FALL 2006 EXAM C SOLUTIONS

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1 FALL 006 EXAM C SOLUTIONS Question # Key: E With n + = 6, we need the 0.3(6) = 4.8 and 0.65(6) = 0.4 smallest obsevations. They ae 0.(80) + 0.8(350) = 336 and 0.6(450) + 0.4(490) = 466. The equations to solve ae: γ γ θ θ 0.3 = and 0.65 γ γ = γ γ θ θ / γ / γ (0.7) = + (336 / θ ) and (0.35) = + (466 / θ ) -/ γ (0.7) (336 / θ ) = / γ (0.35) (466 / θ ) γ = (336 / 466) ln(0.884) = γ ln(336 / 466) γ = Question # Let E be the even of having claim in the fist fou yeas. In fou yeas, the total numbe of claims is Poisson(4λ). P( E TypeI) P( TypeI) e (0.05) P( Type I E) = = = = P( E) P( E) P( E) e ()(0.) P( Type II E) = = = P( E) P( E) 4 e (4)(0.75) P( Type III E) = = = P( E) P( E) Note : P( E) = = The Bayesian estimate of the numbe of claims in Yea 5 is: 0.447(0.5) (0.5) () =

2 Question #3 The sample mean is 0.(400) + 0.7(800) + 0.(600) = 800. The sample vaiance is 0.( ) + 0.7( ) + 0.( ) = 96, 000. The sample thid cental moment is ( ) + 0.7( ) + 0.( ) = 38, 400, 000. The skewness coefficient is.5 38,400,000 / 96,000 =.9. Question #4 Because < < 0.773, the simulated numbe of losses is 4. To simulate a loss by invesion, use F( x) = e u = e ln( u) = ( x / θ ) x = θ ( ln( u)) 3 4 τ ( x / θ ) τ ( x / θ ) u = 0.738, x u = 0.55, x u = , x u = 0.648, x 3 4 = u τ / τ = 00( ln( u)) = 3. = 70.8 = = / With a deductible of 50, the fist loss poduces no payments and 3. towad the 500 limit. The second loss poduces a payment of 0.8 and the insued is now out-of-pocket 63.. The thid loss poduces a payment of and the insued is out 43.. The deductible on the fouth loss is then fo a payment of = 7.5. The total paid by the insue is = 4.44.

3 Question #5 / x The density function is f ( x) = θ x e θ and the likelihood function is θ/86 θ /9 θ / 66 θ / 60 7 L( θ) = θ(86 ) e θ(9 ) e θ(66 ) e ( e ) θ θ e l( θ) = ln L( θ) = 3ln( θ) θ l ( θ) = 3θ = 0 θ = 3/ = 0.5. The mode is θ / = 0.5 / = 0.5. Question #6 We have μ( θ) = 4 θ and μ = 4E( θ) = 4(600) = 400. The aveage loss fo Yeas and is 650 and so 800 = Z(650) + ( Z)(400) which gives Z = 0.8. Because thee wee two yeas, Z = 0.8 = /( + k) which gives k = 0.5. Fo thee yeas, the evised value is Z = 3/(3+ 0.5) = 6/7 and the evised cedibility estimate (using the new sample mean of 0), (6 / 7)(0) + (/ 7)(400) = Question #7 The uncensoed obsevations ae 4 and 8 (values beyond ae not needed). The two values ae 0 and 5 and the two s values ae and. The Kaplan-Meie estimate is S ˆ() = (8/0)(4 / 5) = 0.64 and Geenwood s estimate is (0.64) + = (8) 5(4)

4 Question #8 Thee ae two ways to appoach this poblem. One is LaGange s fomula: (3 4)(3 5) (3 )(3 5) (3 )(3 4) f (3) = = ( 4)( 5) (4 )(4 5) (5 )(5 4) O, if the equation is f ( x) = a+ bx+ cx then thee equations must be satisfied: 5 = a+ b+ 4c 0 = a+ 4b+ 6c 30 = a+ 5b+ 5c The solutions is a = , b = -7.5, and c = The answe is (3) (9) = Question #9 Key: E S Q~ bin( m, Q ) and Q~ beta (,99). Then m E( Sm) = E[ E( Sm Q)] = E( mq) = m = 0.0m. Fo the mean to be at least 50, m must be at + 99 least 5,000. Question #0 The posteio distibution is 4 50λ λ 90 λ 7 λ 3 λ λ e 7 50λ πλ ( data) ( e ) ( λe ) ( λe ) ( λe ) = λ e which is a gamma distibution λ with paametes 8 and /50. Fo one isk, the estimated value is the mean, 8/50. Fo 00 isks it is 00(8)/50 =. Altenatively, The pio distibution is gamma with α = 4 and β = 50. The posteio will be continue to be gamma, with α / = α + no. of claims = = 8 and β / = β + no. of exposues = = 50. Mean of the posteio = α / β = 8/50 = 0.. Expected numbe of claims fo the potfolio = 0. (00) =.

5 Question # Key: E 0.95 = P(0.95μ < X <.05 μ) ~ ( μσ, / =.44 μ / ) X N n n 0.95μ μ.05μ μ 0.95 = P < Z <. μ/ n. μ/ n 0.95 = P( 0.05 n/. < Z < 0.05 n/.) 0.05 n /. =.96 n =.76. Question # Lq ( ) = ( q) q( q) = q ( q) 0 lq ( ) = 5000 ln() ln( q) ln( q) l q = q q = ( ) ( ) qˆ = 0.5 l(0.5) = 5000 ln() ln(0.5) ln(0.75) = Question #3 The estimate of the oveall mean, μ, is the sample mean, pe vehicle, which is 7/0 = 0.7. With the Poisson assumption, this is also the estimate of v = EPV. The means fo the two insueds ae /5 = 0.4 and 5/5 =.0. The estimate of a is the usual non-paametic estimate, 5( ) + 5(.0 0.7) ( )(0.7) VHM = aˆ = = (5 + 5) 0 (The above fomula: Loss Models page 596, Hezog page 6, Dean page 5) Then, k = 0.7/0.04 = 7.5 and so Z = 5/(5+7.5) = /9. The estimate fo insued A is (/9)(0.4) + (7/9)(0.7) =

6 Question #4 Item (i) indicates that X must one of the fou given values. Item (ii) indicates that X cannot be 00 Item (iii) indicates that X cannot be 400. Fist assume X = 00. Then the values of ae 5, 3,, and and the values of s ae,,, and. Then H ˆ (40) = =.3 and thus the answe is 00. As a check, if X = 300, the 5 3 values ae 5, 4, 3, and and the s values ae,,, and. Then, H ˆ (40) = = Question #5 The estimato of the Poisson paamete is the sample mean. Then, E( ˆ λ) = E( X) = λ Va( ˆ λ) = Va( X ) = λ/ n cv.. = λ / n/ λ = / nλ It is estimated by / nλ = / 39 = Question #6 Key: E P( X = 5 θ = 8) P( θ = 8) P ( θ = 8 X = 5) = P( X = 5 θ = 8) P( θ = 8) + P( X = 5 θ = ) P( θ = ) 5(0.5) 0.5 e (0.8) = = (0.5) 5(0.5) 0.5 e (0.8) e (0.) Then, EX ( X= 5) = E( θ X= 5) = (8) () = 7.0. Question #7 ( T) ( T) ( T) () q q q 0.05 = = = () ( T ) () () We have q = ( q )( q ) and so q q () () q = 0.05/ 0.95 = , q = 0.3 / 0.95 = 0.389, and Then, () 40 p 0 = (0.86) = Out of 000 at age 0, 86 ae expected to suvive to age 40.

7 Question #8 ω 4 p ωωω ω ( ω 4 p) L( ω) = = 5 5 ω 4 ( ω 4) ω l( ω) = ln( ω 4 p) 5ln( ω 4) 5 l ( ω) = = 0 ω 4 p ω = l (9) = 5 p 5 p = 5. The denominato in the likelihood function is S(4) to the powe of five to eflect the fact that it is known that each obsevation is geate than 4. Question #9 μ( λ) = v( λ) = λ μ = v= E( λ) = 0. Γ ( + / ) = VHM = a = Va( λ) = (0.) Γ ( + / ) = Z = = / The estimate fo one insued fo one month is 0.937(35/ 500) ( ) = Fo 300 insueds fo months it is (300)()(0.074) = 57.. Question #0 With no censoing the values ae, 9, 8, 7, 6, 4, and 3 and the s values ae 3,,,,,, (the two values at 7500 ae not needed). Then, 3 H ˆ (7000) = = With censoing, thee ae only five uncensoed values with values of 9, 8, 7, 4, and 3 and all five s values ae. Then, H ˆ (7000) = = The absolute diffeence is

8 Question # The simulated paid loss is exp[0.494 Φ ( u) ] whee Φ ( u) is the invese of the standad nomal distibution function. The fou simulated paid losses ae 450,6, 330,04, 939,798, and 688,45 fo an aveage of 60,3. The multiplie fo unpaid losses is ( ) 0.80( ) e = and the answe is (60,3) = 8,50 Question # The deduction to get the SBC is ( / )ln( n) = ( / )ln(60) =.78 whee is the numbe of paametes. The SBC values ae then , , , -40., and The lagest value is the fist one, so model I is to be selected. Question #3 Key: E P( X = λ = )P( λ = ) P ( λ = X = ) = P( X = λ = )P( λ = ) + P( X = λ = 3) P( λ = 3) e (0.75)! = =. 3 e e (3 ) (0.75) + (0.5)!! Then, (3 ).98 = () + (3) (3 ) (3 ) (3 ) = (3 ) Reaange to obtain (3 ) = (3 ) = (3 ) = 7. Because the isks ae Poisson, (μ = EPV, a = VHM): μ = v= E( λ) = 0.75() + 0.5(3) =.5 a= Va( λ) = 0.75() + 0.5(9).5 = 0.75 Z = = /3 +.5/ 0.75 and the estimate is (/3)(7) + (/3)(.5) = 3.33.

9 Question #4 Key: E The unifom kenel speads the pobability of 0. to 0 units each side of an obsevation. So the obsevation at 5 contibutes a density of fom 5 to 35, contibuting nothing to suvival past age 40. The same applies to the point at 30. Fo the next 7 points: 35 contibutes pobability fom 5 to 45 fo 5(0.005) = 0.05 above age contibutes pobability fom 5 to 45 fo 5(0.005) = 0.05 above age contibutes pobability fom 7 to 47 fo 7(0.005) = above age contibutes pobability fom 9 to 49 fo 9(0.005) = above age contibutes pobability fom 35 to 55 fo 5(0.005) = above age contibutes pobability fom 37 to 57 fo 7(0.005) = above age contibutes pobability fom 39 to 59 fo 9(0.005) = above age 40. The obsevation at 55 contibutes all 0. of pobability. The total is Question #5 f (3) = + (3/ )(4) (/ 4)(8) = 6 f (3) = (3/ )()() (/ 4)(3)(4) = 3 f(4) = f(3) + (4 3) f (3) = 6+ (3) = 9. Question #6 X ~ Exp( θ ) n i= X i ( ) ~ Γ( n, θ ) X ~ Γ( n, θ / n) ( θ / ) ( )( ) ( ) θ /. E X = n n n+ = n+ n The second line follows because an exponential distibution is a gamma distibution with α = and the sum of independent gamma andom vaiables is gamma with the α paametes added. The thid line follows because the gamma distibution is a scale distibution. Multiplying by /n etains the gamma distibution with the θ paamete multiplied by /n.

10 Question #7 The sample means ae 3, 5, and 4 and the oveall mean is 4. Then, vˆ = = 3(4 ) 9 (3 4) + (5 4) + (4 4) 8 / 9 7 aˆ = = = Question #8 The odeed values ae: t, 5t, 7h, 8t, 3h, 33t, 35h, 39t, 4h, and 45h whee t is a taditional ca and h is a hybid ca. The s values ae all because thee ae no duplicate values. The c values ae fo taditional cas and e fo hybid cas. Then Hˆ 0 (3) = = e 4+ 5e 3+ 5e 3+ 4e + 4e Question #9 3 3 π ( q ) = 6 q ( q) 6 q( q) q ( q) The mode can be detemined by setting the deivative equal to zeo. 3 3 π ( q ) 3 q ( q) 3 q ( q) = 0 ( q) q= 0 q = 0.5. Question #30 Fo the seveity distibution the mean is 5,000 and the vaiance is 0,000 /. Fo cedibility based on accuacy with egad to the numbe of claims, z 000 =, z = whee z is the appopiate value fom the standad nomal distibution. Fo cedibility based on accuacy with egad to the total cost of claims, the numbe of claims needed is z 0000 / =

11 Question #3 Hˆ ˆ(0) = e (0) = S Hˆ (0) = ln(0.575) = = k The solution is k = 36. Question #3 The annual dental chages ae simulated fom x /000 u = e x = 000ln( u). The fou simulated values ae , 55.73, 03.97, and The eimbusements ae (80% of 56.67), 000 (the maximum), (80% of 03.97), and 0. The total is and the aveage is 5.3. Question # θ θ θ θ L( θ ) = (0 θ) θ l( θ) = 9ln(0 θ) + ln( θ) 9 l ( θ ) = + = 0 0 θ θ (0 θ) = 9θ 0 = 0θ θ = 0 / 0 =

12 Question #34 The maximum likelihood estimate is ˆ θ = x = 000. The quantity to be estimated is S( θ ) = exp( 500 / θ ) and S ( θ ) = 500θ exp( 500 / θ). Fo the delta method, Va[ S( ˆ θ)] [ S ( ˆ θ)] Va( ˆ θ) = [500(000) exp( 500/000)] (000 / 6) = This is based on Va( ˆ θ) = Va( X ) = Va( X )/ n = θ / n. Question #35 Based on the infomation given x 0. = + n n 36 x 0.6y 0.5 = + + n n n n= 00 + x+ y. Then, 0.(00 + x+ y) = x 0.5(00 + x + y) = 36 + x+ 0.6y and these linea equations can be solved fo x = 9.37.