MAY 2007 SOA EXAM MLC SOLUTIONS

1 : œ : : p : œ Þ*& ( ( ( ( Þ*' (& ' B Þ( %:( œ / ( B œ / Þ*& Þ( & ( ( % ( MAY 2007 SOA EXAM MLC SOLUTIONS : œ : : œ Þ*' / œ Þ))* Answer: E 2 Z+<Ð+ XÐBÑlÑœ Ð EB EBÑ - - Since the force of mortality is constant at -, we have EB œ - and EB œ - - Therefore, from EB œ Þ%% œ Þ) and then E œ œ Þ(*, we get - œ Þ%, Þ% B ÐÞ)Ñ Þ% XÐBÑl Z +<Ð+ Ñ œ ÐÞ)Ñ ÒÞ(* ÐÞ%%Ñ Ó œ Þ*' Answer: B 3 Z œ E T + (since we are past the select period of 3 years, the insurance annuity & Ò'Ó '& Ò'Ó '& reverts to ultimate values) We can find E E Ò'Ó + E Ò'Ó ÐÞ&*Ñ Þ&* T Ò'Ó Þ' Þ' from TÒ'Ó œ œ œ œ Þ( Ò'Ó Ò'Ó From the Illustrative Table we have E œ Þ%*) and + œ *Þ)*'*, so that the reserve for face amount 1 is Z œ Þ%*) ÐÞ(ÑÐ*Þ)*'*Ñ œ Þ' & Ò'Ó '& Multiplying by 1000 gives the reserve for the face amount 1000 Answer: D '& 4 Since this is a fully discrete whole life insurance, for face amount 1, the variance of P is Ð E B Ñ Ð EB EBÑ œ Þ%)(, and the standard deviation is ÈÞ%)( œ Þ((( For face amount 150,000, the standard deviation is scaled up by a factor of &ß to &ß ÐÞ(((Ñ œ 'ß (( Answer: E 5 The exponential interarrival times with mean time between arrivals is equivalent to arrivals following a Poisson process with a mean of per unit time We are given that the average interarrival time is 1 month, so the average number of arrivals per month is 1 Because of the independence of arrivals in disjoint intervals of time, the fact that there have been no arrivals by the end of January has no effect on how many arrivals will occur in February and March The number of arrivals in Feb and Mar is Poisson with a mean of 2 The probability of at least 3 arrivals in Feb and Mar is the complement of the probability of at most 2 arrivals This is / / Ð/ Ñ œ Þ Answer: C x x S Broverman 2007 wwwsambrovermancom

6 The units donated and the units withdrawn are independent of one another The units donated follows a compound Poisson process and so do the units withdrawn The mean of a compound Poisson distribution is IÒRÓ IÒ\Ó and the variance is IÒRÓ IÒ\ Ó, where IÒRÓ is the Poisson mean, and \ is the amount of an individual deposit (or withdrawal for the withdrawal process) For the deposits in one week, R has a mean of (ÐÑÐÞ)Ñ œ &' (since 80% of food bank visitors make a deposit) and \ H H has mean 15 and variance 75 For the withdrawals in one week, R [ has a mean of (ÐÑÐÞÑ œ % (since 20% of food bank visitors make a withdrawal) and has mean 40 and variance 533 The expected amount deposited in one week is \ H IÒWHÓ œ IÒRHÓ IÒ\ H Ó œ Ð&'ÑÐ&Ñ œ )% and the variance of the amount deposited is Z +<ÒWHÓ œ IÒRHÓ IÒ\ H Ó œ Ð&'ÑÐ(& & Ñ œ 'ß ) (since IÒ\ ÓœZ+<:\ Ó IÒ\ Ó ) H H H Similarly, the expected amount withdrawn in one week is IÒW[ Ó œ IÒR[ Ó IÒ\ [ Ó œ Ð%ÑÐ%Ñ œ &' and the variance of the amount withdrawn is Z +<ÒW[ Ó œ IÒR[ Ó IÒ\ [ Ó œ Ð%ÑÐ& % Ñ œ *ß )' The net amount deposit in the week is WH W[, which has a mean of )% &' œ ) and a variance of 'ß ) *ß )' œ %'ß '' (because of independence of WH and W[ ) The probability that the amount of food units at the end of 7 days will be at least 600 more than at the beginning of the week is T ÐWH W [ 'Ñ Using the normal approximation, this WH W[ ) T Ð ' ) ' ) Ñ œ FÐ Ñ œ F ÐÞ%)Ñ œ Þ*' œ Þ'*% È%'ß'' È%'ß'' È%'ß'' Answer: A 7 The earlier premium is paid, the higher the reserve will be This can be seen retrospectively, since the accumulated cost of insurance is the same in all cases (level benefit of 1000), so the reserves differ because of different premium payment patterns Earlier premium payment results in greater accumulation to time 5 Pattern E has the most premium paid earliest E has the same total in the first 3 years as A and C and the same premium in years 4 and 5, so E's accumulated premium will be greater than that of A and C The difference between E and D is that E has premium of 1 more than D in the first year and 1 less than D in the 3rd year, but D has one more than E in the 4th year and 1 less than E in the fifth year Since E's excess differential with D occurs earlier (years 1 and 3, vs years 4 and 5), the accumulation of E's premium is greater than that of D From the diagram, it can be seen that D's accumulated premium is greater than that of B Answer: E wwwsambrovermancom S Broverman 2007

8 The expected number of points that Kira will score is Prob that Kira gets to play Expected number of points Kira scores given that she starts to play If Kira gets to play, the expected time until she will be called is / œ œ œ B O3<+ & O3<+ Þ' & hours The expected number of points she would score in that time is ß Ð Ñ œ ''ß ''( The probability that Kira will get to play is the probability that Kevin gets called first This is ;, where Bis Kevin and C is Kira This probability is BC ' Þ(> Þ'> BC > B B > C ' Þ( Þ ; œ : Ð>Ñ : > œ / ÐÞ(Ñ/ > œ œ Þ&)%' The expected number of points Kira will score before she leaves is ÐÞ&)%'ÑÐ''ß ''(Ñ œ )*ß (%% Answer: E ÐÑ 9 We first find ; &, the decrement probability for the continuous decrement ÐÑ ÐÑ wðñ wðñ ÐÑ wðñ wðñ ; œ ' : Ð>Ñ> œ ' : : Ð>Ñ> œ ; ' : > & > & & > & > & & & > & The last inequality follows from UDD in associated single tables for decrement 1 wðñ > & & & % & & % & : œ for Ÿ >, since decrement 2 does not occur until time wðñ > & Then : œ ÐÞÑÐ Ñ œ Þ* for Ÿ >, because of decrement 2 occurs at time and no more of decrement 2 occurs until time & Þ wðñ > & & ÐÑ wðñ wðñ & & > & & Then : œ ÐÞÑœÞ)) for Ÿ>Ÿ, because the rest of decrement 2 occurs at time Then ; œ ; ' : > œ ÐÞÑÒÐÑÐ Ñ ÐÞ*ÑÐ Ñ ÐÞ))ÑÐ ÑÓ œ Þ*' We know that for a 2-decrement model & & & wðñ wðñ wðñ wðñ & & & & & ÐÑ ÐÑ ÐÑ ;& œ ;& ;& Þ) œ Þ*' ;& ÐÑ ;& œ Þ'% ; œ ; ; ; ; œ Þ Þ ÐÞÑÐÞÑ œ Þ), and we also know that, so that, from which we get Answer: E ) 10 We are given that œ E '', where 08 refers to valuation at the start of 2008 ) We wish to find E '& Using the recursive insurance relationship, ) * * E'& œ @ Þ ;'& @ Þ :'& E'', so we need to find E'' * Again using the recursive relationship, we have E œ @ ; @ : E, '' Þ' '' Þ' '' '( E'( * E'' ) E'& ) E '', we get ) * * '' Þ '' Þ '' '(, so that Þ Þ '(, so if we can find then we can get, and then get Applying the recursive relationship to E œ @ ; @ ; E Þ œ ÐÞÑ ÐÞ*))ÑE S Broverman 2007 wwwsambrovermancom

10 continued and we get E'( * * '( '( '( 10 * '' Þ' '' Þ' '' '( ) * '& Þ '& Þ '& '' œ Þ)' Since the interest remains at 6% for 2009 and thereafter, it follows that E is the same as E, so E œ Þ)' Þ Þ' Þ Þ Þ*)) Þ' Þ** Þ Then E œ @ ; @ : E œ ÐÞ)'Ñ œ Þ Finally, E œ @ ; @ : E œ ÐÞÑ œ Þ)* Answer: C 11 K+ œ E Þ*K Þ*K+ &+ %Àl %À*l %À*l %Àl Using the relationship + % œ + %Àl I % +&, from the Illustrative Table, we get %Þ')'% œ + %Àl ÐÞ&''(ÑÐÞ'')Ñ, so that + %Àl œ (Þ&''& Using the relationship + % œ + %Àl I % +', from the Illustrative Table, we get %Þ')'% œ + ÐÞ(%%ÑÐÞ%&%Ñ, so that + œ Þ' We also use %Àl E œ E I E % % ' %Àl %Àl, so from the Illustrative Table we get Þ' œ E ÐÞ(%%ÑÐÞ'*Ñ, so that E œ Þ' %Àl %Àl Then, + %À*l œ + %Àl œ 'Þ&''& and + %À*l œ + %Àl œ Þ' Substituting these values into the original equation results in (Þ&''&K œ 'Þ Þ*K Þ*KÐ'Þ&''&Ñ &ÐÞ'Ñ K œ )Þ% Answer: A, and solving for K results in 12 Given that OÐ&&Ñ (means that (55) is still alive at age 56) P is a 5-point random variable as of age 56 The 5 possible values for P are ÐÑ @ & œ )' if death is acc at age 56, prob ;&' œ Þ& ÐÑ @ & œ )* if death is not acc at age 56, prob ;&' œ Þ% Pœ @ ÐÑ &Ð @Ñ œ ') if death is acc at age 57, prob l;&' œ ÐÞ*&&ÑÐÞ)Ñ ÐÑ @ &Ð @Ñ œ (* if death is not acc at age 57, prob l; 2 &' œ ÐÞ*&&ÑÐÞ'Ñ &Ð @Ñ œ *( if (56) survives to age 58, prob : œ ÐÞ*&&ÑÐÞ*Ñ From this table, we see that T Ò P Ÿ loð&&ñ &' Ó œ ÐÞ*&&ÑÐÞ*Ñ œ Þ)* (only on survival to age 58), and T Ò P Ÿ (*loð&&ñ Ó œ Þ)* ÐÞ*&&ÑÐÞ'Ñ œ Þ*%(' (still not Þ*& ), and T Ò P Ÿ )*loð&&ñ Ó œ Þ*%(' Þ% œ Þ*)(' Þ*& Answer: D wwwsambrovermancom S Broverman 2007

13 We can use the recursive relationship Z+<Ò2PlOÐBÑ 2ÓœÒ@Ð, 2 2 ZÑÓ : B 2 ; B 2 @: B 2Z+<Ò2 PlOÐBÑ 2 Ó to find Z+<Ò PlOÐBÑ Ó Since the policy terminates at time 3, Z+<ÒPlOÐBÑ Óœ Then B B B Z+<ÒPlOÐBÑ ÓœÒ@Ð, Z ÑÓ : ; @ : Z +<Ò PlOÐBÑ Ó œ Ò@Ó ÐÞ&ÑÐÞ&Ñ œ 'ß ' (since Z œ and, œ ) Then, Z+<ÒPlOÐBÑ ÓœÒ@Ð, Z ÑÓ : ; @ : Z +<Ò PlOÐBÑ Ó œ Ò@Ð Þ)ÑÓ ÐÞ'ÑÐÞ%Ñ @ ÐÞ'ÑÐ'ß 'Ñ œ &&ß (' Answer: C 14 In order for (30) to die second and within 5 years the death of (35) it must be true that (35) dies first and (30) dies within 5 years after that This probability is ' > : & & Ð>Ñ > : & ; > > The integral is set up based on the density of (35)'s death at time >, and (30) being alive at the time but dying in the next 5 years We can write > : &; > in the form > : &; > œ > : Ð &: > Ñ œ > : > & : œ > : &: > : & Þ The integral becomes ' > : & & Ð>ÑÐ > : & : > : & Ñ> ' ' > & & > & > & & > & œ : Ð>Ñ : > : : Ð>Ñ : > The first integral is the probability that 35 will die before 30, which is, The second integral is, because it is the probability that one of two people of equal age 35 will be the first to die We are also given &: œ + The overall probability is, + Answer: E ÐßÑ 15 The probability of transform from state 1 to state 1 in the first year is U œ 3 The probability of transferring from state 1 to state 2 in the second year is ÐßÑ ÐßÑ U U œ ÐÞ'ÑÐÞ%Ñ œ Þ% The probability of transferring from state 1 to state 2 in the third year is ÐßÑ ÐßÑ ÐßÑ U U2 œ [ ÐÞ'ÑÐÞ%ÑÓÐÞÑ œ Þ% (we can see from U that since U œ, the only way to still be in state 1 at the start of the third year is to stay in state 1 from year 1 to year 2 and the stay from year 2 to year 3) The actuarial present value of the payments made because of transfer from state 1 to state 2 is ÒÞ@ Þ%@ Þ%@ Ó œ %&Þ)* The fee T is paid if in state 1 The fee will be paid at the star of the first year S Broverman 2007 wwwsambrovermancom

15 continued The fee will be paid with probability U The fee will be paid with probability ÐßÑ ÐßÑ U œ Þ' œ Þ% The APV of fees is T Ò Þ'@ Þ%@ Ó œ Þ''T at the start of the second year at the start of the third year According to the equivalence principle, we have Þ''T œ %&Þ)*, so T œ %)Þ%' Answer: D 16 In order for Tom to find at least 3 coins in the next two blocks, Tom must find either 1,2 or 2,1 or 2,2 in the next two blocks The probabilities of these combinations are ÐßÑ ÐßÑ ÐßÑ ÐßÑ U U œðþ'ñðþñœþ), U U œðþñðþ&ñœþ&, and ÐßÑ ÐßÑ U U œ ÐÞÑÐÞ%Ñ œ Þ The total probability is Þ) Þ& Þ œ Þ%& Answer: B 17 The probability of getting 100 at the end of the first year is 8 The probability of getting 100 at the end of the second year is ÐÞ)ÑÐÞ)Ñ ÐÞÑÐÞ(Ñ œ Þ() (these are the 2 combinations of RR and ]R, where R and ] denote the events of no accident and accident) The probability of getting 100 at the end of the third year is ÐÞ)ÑÐÞ)ÑÐÞ)Ñ ÐÞ)ÑÐÞÑÐÞ(Ñ ÐÞÑÐÞ(ÑÐÞ)Ñ ÐÞÑÐÞÑÐÞ(Ñ œ Þ(() (these are the probabilities of the combinations of RRRß R]Rß ]RRß ]]R) The actuarial present value of the payments is ÒÞ)@ Þ()@ Þ(()@ Ó œ )Þ The probability of getting V at the end of 3 years is ÐÞ)Ñ œ Þ& The actuarial present value of that payment is Þ&V@ œ Þ%&&''V The two choices are actuarially equivalent if they have the same actuarial present value Solving for V from Þ%&&''V œ )Þ results in V œ %(* Answer: D wwwsambrovermancom S Broverman 2007

(( ') ) ') ) '' Ò'' Ó ) Ò'(Ó 18 From the ultimate column, we have : œ, so that ; œ Then from (ii), we get %; œ&;, so that %Ð Ñœ&;, from which we get ; œ Þ, and then : œ Þ*( Ò'(Ó j') jò'(ó Ò'(Ó Then from : œ œ œ Þ*(, we get j œ (*)Þ% Ò'(Ó (( j Ò'(Ó Ò'(Ó We continue in a similar way From (ii) again we get %; œ &; '& Ò''Ó ) ) ) But from the ultimate table, we have ; œ : œ œ '( ) Ò''Ó Ò''Ó Then, %Ð Ñ œ &; so that ; œ Þ*& Then from (i), we get ; œ %;, so that ; œ Þ%'%, '( Ò''Ó Ò'' Ó Ò'(Ó jò'(ó (*)Þ% Ò'(Ó jò'(ó jò'(ó and : œ Þ*)&'' Since : œ œ œ Þ*)&'', Ò'(Ó we get j œ )&' Answer: C Ò'(Ó 19 For fully discrete whole life reserves, we have Z œ + B > + + & + & Z% œ + Z% œ % + % > B and Since Z% œ Z %, it follows that + & œ + & œ Þ Using the relationship + œ @: @ : @ : + & & & & & B, we get œ @: @ : @ :, where : is the common value of :, : and : & & & By trial and error we try each of the possible values of : The value : œ Þ*&% satisfies the equation Answer: D 20 In order for Derek and A-Rod to survive two years, they must both survive the first year in which the are subject to a single common shock, and they must both survive the second year as independent lives The probability is T U V, where Tis the probability that they both survive the first year, Uis the probability that Derek survives the 2nd year, and V is the probability that A-Rod survives the 2nd year Uand V are both / Þ (they are each subject to Þ H E the total force of mortality of 001) T is : : / (this is the probability that Derek does not die to causes other than the common shock, and A-Rod also doesn't die to causes other than common shock, and the common shock doesn't occur in the first year) Derek' force of mortality due to causes other than common shock is Þ Þ œ Þ), and same for A-Rod Þ) Þ) Þ Þ) Therefore, T œ / / / œ / The total probability we are looking for is Þ) Þ Þ / / / œ Þ**' Answer: C S Broverman 2007 wwwsambrovermancom

21 The original model is a DeMoivre model Survival under the new model is based on a Generalized DeMoivre model The new model has a new α, but the same = = % = α = ' α * ( ( (ii) tells us that œ It follows that α œ (iii) tells us that œ, so that = ' œ Ð Ñ œ, and = œ * Under the original DeMoivre model, we have / œ œ Þ Answer: B ÞX 22 ^ œ / if death occurs with X Ÿ years, and ^ œ &/ if X 68ÐÞ(Ñ Þ 68ÐÞ)Ñ Þ Þ> / ( if Þ> 68ÐÞ(Ñ, or equivalently, > œ Þ&( Þ> &/ ( if Þ> 68ÐÞ)Ñ, or equivalently, > œ Þ(, but in this case, we must also have > (for the benefit to be 2500) The total probability is the combination of T ÐX Þ&(Ñ and T Ð X Þ(Ñ This is ; ; œ œ Þ& Answer: C Þ&( % lþ( % Þ&( % Þ( % ÞX ÐÑ ÐÑ 23 l B B ; œ : ; wðñ wðñ For a 2 decrement table, : œ : : B B B wðñ ; B œ Þ wðñ : B œ Þ* ÐÑ B wðñ B Þ B Þ wðñ wðñ, and œ Þ& Þ)(( œ Þ%(( ÐÑ B Ð Ñ ; œ 7 ; B B Þ%(( ÐÑ Þ)(( ; œ Þ%(( ÐÞ&%&Ñ œ Þ ÐÑ l; B œ : B ; œ ÐÞ('*ÑÐÞÑ œ Þ( We are given, so that From constant force œ Þ, we get : œ /, so that : œ Þ*/ œ Þ('* From ; œ Þ& we get : œ Þ(&, and then from constant force of decrement we get ÐÑ œ 68ÐÞ(&Ñ œ Þ)(( Then, also from constant force of decrement, From constant force,we have : œ / œ / œ Þ'%&& so that ; œ Þ&%&, and then Finally, Answer: B 24 + œ @: @ : (&Àl (& (& ' (& (& 8 Þ Þ From (i) we get : œ / œ / Therefore, (& Ð > Ñ > ÞÒÐ(& 8Ñ (& Ó 8 (& Þ Þ ÞÒ(' (& Ó :(& œ / œ Þ*(* and Þ Þ ÞÒ(( (& Ó : œ / œ Þ*%%& Þ*(* Þ*%%& Þ ÐÞÑ The APV of the annuity is œ Þ'% Answer: A wwwsambrovermancom S Broverman 2007

25 The expected number of trains that will arrive between 7:00 AM and 7:25AM is & ' ' ' > & ' -Ð>Ñ > œ ÐÞ&Ñ > Ð Ñ> ÐÞÑ > œ Þ& Þ(& Þ& œ Þ(& The probability of exactly four trains in the time interval is the Poisson probability Þ(& % œ Þ( Answer: B / ÐÞ(&Ñ %x 26 For a Poisson process with rate - per unit time, the time of the 8-th event, W 8, has a gamma 8 8 distribution with mean - and variance - In this problem, - œ% and X œw)*, so Xhas mean )* and variance )* % œ (Þ& ' œ )Þ'& Applying the normal approximation to X, X (Þ& ') (Þ& we get T ÐX ')Ñ œ T Ð Ñ œ Ð Ñ œ Þ)% Answer: A È)Þ'& È F )Þ'& ' > > > 27 IÒ^Ó œ / / / > œ œ Þ'' Since œþ', it follows that œþ% Then, IÒ^ Ó œ ' > > > / / / > œ œ Þ)& Z +<Ò^Ó œ Þ)& ÐÞ''Ñ œ Þ(, Answer: A 28 The recursive relationship for assets shares is Ò ÐÑ ÐÑ EW KÐ - Ñ / ÓÐ 3Ñ,; GZ; œ: EW 2 2 2 B 2 2 B 2 B 2 2 Using this, we have ÐÑ ÐÑ Ò& EW KÐ -& Ñ / & ÓÐ 3Ñ,; % & ' GZ; % & œ: % & ' EW, which becomes Ò& *ÐÞ*&ÑÓÐÞ)Ñ ß ÐÞ%Ñ ' GZ ÐÞ&Ñ œ ÐÞ*%'ÑÐÑ Solving for GZ results in GZ œ *Þ% Answer: C ' ' 29 The equivalence principle equation is U œ + UE l & ' l Àl + œ I I + œ ÐÞ*(%ÑÐÞ&)ÑÐÞ%&%Ñ œ Þ'( E œ E I I E, so that & ' Àl E œ Þ) ÐÞ*(%ÑÐÞ&)ÑÐÞ'*Ñ œ Þ%'** Àl ÐÞ'(Ñ Þ%'** Solving for U results in U œ œ & Answer: A S Broverman 2007 wwwsambrovermancom

30 The benefit premium is U The equivalence principle equation is U ' > / : > œ ' > ÐÑ / : Ð>Ñ> ' > ÐÑ / : Ð>Ñ> > B > B B > B B ÐÑ ÐÑ B Ð>Ñ œ Þ B Ð>Ñ œ Þ% B Ð>Ñ œ Þ' > B Ð 7 Ñ Þ'> Þ Þ% : œ / The equation becomes U ' œ Þ' Þ' Since and, it follows that, and which becomes U œ ÐÞÑ Þ% œ Þ Answer: D wwwsambrovermancom S Broverman 2007