course4solutions_1103 - Society of Actuaries Course 4 Exam...

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Society of Actuaries Course 4 Exam Solutions Fall 2003 Question #1 Key: A The Yule-Walker equations are: 12 .5 .5 .1 .5 φ =+ The solution is 2 .2 =− . Question #2 Key: E The 40 th percentile is the .4(12) = 4.8 th smallest observation. By interpolation it is .2(86) +.8(90) = 89.2. The 80 th percentile is the .8(12) = 9.6 th smallest observation. By interpolation it is .4(200) +.6(210) = 206. The equations to solve are (89.2/ ) (206/ ) .4 and .8 1 (89.2/ ) 1 (206/ ) γγ γ θθ θ == ++ . Solving each for the parenthetical expression gives 2 (89.2/ ) and 4 3 . Taking the ratio of the second equation to the first gives 6 (206/89.2) = which leads to ln(6) / ln(206/89.2) 2.1407 . Then 1/ 2.1407 4 206/ = for 107.8 = . Question #3 Key: E The standard for full credibility is 2 2 1.645 ( ) 1 .02 ( ) Var X EX ⎛⎞ + ⎜⎟ ⎝⎠ where X is the claim size variable. For the Pareto variable, ( ) .5/5 .1 and 2 2 2(.5) ( ) (.1) .015 5(4) Var X = . Then the standard is 2 2 1.645 .015 1 16,913 .02 .1 += claims.
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Question #4 Key: B The kernel is a triangle with a base of 4 and a height at the middle of 0.5 (so the area is 1). The length of the base is twice the bandwidth. Any observation within 2 of 2.5 will contribute to the estimate. For the observation at 2, when the triangle is centered at 2, the height of the triangle at 2.5 is .375 (it is one-quarter the way from 2 to the end of the triangle at 4 and so the height is one-quarter the way from 0.5 to 0). Similarly the points at 3 are also 0.5 away and so the height of the associated triangle is also .375. Each triangle height is weighted by the empirical probability at the associated point. So the estimate at 2.5 is (1/5)(3/8) + (3/5)(3/8) + (1/5)(0) = 12/40. Question #5 Key: D The standard error is 92/8 . With 4 of the X s equal to 1 and 6 equal to 0 the average is 0.4. Then 2 2 ˆ 222 92/8 4.7917 ( ) 4(1 .4) 6(0 .4) i s s XX β == = −− + and ˆ ˆ 4 1.83 4.7917 t s = . Question #6 Key: A The distribution function is 1 1 1 () 1 x x Fx t d t t x α αα = . The likelihood function is 2 11 1 2 3 (3) (6) (14)[1 (25)] 361 4 ( 2 5 ) [3(6)(14)(625)] . Lf f f F ααα =− = Taking logs, differentiating, setting equal to zero, and solving: 1 ln 3ln ln157,500 plus a constant (ln ) 3 ln157,500 0 ˆ 3/ ln157,500 .2507. L L =
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Question #7 Key: C 35 2 1 52 0 ( | 1 , 1 ) ( 1 | ) ( 1 | ) ( ) 2 ( 1) 2 ( 4 ( (1 ) 1/168, ( |1,1) 168 ) . q p q p q q qq q d q q ππ π ∝= −= = The expected number of claims in a year is ( | ) 2 EX q q = and so the Bayesian estimate is 1 0 (2 |1,1) 2 (168) ) 4/3. Eq q q qd q =− = The answer can be obtained without integrals by recognizing that the posterior distribution of q is beta with a = 6 and b = 3. The posterior mean is ( |1,1) /( ) 6/9 2/3. a a b = += = The posterior mean of 2 q is then 4/3. Question #8 Key: D For the method of moments estimate, 22 .5 2 2 2 386 , 457,480.2 5.9558 .5 , 13.0335 2 2 ˆˆ 5.3949, 1.1218. ee µσ µ σµ σ ++ == =+ = + Then 5.3949 .5(1.1218) ln500 5.3949 1.1218 ln500 5.3949 ( 500) 500 1 1.1218 1.1218 386 ( .2853) 500[1 (.7739)] 386(.3877) 500(.2195) 259. EX e + −− ⎛⎞ ∧= Φ + Φ ⎜⎟ ⎝⎠ + Φ =+= Note-these calculations use exact normal probabilities. Rounding and using the normal table that accompanies the exam will produce a different numerical answer but the same letter answer.
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course4solutions_1103 - Society of Actuaries Course 4 Exam...

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