北美精算师(soa)考试p 2001年真题和注解.pdf

Course 1 Solutions1May 2001Course 1 May 2001 Answer Key1E21E 2C22C 3C23A 4E24E 5C25B6D26B 7B27A 8A28C 9C29D 10E30B11D31C 12D32E 13E33D 14A34A 15D35C16A36B 17B37E 18D38D 19B39A 20D40BA7 B8 C8 D9 E8Course 1 Solutions2May 20011.E We are given that56 5656bbeppe−−=== It follows that6 6565 55 6 5ln656lnln65b bbbbb beeeeeebb− −−+− −===== − = −=2.C First, solve for m such that()()()()() ()() () ()111.071.07150088 1.07...8 1.07881 1.070.075.3751.07ln 5.375 ln 1.07ln 5.37524.86ln 1.07mm mmmm−−−=+++==−====We conclude that 25m = .3.C Observe that()2 cos /2 and 2 cost 2 sin dxdytttdtdt==−Therefore,()()()/2/222 cos /4222 cos /2 sin /2ttdx dtdy dtπππππππ======−= −It follows that the length of the velocity vector at time 2tπ= is given by()()22222ππ+ −=+ .Course 1 Solutions3May 20014.E Let X1, X2, X3, and X4 denote the four independent bids with common distribution function F. Then if we define Y = max (X1, X2, X3, X4), the distribution function G of Y is given by ( )[] ()()()() [][][][]( )()1234123444PrPrPrPrPrPr1351 sin , 1622G yYyXyXyXyXyXyXyXyXyF yyyπ=≤=≤∩≤∩≤∩≤ =≤≤≤≤= =+≤≤It then follows that the density function g of Y is given by ( )( )() ()()33'11 sincos4 35cos1 sin , 422g yGyyyyyyππππππ==+=+≤≤Finally, [ ]( )()5/23/25/233/2cos1 sin4E Yyg y dyyyydyπππ==+∫∫Course 1 Solutions4May 20015.C The domain of X and Y is pictured below. The shaded region is the portion of the domain over which X0 for 010 1( )0.02 for 10205 30.04 for 205xT xxxxx ≤≤ =−≤  −Therefore, ( )220 for 010 1' for 10205 3for 205xTxxxxx  =   and ( )330 for 010 2“ for 10205 6for 205xTxxxxx  = −  − We can infer the following about T(x): ( ) ( ) ( ) ( ) ( )i) 0 for 010ii) is strictly increasing for 1020 and 20 since '0 on both of these intervals.iii) is concave down for 1020 and 20 since “0 on both of these T xxT xxxTxT xxxTx=≤>>This implies that n should satisfy the following inequality: 4032n n−− ≥To find such an n, let’s solve the corresponding equation for n:()()403224033240031040416n nnnnnnnn n−− =−=−−−=+−== =20.D The density function of T is( )/31, 03tf tet−=∞and ( )/31 , 0yG yey−= −> Therefore, ()()[][] ( ) ( ) ()()1/22/32/31/27/6Pr32Pr3 Pr232111XYXYFGeeeee−−−−−≤∩≤=≤≤ ==−−= −−+23.A Let C = Event that shipment came from Company X I1 = Event that one of the vaccine vials tested is ineffectiveThen by Bayes’ Formula, [][] [ ] [] [ ]1 1 11||||ccP IC P CP C IP IC P CP ICP C=+ Now[ ][ ][]( )()()( )()()2930 112930 111 5 141155|0.100.900.141|0.020.980.334ccP CP CP CP ICP IC= = −= −=== == Therefore,[]()() ()() ()()10.141 1/5|0.0960.141 1/50.3344/5P C I==+Course 1 Solutions16May 200124.E The domain of s and t is pictured below.Note that the shaded region is the portion of the domain of s and t over which the device fails sometime during the first half hour. Therefore,()()1/2111/201/20011Pr,,22STf s t dsdtf s t dsdt≤∪≤=+∫ ∫∫ ∫(where the first integral covers A and the second integral covers B).25.B Note that V, S and r are all functions of time t. Therefore,24dVdrrdtdtπ=and8dSdrrdtdtπ=We are given that 660 when 32dVrdt===.It follows that( )( )260435 3 583403dr dt dr dt dS dtππππ====Course 1 Solutions17May 200126.B Let u be annual claims, v be annual premiums, g(u, v) be the joint density function of U and V, f(x) be the density function of X, and F(x) be the distribution function of X. Then since U and V are independent,()()/2/211, , 0 , 022uvuvg u veee euv−−−−==∞∞Next, let X1, X2, and X3 denote the three claims made that have this distribution. Then if Y denotes the largest of these three claims, it follows that the distribution function of Y is given by ( )[][][]12333PrPrPr11 , 1G yXyXyXyyy=≤≤≤=−>while the density function of Y is given by( )( )2234431391'3 11 , 1g yGyyyyyy ==−=−  >Therefore,[ ]233336113692581 191921119189918925812192.025 (in thousands)258E Ydydyyyyyydyyyyyyy∞∞∞ ∞=−=−+=−+= −+−=−+=∫∫∫Course 1 Solutions19May 200128.C Since ( )( )0 and '0 for 0f tftt ≥>Applying these inequalities, we see that( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )()( )( )( )( )( )2022021220120101201200012011911192012212011010011111kkkkkkkkfff t dtff t dtf t dtff t dtf t dtf t dtf t dtf t dtf t dtff t dtff t dtff kff kf kf t dtf t+===+=++++=++==++=+++=+==∫∫∫∫∫∫∫∫∫∑∫∫∑∑ ∑∫>>>>>( )1201201kdtf t dt=∑ ∫∫>We conclude that ( )( )2011ff t dt+∫ produces the smallest number that exceeds( )201kNf k==∑.Course 1 Solutions20May 200128.Note a more heuristic approach to the result that (E) > (B) > (A) > (C) > ( )201kf k=∑> (D)can be obtained from diagrams of the following sort:gives ( )( )( )( )( )( )( )202021010Efff t dtff t dtB=+++=∫∫>andgives ( )( )( )2020111kff t dtf t=+∑∫>29.D Let X = number of low-risk drivers insured Y = number of moderate-risk。