + µ + + µ [ Z ( Z).5] [ Z ( Z).5] [3 Z ( Z) 3]. Settig partial derivatives equal to zero will give the values. However, it should be clear that µ is the average of the Bayesia estimates, that is, µ = ( ) =. 3 The derivative with respect to Z is (deletig the coefficiets of /3): ( Z +.5)( ) + (.5)() + ( Z )() = Z =.75. The aswer is.75() +.5() =.5. Questio # 8 Aswer: E / The cofidece iterval is ˆ θ ˆ θ ( St ( ), St ( ) ). Takig logarithms of both edpoits gives the two equatios l.695 = = l St ˆ( ) θ l.843 =.779 = θ l St ˆ( ). Multiplyig the two equatios gives.64 = [l St ˆ( )] l St ˆ( ) =.498 St ˆ( ) = The egative square root is required i order to make the aswer fall i the iterval (,).

4 Questio # 9 Aswer: E Because k ρk = φ there are a umber of ways to get the value of φ. / /3. φ =.5 =.46368; φ = (.) =.4646; φ = = Also, because the mea is zero, δ must be zero. The (usig the first choice for φ), y ˆ + =.46368(.43) + =.998. T Questio # Aswer: B The likelihood is: α α α α5 α5 α5 L = (5 + 5) (5 + 55) (5 + 95) 3 3α α 5 =. α + (375i675i) The loglikelihood is: α+ α+ α+ l = 3lα + 3α l5 ( α + ) l(375i675i) 3 3 l = + 3l5 l(375i675i ) = 4.48 α α ˆ α = 3/ 4.48 = Questio # Aswer: D For this problem, r = 4 ad = 7. The, 33.6 v ˆ = =.4 4(7 ) The, ad a ˆ = = k = = ; Z = = = (4 / 9) 77

5 Questio # Aswer: A r r r YX YX 3 X X 3 YX. X = = =.54 3 ryx r X X 3 r Questio # 3 Aswer: C For a mixture variable, raw momets are weighted averages of the idividual momets. Thus, EX ( ) =.5 m+.5m ad EX ( ) =.5( m) +.5( m). The square of the coefficiet of variatio is VarX ( ) EX ( ) EX ( ) m + m = =. EX ( ) EX ( ).5( m m) + Divide umerator ad deomiator by variatio becomes m ad let r m/ m =. The square of the coefficiet of r +. Settig the derivative equal to zero yields r =, however, this value miimizes.5( r + ) the fuctio (at a value of ). There are o other critical poits. Lookig at the edpoits (r = ad r = ifiity) the limitig value is 3, which is the maximum. Therefore, the least upper boud for the coefficiet of variatio is the square root of 3. Questio # 4 Aswer: B X is the radom sum Y + Y Y N. N has a egative biomial distributio with r = α = 5. ad β = θ =.. ( ) = rβ =.3 ( ) rβ( β) E N Var N bg = + =.36 EY = 5 Var bg Y = 5,,

6 ( ) =.3 5 = 5 E X Var( X ) =.3 5,, ,, = 6,5, Number of exposures (isureds) required for full credibility ( ) FULL = (.645/.5) 6,5, / 5 = Number of expected claims required for full credibility EN ( ) FULL = = 38. Questio # 5 Aswer: C The estimated relative risk is b ht ( Z= ) h () t e b = = e =.8 b=.6. ht ( Z= ) h() t For the sigle covariate case, the Wald test for testig H : β = reduces to: ( b ) I( b) = (.6) (3.968) =.43. Questio # 6 Aswer: D See pages 535-7, the bottom of page 567 ad the top of page 568. The oly correct statemet ad the correct aswer is (D). Questio # 7 Aswer: E X Fb xg d i ( ) F x b g b g F x F x F x F x F x d i where: ˆ θ = x = ad ( ) x / F x = e. The maximum value from the last two colums is.73. bg

7 Questio # 8 Aswer: E µ = E λ = v= E σ = a= Var λ = k = v/ a= 5; Z = = ( ) ; ( ).5; ( ) /. Thus, the estimate for Year is 5 () + () = Questio # 9 Aswer: B Time must be reversed, so let T be the time betwee accidet ad claim report ad let R = 3 T. The desired probability is Pr( T < T 3) = Pr(3 R< 3 R 3) = Pr( R> R ) = Pr( R> ). The product-limit calculatio is: R Y d The estimate of survivig past (reversed) time is (3/4)(3/55) =.459. Questio # Aswer: E The solutio depeds o idetifyig the parameter a with the overall average, which requires the covetio stated i Pidyck ad Rubifeld uder the table o page 4. The covetio also applies to subgroups. With this covetio, the solutio is the followig coditioal expectatio: (,,, ) E Y E = F = G = H = a= b b c c.

8 Questio # Aswer: E The posterior desity, give a observatio of 3 is: θ 3 f (3 θπθ ) ( ) (3 + θ) θ πθ ( 3) = = θ πθ θ θ θ 3 f (3 ) ( ) d (3 + ) d 3 (3 + θ ) = = + > (3 + θ ) 3 3(3 θ), θ. The, Pr( Θ> ) = 3(3 + θ) dθ = 6(3 + θ) = =.64. Questio # Aswer: B Because all previous values are, previous sigle ad double smoothed values are also zero. The, y =.6() +.4() =.6 y =.6(.) +.4(.6) =.96 y =.6(.3) +.4(.96) =.64 y =.6(.6) +.4() =.36 y =.6(.96) +.4(.36) =.7 y =.6(.64) +.4(.7) = Questio # 3 Aswer: B L= F() [ F() F()] [ F()] = ( e ) ( e e ) ( e ) / θ 7 / θ / θ 6 / θ 7 = ( p) ( p p ) ( p ) = p ( p) where p θ = = / θ = e. The maximum occurs at p = /33 ad so ˆ / l( / 33)

9 Questio # 4 Aswer: A EXθ ( ) = θ /. 6 3(6 ) EX ( 3 4,6) EX ( ) f( 4,6) d θ θ = θ θ θ = 3 dθ 6 = 6 4 θ 3 3(6 )(6 ) = = Questio # 5 Aswer: D The data may be orgaized as follows: t Y d St ˆ( ) (9/) = (7/9) = (6/7) = (4/5) = (3/4) = (/) =.8 Because the product-limit estimate is costat betwee observatios, the value of S ˆ(8) is foud from S ˆ(7) =.36. Questio # 6 Aswer: D As of time there were 7 observed paymets, so O = 7. For the Weibull distributio, the cumulative hazard fuctio is H( x) = l S( x) = x /5. The xi EZ ( ) = VarZ ( ) = = ( ) = i= 5 5 The chi-squared test statistic (with oe degree of freedom) is (7 5.48) /5.48 = From the tables, this leads to rejectio at the 5% level, but ot at the.5% level.

10 Questio # 7 Aswer: D ESSR = 5, 5,565 = 9, 435 ESSUR RUR =.38 = ESSUR =.6(5,) = 9,3 TSS (9, 435 9,3) / 3 F = = ,3 / 3,4 Questio # 8 Aswer: C The maximum likelihood estimate for the Poisso distributio is the sample mea: ˆ 5() + () + () + 9(3) λ = x = = The table for the chi-square test is: Number of days Probability Expected* Chi-square.6438 e = e = e = ** *365x(Probability) **obtaied by subtractig the other probabilities from The sum of the last colum is the test statistic of Usig degrees of freedom (4 rows less estimated parameter less ) the model is rejected at the.5% sigificace level but ot at the % sigificace level.

11 Questio # 9 Aswer: D.4() +.() +.().() +.() +.() µ () = =.5; µ () = =.6.4 µ =.6(.5) +.4() =.7 a =.6(.5 ) +.4( ).7 =.6.4() +.() +.(4) 7.() +.() +.(4) v.6.4 v =.6(7 /) +.4(.5) = / 6 k = v/ a= 55/6; Z = = + 55/ 6 5 () =.5 = ; v() = =.5 Bühlma credibility premium = (.7) = Questio # 3 Aswer: A All the statemets about R are true, but oly (A) is ot raised as a objectio to R. Questio # 3 Aswer: C µ =.5() +.3() +.() +.(3) =.8 σ =.5() +.3() +.(4) +.(9).64 =.96 3 ES ( ) = σ = (.96) =.7 4 bias =.7.96 =.4.

12 Questio # 3 Aswer: C The four classes have meas.,.,.5, ad.9 respectively ad variaces.9,.6,.5, ad.9 respectively. The, µ =.5( ) =.45 v =.5( ) =.475 a =.5( ).45 = k =.475 / = Z = = The estimate is [.743( / 4) +.757(.45)] 5 =.4. Questio # 33 Aswer: D The lower limit is determied as the smallest value such that St ˆ( ) VSt ˆ[ ˆ( )]. At t = 5 the two sides are.36 ad (.47) =.34 ad the iequality does ot hold. At t = 54 the two sides are.93 ad (.456) =.339 ad the iequality does hold. The lower limit is 54. Questio # 34 Aswer: A (( ) ) K ˆk k = Q= T r = = Questio # 35 Aswer: A The distributio used for simulatio is give by the observed values.

13 Questio # 36 Aswer: B First obtai the distributio of aggregate losses: Value Probability /5 5 (3/5)(/3) = /5 (/5)(/3)(/3) = 4/45 5 (3/5)(/3) = /5 5 (/5)()(/3)(/3) = 4/45 4 (/5)(/3)(/3) = /45 µ = (/ 5)() + (/ 5)(5) + (4 / 45)() + ( / 5)(5) + (4 / 45)(5) + (/ 45)(4) = 5 σ = (/ 5)( ) + (/ 5)(5 ) + (4 / 45)( ) + ( / 5)(5 ) + (4 / 45)(5 ) + (/ 45)(4 ) 5 = 8,. Questio # 37 Aswer: A Loss Rage Cum. Prob At 4, F(x) =.8 = e 4 θ ; solvig gives θ = Questio # 38 Aswer: D The sum of the squared values of the idepedet variable is 66 + () =,66. The value of s is 5348/8 = 97.. The, ΣX i, 66 sˆ ˆ α = s = 97. = 36. N x Σ i (66) The 97.5 th percetile of a t-distributio with 8 degrees of freedom is., so the lower limit of the symmetric 95% cofidece iterval for α is = 7.78.

14 Questio # 39 Aswer: B (/ )(/ 3) 3 Pr( class ) = = (/ )(/ 3) + (/ 3)(/ 6) + (/ 6)() 4 (/ 3)(/ 6) Pr( class ) = = (/ )(/ 3) + (/ 3)(/ 6) + (/ 6)() 4 (/ 6)() Pr( class3 ) = = (/ )(/ 3) + (/ 3)(/ 6) + (/ 6)() because the prior probabilities for the three classes are /, /3, ad /6 respectively. The class meas are µ () = (/3)() + (/3)() + (/3)() = µ () = (/ 6)() + (/3)() + (/ 6)(3) =. The expectatio is EX ( ) = (3/ 4)() + (/ 4)() =.5. Questio # 4 Aswer: E The first, secod, third, ad sixth paymets were observed at their actual value ad each cotributes f(x) to the likelihood fuctio. The fourth ad fifth paymets were paid at the policy limit ad each cotributes F(x) to the likelihood fuctio. This is aswer (E).