NOVEMBER 2003 SOA COURSE 3 EXAM SOLUTIONS

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1 NOVEMER 2003 SOA COURSE 3 EXAM SOLUTIONS Prepared by Sam roverman 2brove@rogerscom sam@utstattorontoedu 1 l; $!À$ œ $ ; $!À$ ; $!À$ œ $ ; $! $ ; $ ; $! ; $ (the second equality is true as a result of the independence of X Ð$!Ñ and X Ð$Ñ $ ; $! œ $ : $! œ : $! : $ : $ œ ÐÞ*ÑÐÞ)ÑÐÞ(Ñ œ Þ*' Similarly, $ ; $ œ ÐÞ&ÑÐÞÑÐÞ$Ñ œ Þ* ß ; $! œ ÐÞ*ÑÐÞ)Ñ œ Þ) ß and ; $ œ ÐÞ&ÑÐÞÑ œ Þ) Then l; $!À$ œ ÐÞ*'ÑÐÞ*Ñ ÐÞ)ÑÐÞ)Ñ œ Þ Answer: E ' > $! > 2 The single benefit premium is!!! / : Ð>Ñ > This can be formulated as!!!ò'! $ / >! $ : Ð>Ñ > / : ' $ > / : Ð! >Ñ >Ó! >!! >! In this expression, $ œþ! is the force of interest during the first 10 years and $ œþ!& is the force of interest after 10 years For the constant force of mortality of 06 during the first 10 years, we have : œ / Þ!'>, and for the constant force of mortality of 07 after 10 years, we have > Þ!(> > :! œ / The single benefit premium becomes!!!ò'! Þ!> Þ!'>!ÐÞ!Ñ!ÐÞ!'Ñ / / ÐÞ!'Ñ > / / ' Þ!&> Þ!(> / / ÐÞ!(Ñ >Ó!! / Þ Þ œ!!!òðþ!'ñð Ñ / ÐÞ!(ÑÐ ÑÓ œ &* Answer: E -Ð!! \Ñ \ Ÿ!! 3 Suppose that \ denotes total claims Then the bonus is F œ œ!!! >œ\ This can be written in the form F œ -Ò!! Ð\!!ÑÓ, and the expected bonus is IÒFÓ œ!! œ -Ò!! IÐ\!!ÑÓ Þ For the Pareto distribution with α œ and ) œ$!! (from the Tables for the Course 3 Exam) $!! $!!!! $!! we have IÐ\!!Ñ œ Ò Ð Ñ Ó œ (Þ$ Then!! œ -Ò!! (Þ$Ó p - œ Þ$(& Answer: A 2brove@rogerscom sam@utstattorontoedu

2 4 The annual maintenance cost for one computer, say [, has a compound Poisson distribution with a Poisson frequency distribution R with a mean of 3 and severity distribution \ with mean 80 and variance 40,000 Then IÒ[ÓœIÒRÓ IÒ\ÓœÐ$ÑÐ)!Ñœ! and Z +<Ò[ Ó œ IÒRÓ Z +<Ò\Ó Z +<ÒRÓ ÐIÒRÓÑ œ Ð$ÑÐ!ß!!!Ñ Ð$ÑÐ)!Ñ œ $*ß!! With 8computers, the annual maintenance cost will be X8 œ[ â [ 8, with mean IÒX8Ó œ!8 and variance Z +<ÒX8Ó œ $*ß!!8 The probability that annual maintenance costs exceed 120 of expected costs is T ÒX8 ÞIÒX8Ó Ó œ T ÒX8 ))8Ó Applying the normal approximation, this probability is X8!8 ))8!8 T Ò È Ó œ ÐÞ)( 8Ñ Þ $*ß!!8 È F È $*ß!!8 A maintenance contract will be avoided if this probability is less than 10 The probability is less than 10 if FÐÞ)( È8Ñ Þ* From the normal table, the 90th percentile of the standard normal distribution is 1282, so that the maintenance contract will be avoided if Þ)( È8 Þ), or equivalently, if 8 **Þ$ Answer: C 5 Since small random numbers correspond to small values of R, we use the distribution function of R to determine the simulated value R has a discrete integer distribution with probability function and distribution function: R $ Þ :Ð8Ñ Þ* Þ!* Þ!!* Þ Þ Þ J Ð8Ñ Þ* Þ** Þ*** Þ Þ Þ The random number?œþ!& is less than 9, so according to the inverse transformation method, the simulated value if R is 1 There is 1 accident Since small random numbers correspond to small values of \, according to the inverse transformation method, we simulate \ by solving the œ JÐ ÑÞ Þ! \ has an exponential distribution with mean 100, so that JÐÑ œ / Þ! Then Þ$ œ /, from which we get œ $&Þ( Þ This is the simulated total amount of claims during the year (one claim of amount 357) Answer: 2brove@rogerscom sam@utstattorontoedu

3 6 The loss-at-issue is the present value at time 0 of the benefit paid minus premium received We must simulate the time of death and cause of death In this example, the combined force of decrement is all causes of mortality, so what is referred to as > ; in Chapter 3 of Actuarial Mathematics, is > ; Ð7Ñ in this example The distribution function for XÐÑis J Ð>Ñ œ ; Since the total force of mortality decrement is constant at XÐÑ > 02, the time of death random variable XÐÑhas distribution function Ð7Ñ Þ!> JXÐÑ Ð>Ñœ > ; œ > : œ / We are told that for the simulation of the time of death, low random numbers correspond to long times of death This means that the simulated value of >, time until death, is found by solving the equation <œ JXÐÑÐ>Ñ, where is the random number (note that the standard form of the inverse transformation method has small random numbers corresponding to small values of the random variable being simulated, so we would solve for > from the equation < œ J Ð>Ñ in that case) We solve Þ$& œ Ð / Þ!> Þ!> Ñ œ /, which results in > œ &Þ& XÐÑ In general, for a multiple decrement model, if decrement occurs at time >, the conditional probability that the cause of decrement is cause 4 is T ÒN œ 4lX ÐÑ œ >Ó œ Þ The total force of morality is 002, and the force of mortality due to accidental death is 005 If death occurs at time >, then the probability that death is due to accidental causes is Þ! œ Þ&Þ To say that high random numbers correspond to death by accidental means indicates that for Þ(& Ÿ < Ÿ, the simulated cause of death is accident (we must choose a region for the uniform random number that has the same probability as the probability of the event being simulated, 25 for accidental death in this case) The given random number of 775 results in accidental means being the simulated cause of death Therefore, the double indemnity provision results in a total death benefit of 2 being paid at time 525 The simulated value of the loss-at-issue is then PV of benefit paid PV of premium received œ / Þ!&Ð&Þ&Ñ ÐÞ!&Ñ + / Þ!& Þ!&Ð&Þ&Ñ Þ!&Ð&Þ&Ñ œ / ÐÞ!&Ñ œ Þ$* Answer: D &Þ&l Ð4Ñ Ð7Ñ Ð>Ñ Ð>Ñ Þ!!& 2brove@rogerscom sam@utstattorontoedu

4 Ð 7 7 Ñ ÐÑ ÐÑ Ð$Ñ Ð>Ñ œ Ð>Ñ Ð>Ñ Ð>Ñ œ Þ!!!!!!& Þ!!!!! Þ!!! œ Þ!!!!& Þ 7 Since the force of decrement for all causes is constant, it follows that : œ / The single benefit premium is ß!!!ß!!!' / Þ!'>! / Þ!!!!&> ÐÞ!!!!!!&Ñ > &!!ß!!!' / Þ!'>! / Þ!!!!&> ÐÞ!!!!!Ñ > '! Þ!'> Þ!!!!&> &!ß!!! / / ÐÞ!!!Ñ > Þ!!!!!!& Þ!!!!! Þ!!! œ ß!!!ß!!!Ð Þ!'!!& Ñ &!!ß!!!Ð Þ!'!!& Ñ &!ß!!!Ð Þ!'!!& Ñ œ &) Answer: > Ð Ñ Þ!!!!&> 8 Since premiums are benefit premiums, they are based on the equivalence principle, so on the issue date we have the equivalence principle relationship: actuarial present value of benefit œ actuarial present value of premium APV of benefit is!!!e œ!!! ; œ!!!òe ; Ó! 5l! 5l! 5œ!!À!l 5œ! 5!À!l! 5 5! 5 5l ;! 5œ! APV of premiums is 1+ Ð!!!@; : Þ The second factor in the APV of premiums is second factor in the expression for the APV of benefit, which is the same as the When we set APV of benefit equal to APV of premium we get!!!òe 5 ; Ó œ 1 + Ð!!!@; :, which simplifies to!à!l 5œ! 5l!!À!l! 5 5! 5œ!!!!E!À!l +!À!l!!!E œ 1+ Therefore, 1 œ!à!l!à!l 8 To find E we use the relationship E œ : E, so that 8 8!À!l À8l! E œ :! E '! Þ From the Illustrative Table we get 40!À!l Þ$ œ ÐÞ$'*$Ñ œ E ÐÞ$)!&Ñ ÐÞ$'*$Ñ Solving for! j!à!l j '!!!À!l E results in E œ Þ!'! Þ!À!l!À!l )ß))ß!( *ß$$ß' 2brove@rogerscom sam@utstattorontoedu

5 8 continued We use a similar relationship for annuities to find + : + œ : +!À!l 40!À!l Again, from the Illustrative Table we get )ß))ß!( Þ)' œ +!À!l ÐÞ$)!&Ñ *ß$$ß' ÐÞ&Ñ so that +!À!l œ Þ(' Then 1 œ œ &Þ Þ!!!ÐÞ!'!Ñ Þ('!!! '! The problem has been constructed so that the premiums starting at age 60 exactly match the benefit cost from age 60 on Therefore, for the equivalence principle, we must match the first 20 years of premium with the first 20 years of benefit Answer: Ð7Ñ 9 Under the UDD assumption, + œ α Ð7Ñ + Ð7Ñ ÐÑ Therefore, + œ α ÐÑ + ÐÑ '* '* Since the interest rate is 6, from the Illustrative Tables we have α ÐÑ œ Þ!!! and ÐÑ œ Þ&($* E The relationships we will use to find + '* are + œ, + C C + C (both of which 3 are always true) and E œ E (which is true under the UDD assumption) $ We are given E(! œ Þ&$! Since 3 œ Þ!', we have $ œ 68ÐÞ!'Ñ œ Þ!&)'* and 3 $ E(! œ 3 œ Þ!&'' Þ Then E(! œ 3 E(! œ Þ&(, and + (! œ œ )Þ&(, Þ*( and + '* + (! œ Þ!' Ð)Þ&(Ñ œ )Þ)', and finally, ÐÑ + œ α ÐÑ + ÐÑ œ ÐÞ!!!ÑÐ)Þ)'Ñ Þ&($* œ )Þ&* Answer: C '* '* 10 From the recursion relationship we have ; :'* 3 :'* 1 '*Àl '*Àl '*?Ð(!Ñ œ α (70) Ð(!Ñ?Ð'*Ñ p œ?ð'*ñ œe E œe '* '* '*Àl Going back one more year, we have ; 3 : : ')?Ð'*Ñ œ α ('*) Ð'*Ñ?Ð')Ñ p E'*Àl œ?ð')ñ ') ') ') ') '*Àl ')Àl ') '*Àl ')Àl E œ E œ E 2brove@rogerscom sam@utstattorontoedu

6 10 continued In general, this recursion will be of the form?ð5ñ œ E 5À(! 5l (endowment insurance) ; 5 3 with?ð5ñ œ α Ð5Ñ Ð5Ñ?Ð5 Ñ p E5À(! 5l œ : :?Ð5 Ñ 5 5 p?ð5 E E œe 5 5 5À(! 5l 5 5Àl 5À(! 5l 5 À(! 5 l Therefore,?Ð!Ñ œ E!À$!l (Note that if we had been given?ð(!ñ œ!, then?ð'*ñ would be?ð'*ñ œ E Continuing with the recursion would result in?ð5 Ñ E œ E, and then?ð!ñ œ E ) Answer: C 5À(! 5l 5 À(! 5 l!à$!l 5 '* '*Àl 11 If the next train is an express train the you and your co-worker arrive at the same time In order for you to arrive at work earlier than your co-worker, the next train must be a local train, and the next express train must arrive more than 12 minutes after the next local train If and X / denote the times (in minutes) until the next local and express trains respectively, then the probability that you arrive before your co-worker is T ÒX/ X6 Ó The number of local! trains arriving per minute has a Poisson distribution with a mean of '! Þ(& œ, and the number of express trains arriving per hour has a Poisson distribution with a mean of! '! Þ& œ We are told that the type of a train is independent of types of preceding trains, so that the numbers of local and express trains arriving per hour are independent of each other X 6 The Poisson distribution is related to the exponential distribution If R is the number of events of a certain type occurring per hour and if R has a Poisson distribution with mean -, we define X to be the time between events Then X has an exponential distribution with mean - Therefore, the time until the next local train, has an exponential distribution with a mean of 4 minutes X 6 and the time until the next express train, X / has an exponential distribution with a mean of 12 minutes Since the numbers of local and express trains are independent, the arrival times of the next local and express trains X and X are also independent The density functions of X and X 6 / 6 / =Î >Î 6 / 6 / =Î >Î (since 6 and / are independent, the joint density is the product of are 0Ð=Ñœ / and 0Ð>Ñœ /, and the joint density function of X and X is 0Ð=ß>Ñ œ / / X X the two individual densities) Then, TÒX/ X6 Ó œ ' ' =Î >Î! = Ð / / Ñ > = œ ' =Î Ð= ÑÎ! / / = $ œ / œ Þ(' Þ Answer: A 2brove@rogerscom sam@utstattorontoedu

7 12 The three states and transition probabilities can be modeled as a Markov chain The onestep transition probability matrix is Þ Þ& Þ$, where the first row and column correspond Ô Þ' Þ$ Þ Õ!! Ø to the state acutely ill (state1), the second row and column correspond to the state in remission (state 2), and the third row and column correspond to the state cured or dead (state 3) If in state 1, it is possible to transfer to state 3 and never return to state 1, and the same is true if in state 2 Therefore, when in state 1, the probability of returning to state 1 in some later transition (denoted 0 and also 0 in the Ross text) is less than 1, and the same is true of state 2 Therefore, states 1 and 2 are transient states The process is currently in state 1 If we can find the expected number of times the process will be in state 1 (say E, including the current time) and if we can find the expected number of times the process will visit state 2 (say F), then the expected total treatment costs for the current and future years will be!e F The method for finding E and F is outlined in the Ross text Suppose that in a finite state Markov chain, the transient states are numbered X œ Öß ß Þß > Given transient states 3 and 4, we define the factor $ œ 3œ4 3ß4 š, and! 3Á4 = 34 œ expected number of times the process is in (visits) state 4given that it starts in state 3 In this definition of = 34, for the special case of 3œ4, = 33 is the expected number of times the process will be in state 3now and in the future (we are assuming the process is in state 3now), so that = 33 is 1 plus the expected number of returns to state 3 from state 3 Therefore, E œ = and Fœ= The set of = 34 values can be found by the following matrix approach Ô T T â T> T T â T> T X œ Ö Ù ã ã ã ã ÕT T â T Ø > > >> is the > > matrix of transition probabilities for the set of transient states Ô = = â = > = = â = > W œ Ö Ù is the > > matrix of the = 34's ã ã ã ã Õ= = â = Ø > > >> œ X X W ÐM T Ñ (matrix inverse), where M is the identity matrix of the same size as T 2brove@rogerscom sam@utstattorontoedu

8 12 continued In this example, the transient states are 1 and 2, and Þ' Þ$ TX œ, and Þ Þ& Þ& Þ$! Þ' Þ$ Þ Þ$ W œ ÐM TX Ñ Þ Þ œ Œ œ œ! Þ Þ& Þ Þ& Þ Þ Þ& Þ$ Þ Þ From the final matrix we get = œ œ $Þ&( œ E, and = œ œ Þ œ F The expected total treatment costs for the current and future years will be!e F œ!ð$þ&(ñ Þ œ $(Þ) Þ (Note that for transient states 3and 4we could also find 0 œ probability that the process ever enters state 4 given that it starts in state 3 This is found from 0 œ = $ = 34 3ß4 44 Þ Þ ) Answer: E 13 The probability of ruin occurring by time 2 can be broken into two components TÒruin by time 2ÓœTÒruin occurs by time 1Ó TÒruin occurs between times 1 and 2 Ó The expected aggregate amount of claims in the first period is IÒWÓ œ IÒRÓ IÒ\Ó œ ÐÞ)ÑÐ!!!Ñ œ )!! With a relative security loading of 25, the premium per period is ÐÞ&ÑÐ)!!Ñ œ!!! Since surplus starts at 1000 and premium for the period will be 1000 by the end of the period, the only way ruin can occur in the first period is if there are 2 or more claims Therefore, TÒruin occurs by time 1 ÓœTÒR Ó Þ) Þ) œ T ÒR œ! or Ó œ Ò/ / ÐÞ)ÑÓ œ Þ* ( R is the number of claims in the first period and has Poisson distribution with mean 8) If ruin does not occur by time 1, then it might occur between time 1 and time 2 The probability will depend on the surplus at the start of the second period, which in turn depends upon whether there was 0 or 1 claim in the first period We can formulate TÒruin occurs between times 1 and 2 Ó as T Òruin occurs between times 1 and 2 lr œ!ó T ÒR œ!ó TÒruin occurs between times 1 and 2 lr œó TÒR œó If R œ! then the surplus at the start of period 2 is 2000, and ruin will occur during period 2 if there are 3 or more claims during period 2 We will denote by the number of claims in the second period Then, TÒruin occurs between times 1 and 2 lr œ!óœtòr $Óœ TÒR œ! or or Ó Þ) Þ) Þ) / ÐÞ)Ñ œ Ò/ / ÐÞ)Ñ ÓœÞ!( R 2brove@rogerscom sam@utstattorontoedu

9 13 continued If R œ then the surplus at the start of period 2 is 1000, and ruin will occur during period 2 if there are 2 or more claims during period 2 We will denote by the number of claims in the second period Then, TÒruin occurs between times 1 and 2lR œ1 ÓœTÒR Óœ TÒR œ! or Ó Þ) Þ) œ Ò/ / ÐÞ)ÑÓ œ Þ* Then, TÒruin occurs between times 1 and 2Ó œtòruin occurs between times 1 and 2 lr œ!ó TÒR œ!ó TÒruin occurs between times 1 and 2 lr œó TÒR œó Þ) Þ) œ ÐÞ!(ÑÐ/ Ñ ÐÞ*ÑÐ/ Þ)Ñ œ Þ!* Finally, TÒruin by time 2Ó œ T Òruin occurs by time 1Ó T Òruin occurs between times 1 and 2 Ó œ Þ* Þ!* œ Þ) Answer: C R 14 The original mortality assumption is DeMoivre's law with = œ*!, since the deferred mortality probabilities are constant for 50 more years from age 40 The contract premium is the!! whole life benefit premium!!!t œ œ E œ = + = l *!! *!!l &! &!l &! 3!!!E!! E! Under DeMoivre's law!!!!e!!!e + E!!, so that E œ + œ + œ œ Þ$&!) ( 3 œ œ Þ!&'$ ) Then!!!T œ œ (Þ! IÒ PlOÐ!Ñ!Ó œ!!!e (Þ!! &! + &!, where the insurance and annuity values are calculated under the revised mortality assumption Under the revised mortality assumption at age 50, survival from age 50 follows DeMoivre's law with = œ(& (constant deferred mortality for 25 years from age 50) Therefore, E œ + œ + œ œ Þ&*, &! E &! (& &! (& &!l & &l & 3 and + œ &! œ *Þ! Finally, IÒ! PlOÐ!Ñ!Ó œ!!!ðþ&*ñ Ð(ÑÐ*Þ!Ñ œ $!' Answer: E 2brove@rogerscom sam@utstattorontoedu

10 15 We use the relationship for the variance of a continuous whole life annuity of 1 per year: Z+<Ò+ Óœ Ò E ÐE Ñ Ó Xl $ The survival distribution is a mixture of the survival distributions for smokers and non-smokers smoker non-smoker E œ Þ$E Þ(E œ ÐÞ$ÑÐÞÑ ÐÞ(ÑÐÞ)'Ñ œ Þ$$$ E œ Þ$ E smoker Þ( E non-smoker smoker non-smoker We find E and E from the given annuity variances )Þ)) œ Z +<Ò+ smoker smoker smoker Ó œ Ò E ÐE Ñ Ó œ Ò E ÐÞÑ Ó Xl from which we get $ ÐÞÑ, smoker E œ Þ)&$ Similarly, )Þ&!$ œ Z +<Ò+ non-smoker non-smoker smoker Ó œ Ò E ÐE Ñ Ó œ Ò E ÐÞ)'Ñ Ó, Xl $ ÐÞÑ non-smoker E œ Þ') smoker non-smoker E œ Þ$ E Þ( E œ Þ!, and finally, Xl ÐÞÑ from which we get Then, Z +<Ò+ Ó œ $ Ò E ÐE Ñ Ó œ ÒÞ! ÐÞ$$$Ñ Ó œ *Þ Answer: E 16 This problem is very similar to 13 of the May 2001 exam For a particular radio, the probability that the radio has stopped transmitting by time 225 is Þ&;! The probability that the probe is no longer transmitting by time 225 is TÒ(radio 1 has failed) (radio 2 has failed) (radio 3 has failed) Ó Since the radios are independent of one another, this probability is TÒ(radio 1 has failed) Ó TÒ(radio 2 has failed) Ó TÒ(radio 3 has failed) ӜРÞ&; Ñ $ (we use the rule for independent events E and F that says T ÒE FÓ œ T ÒEÓ T ÒFÓ ) From the given information we have ;! œ!l;! œ Þ ß ;! œ ;! l;! œ Þ Þ œ Þ$ ß $ ;! œ ;! l;! l;! œ Þ' Then :! œ Þ( ß $ :! œ Þ According to the hyperbolic assumption, œð >Ñ > Therefore, jc > jc jc œ Ð Þ&Ñ ÐÞ&Ñ and multiplying by j results in j Þ& j j $ j j j j œ ÐÞ(&Ñ Þ& j ÐÞ&Ñ j, or equivalently, $ : : : Þ&! Þ& $ ÐÞ(&ÑÐ Ñ ÐÞ&ÑÐ Ñ Þ&! œ ÐÞ(&Ñ ÐÞ&Ñ Then, : œ œ Þ&)*, and ; œ Þ Þ( Þ Finally, the probability that the probe is no longer transmitting by time 225 is Ð Þ&; Ñ $ œ ÐÞÑ $ œ Þ!'* Answer: E 2brove@rogerscom sam@utstattorontoedu

11 8 8! ' &!:! &!;!! X Ð!Ñ!!!:!!;!!!!!;! œ! 0X Ð!Ñ Ð>Ñ > œ! 50 Ð>Ñ > œ 5! 0 Ð>Ñ > œ 5!;! M 0 Ð>Ñ the Illustrative Life Table), where!;! M M M j! *ß$$ß'!;! œ!:! œ j œ!ß!!!ß!!! œ Þ!')') Þ! M M j Similarly, &! ;! œ &! :! œ j &! œ Þ!* Þ! &!;! 5 &!;! Then,! :! œ ; œ M Þ!*5 œ Þ!')')5 Þ!! 5!;! M 17 We use the relation ; œ 0 Ð>Ñ>, so that : œ œ Þ ' ' ' (since is the density from is found from the Illustrative Life Table to be We must find 5 to complete the calculation &!!! We know that ' 0, so that ' '! X Ð>Ñ > œ 5 0! Ð>Ñ > Þ 0 &! Ð>Ñ œ &! Since 0 Ð>Ñ is the density from the Illustrative Life Table, we have ' 0 Ð>Ñ > œ ; œ,! &!! M and since 0Ð>Ñis the density from DeMoivre's Law with = œ!!, we have 0Ð>Ñœ œþ! ' &!!! and 0 Ð>Ñ > œ &!ÐÞ!Ñ œ Þ& Then ' &! 5 0 Ð>Ñ > '!! Þ 0 Ð>Ñ œ Þ!*5 ÐÞÑÐÞ&Ñ œ p 5 œ $Þ)! &! Þ!*5!! Þ!')')5 Finally, : œ œ Þ) Answer: A!! Since X ÐÑ and X ÐÑ have constant forces of mortality, their survival probabilities have an = 8 exponential form, : œ / > > 18 The future lifetime is XÐÑ, where XÐÑis a mixture of two distributions The two mixing distributions are X= ÐÑ, the future lifetime of a smoker, and X8ÐÑ, the future lifetime of a nonsmoker The mixing weights are 3 for smokers and 7 for non-smokers The 75-th percentile of X ÐÑ is -, where T ÒX ÐÑ Ÿ -Ó œ Þ(& For a mixture distribution, T ÒX ÐÑ Ÿ -Ó œ ÐÞ$Ñ T ÒX= ÐÑ Ÿ -Ó ÐÞ(Ñ T ÒX8 ÐÑ Ÿ -Ó = = Þ- TÒXÐÑŸ-Óœ = -; œ -: œ /, and = 8 Þ- TÒX8ÐÑŸ-Óœ -; œ -: œ / Þ- Þ- Then, T ÒX ÐÑ Ÿ -Ó œ ÐÞ$ÑÐ / Ñ ÐÞ(ÑÐ / Ñ œ Þ(& Þ- This becomes a quadratic equation with ] œ / : Þ$] Þ(] Þ& œ! The solutions are ] œ Þ$& ß Þ'& Since ] œ / Þ-, we ignore the negative root Þ- Then / œ Þ$& p - œ Þ' Note that once we have formulated the quadratic equation, we can substitute in each of the five possible answers for - to see which one satisfies the quadratic equation Answer: D 2brove@rogerscom sam@utstattorontoedu

12 19 After the deductible is imposed, a payment will only be made if the loss is above the deductible The probability that a loss \ will result in a payment being made is TÒ\ Ó œ,, The expected number of losses that will result in a payment being made is (total expected number of losses) (probability a loss results in a payment) œ -,, Answer: D 20 We will denote by [ the loss-at-issue random variable for one insurance policy sold Then [ œ!ß!!!^ &!!], where ^ is the present value random variable for a discrete whole life insurance of 1 at age 65, and ] is the present value random variable for a discrete whole life annuity-due of 1 at age 65 The number of telephone enquiries in one day resulting in a sale, say R, has a Poisson distribution with a mean of Þ &!œ! Then W œ[ [ â [ R, and W has a compound Poisson distribution Rhas mean and variance IÒRÓœZ+<ÒRÓœ! To find the mean and the variance of [ we note that, so that &!! &!! [ œ!ß!!!^ &!!] œ Ò!ß!!! Ó^ œ )ß )$$^ )ß )$$Þ$$ 3 (note that œ œ Þ!&'' ) Then IÒ[ Ó œ )ß )$$E )ß )$$Þ$$ œ &'(Þ&$, and ]œ ^ 3 &!! Z +<Ò[ Ó œ Ò!ß!!! Ó Z +<Ò[ Ó œ )ß )$$ Ò E'& ÐE'& Ñ Ó œ &ß ß &' The mean of the compound distribution variable W is IÒWÓ œ IÒRÓ IÒ[ Ó œ &ß '(&Þ', and the variance is Z +<ÒWÓ œ IÒRÓ Z +<Ò[ Ó Z +<ÒRÓ ÐIÒ[ ÓÑ œ &ß $$'ß )'( Applying the normal approximation we have W IÒWÓ! Ð &ß'(&Þ'Ñ TÒW!ÓœTÒÈ Óœ ÐÞ'ÑœÞ'( Z+<ÒWÓ È F (this found from the normal table by &ß$$'ß)'( interpolating between FÐÞÑ and FÐÞ&Ñ ) Answer: C '& 2brove@rogerscom sam@utstattorontoedu

13 21 We use the retrospective form of the reserve, as it does not require finding 1 We first find! Z and then apply the recursive reserve relationship to find Z Since the interest rate is 06, the insurance and annuity values can be found from the Illustrative Table Since the benefit for the first 20 years is 1000 and the premium for the first 20 years is!!!t!, it follows that! Z œ!!!! Z! (the fact that the premiums and benefits change after time 20 does not change the fact that the reserve at the end of 20 years is the same as!!! Z ) We!! can use any method available to find!!!! Z!, and the easiest is usually the annuity form + (which is valid for whole life and endowment policies)!!! Z =1000( '! ) œ (Þ()!! +! We now use the recursive reserve relationship ÐZ 1 ÑÐ 3Ñ, ; œ: Z for 5œ! E! +! '! The premium at time 20 (age 60) is &!!!T œ &!!!Ð! Ñ œ &Þ, and ; œ Þ!$(', and the death benefit is 5000 at age 61, so that Ð(Þ() &Þ&&ÑÐÞ!'Ñ &!!!ÐÞ!$('Ñ œ ÐÞ*)'Ñ Z Solving for Z results in Z œ && Þ Note again that this solution did not require us to find the value of 1 Answer: A 22 We use the relationship E to get E Since we are given E E! œ Þ!!), we have Þ!!) Þ**( E Þ!!) œ E p E œ Þ& Þ!& Þ!& Similar relationships hold for E : E : E Then E : E and E E œ Þ!!$$ results in!!!! Þ!!) Þ**( E Þ!!$$ œ ÐÞ!&Ñ ÐÞ!&Ñ E p E œ Þ!(* Finally, Z +<Ò^Ó œ E ÐE Ñ œ Þ!& Answer: C 2brove@rogerscom sam@utstattorontoedu

14 23 The amount by which surplus falls below its initial level when this happens for the first time \ is the random variable P The density function of P is 0 ÐCÑ œ œ P IÒ\Ó : J ÐCÑ T ÐCÑ We are told that \ has a uniform distribution on the interval Ð!ß!Ñ Therefore, : œiò\óœ& and J\ ÐCÑœ! The pdf of P is 0 ÐCÑ œ œ Þ Þ!C,! C! ÞC P & C Since P is the amount by which surplus falls below its initial level when this happens for the first time, to say that surplus falls below 35 is equivalent to saying that P & The probability of that is T ÒP &Ó œ '! 0 ÐCÑ C œ '! ÐÞ Þ!CÑ C œ Þ& Answer: A & P & 24 From the transition matrix we see that after one transition the chain will move to either state 0 or state 2, and then must move back to state 1 Every second transition will be back to state 1 The present value factor for the first year is 94 The present value factor for the second year is either 95 (prob 9) or 93 (prob 1) At the end of two years, the state is again 1, and the present value at the end of two years of the future perpetuity immediate is again IÒ] Ó since the process has started over Year PV factor 94? 94? IÒ]ÓœÞ* Þ* IÒTZ Ó Þ* IÒTZ Ó O where TZ is the present value factor for the 2nd year, and O is the present value at the end of the 2nd year (just after the 2nd payment) of the rest of the perpetuity Since after the 2nd year the state must be 1, it is as if the process has started over, and OœIÒ]Ó Therefore IÒ]ÓœÞ* Þ* IÒTZ Ó Þ* IÒTZ Ó IÒ]Ó Þ* Þ* Þ*) IÒT Z Ó œ ÐÞ*&ÑÐÞ*Ñ ÐÞ*$ÑÐÞÑ œ Þ*) Then, IÒ] Ó œ œ Þ) Þ* Þ*) Alternatively, we see that the present value factor for the 1st, 3rd, 5th, years is always 94 (because the process must return to state 1 every second transition) The expected present value factor in the 2nd, 4th, 6th, years is IÒT Z Ó œ ÐÞ*&ÑÐÞ*Ñ ÐÞ*$ÑÐÞÑ œ Þ*), so that IÒ] Ó œ ÐÞ*Ñ ÐÞ*ÑÐÞ*)Ñ ÐÞ*ÑÐÞ*)ÑÐÞ*Ñ ÐÞ*ÑÐÞ*)ÑÐÞ*ÑÐÞ*)Ñ â $ œ ÒÐÞ* ÐÞ*ÑÐÞ*)ÑÓÒ ÐÞ* Þ*)Ñ ÐÞ* Þ*)Ñ ÐÞ* Þ*)Ñ âó œ Þ* Þ* Þ*) Þ* Þ*) œ Þ) Answer: D 2brove@rogerscom sam@utstattorontoedu

15 25 The year to year behavior of stock prices can be modeled as a Markov chain with 3 states: state 1 - market price strike price for neither stock, state 2 - market price strike price for exactly one stock, state 3 - market price strike price for both stocks We are asked to find the limiting probability that the process is in state 3 (denoted 1 $ ) in the Probability Models book 1000 transitions is far enough into the future to be considered as a limiting point We assume that at time 8, the relationship of the market price of stock A to its strike price is independent of the relationship of the market price of stock to its strike price given the state at time 8 If the process is in state 1 at time 8, then there is a probability of that stock A's market price will be less than its strike price on day 8, and the same is true for stock The probability that at time 8 neither market price is greater than its strike price is œ ; thus, T œ In a similar way, given state 1 at time 8, the probability that both stocks have market price $ $ $ * * that are greater than their strike price (state 3) at time 8 is œ œ T$ Since ' T T T$ œ, it follows that T œ (alternatively, if the state is 1 at time 8, $ then there is a probability that stock A market price is greater than its strike price and $ stock market price is not greater than its strike price, and there is probability that stock market price is greater than its strike price and stock A market price is not greater than its strike price, for a combined probability of than the strike price) ' that exactly one of the two market prices is greater Similar reasoning results in the following one-step transition probability matrix: * ' Ô Ö ÙÞ Õ Ø ) ' * The equations for the limiting probabilities are formed by using coefficients down each column: * ' ) ' * 1 œ 1 1 1$, 1 œ 1 1 1$, 1$ œ 1 1 1$, (the third equation in the previous line is redundant) and 1 1 1$ œ The first and second equation can be written as ( 1 œ 1 1 and ) 1 œ ' 1 ' 1 $ $ If we write the final equation as 1 œ 1 1$ and substitute into the first two earlier equations we get (Ð 1 1 Ñœ 1 1 and ) 1 œ'ð 1 1 Ñ ' 1 $ $ $ $ Solving these two equations for results in Answer: A 1 1 $ $ œ ( 2brove@rogerscom sam@utstattorontoedu

16 26 The number of problems solved in 5 minutes has a Poisson distribution with a mean of 1 Therefore, the number of problems solved within 10 minutes, say R, has a Poisson distribution with a mean of 2 (the number of problems solved by time > forms a Poisson process) We are looking for TÒ3 is solved within 10 minutes Ó This can be formulated as TÒ3 is solved within 10 min 3 is first problem chosenó TÒ3 is solved within 10 min 3 is second problem chosenó TÒ3 is solved within 10 min 3 is third problem chosenó œtò3 is solved within 10 min l3 is first problem chosen Ó TÒ3 is first problem chosenó TÒ3 is solved within 10 min l3 is second problem chosen Ó TÒ3 is second problem chosenó TÒ3 is solved within 10 min l3 is third problem chosen Ó TÒ3 is third problem chosen Ó Since the problems are chosen randomly, there is a 1/3 chance that problem 3 is chosen first, and the same probability of being chosen second and also third TÒ3 is solved within 10 min l3 is first problem chosenó œtòr Óœ TÒRœ!Óœ / œþ)'& TÒ3 is solved within 10 min l3 is second problem chosenó œtòr Óœ TÒRœ!ßÓœ Ò/ / ÐÑÓœÞ&* TÒ3 is solved within 10 min l3 is third problem chosenó / Ð Ñ x œtòr $Óœ TÒRœ!ßßÓœ Ò/ / ÐÑ ÓœÞ$$ $ Then, T Ò3 is solved within 10 minutes Ó œ ÒÞ)'& Þ&* Þ$$Ó œ Þ&* Answer: E 27 We use the exponential formulation for wðñ wðñ : wðñ ' ÐÑ Ð>Ñ > ' Ð Ñ Þ Ð>Ñ > ' 7 Þ5> > Þ5Î$!Þ!œ; œ : œ /! œ /! œ /! œ / Þ5Î$ so that / œ Þ*' ÐÑ Ð7Ñ ÐÑ Ð7Ñ In the two decrement table, since Ð>Ñ œ Þ Ð>Ñ it follows that Ð>Ñ œ Þ) Ð>Ñ, since ÐÑ ÐÑ Ð7Ñ Ð>Ñ Ð>Ñ œ Ð>Ñ We use the integral of density form to represent multiple decrement ÐÑ Ð7Ñ ÐÑ Ð7Ñ Ð7Ñ Ð7Ñ probabilities: ; œ '! > : Ð>Ñ> and ; œ '! > : Ð>Ñ> ÐÑ Ð7Ñ It then follows from Ð>Ñ œ Þ) Ð>Ñ that ÐÑ Ð Ñ ÐÑ Ð Ñ Ð Ñ Ð Ñ ; œ ' 7 : Ð>Ñ> œ ' : ÐÞ) Ð>ÑÑ> œ Þ) ;! >! >, 2brove@rogerscom sam@utstattorontoedu

17 27 continued Ð7Ñ Ð7Ñ Ð7 Ñ ' Ð>Ñ > 5> > )5Î$ We find ; from ; œ : œ / Ð7Ñ '! œ /! œ /, and we note that / )5Î$ œ Ð/ Þ5Î$ Ñ! œ ÐÞ*'Ñ! œ Þ*&, so that ; Ð7Ñ œ Þ)!' ÐÑ Ð7Ñ Finally, ; œ Þ) ; œ Þ' Þ5Î$ We could have solved the problem another (more tedious) way From / œ Þ*' we can Ð7Ñ solve for 5, 5 œ Þ'$, so that Ð>Ñ œ Þ'$>, and therefore ÐÑ 2 ÐÑ 7 = ' Ð7Ñ Ð>Ñ > Þ'$= $ Î$ Ð>Ñ œ ÐÞ)ÑÞ'$> Ñ œ Þ)**> Then, = : œ /! œ / ÐÑ Ð Ñ ÐÑ Then, ; œ ' 7 Þ'$> $ Î$! > : Ð>Ñ > œ '! / Þ)**> > This integration requires a change of variable?œ> to simplify Answer: D 28 The probability function for Ois TÒOœ5Óœ ;, and also we have TÒO 5Óœ : for the curtate future lifetime O $œ O if O Ÿ $ O if œ which is also equal to O Ÿ $ if O $ œ $ if O $ 5l 5 IÒO $ÓœÐ!Ñ TÒO œ!ó ÐÑ TÒO œó ÐÑ TÒO œó Ð$Ñ TÒO $Ó œ l; l; $ $ : œ Ð: : Ñ Ð : $ : Ñ $ $ : œ : : $ : œ Þ* ÐÞ*ÑÐÞ)Ñ ÐÞ*ÑÐÞ)ÑÐÞ(Ñ œ Þ Note that IÒO $Ó œ / IÒO $ ( ) ÓœÐ! Ñ TÒO œ!ó Ð Ñ TÒO œó Ð Ñ TÒO œó Ð$ Ñ TÒO $Ó œ l; l; * $ : œð: : Ñ Ð : $ : Ñ * $ : œ: $ : & $ : œ Þ* $ÐÞ*ÑÐÞ)Ñ &ÐÞ*ÑÐÞ)ÑÐÞ(Ñ œ &Þ&) Z +<ÒO $Ó œ IÒ( O $ ) Ó ÐIÒO $ÓÑ œ Þ!( Answer: A À$l IÒ\ Ó 29 The loss elimination ratio is IÒ\Ó, where is the deductible Þ!! Ÿ)! We see that 0ÐÑ œ Þ!!!&Ð! Ñ )! Ÿ!! otherwise! )!! IÒ\Ó œ ' 0ÐÑ œ ' Þ! '!! )! Þ!!!&Ð! Ñ œ &!Þ'( IÒ\!Ó œ '! 0ÐÑ! T Ò\!Ó œ '!!! Þ!!Ð T Ò\ Ÿ!ÓÑ œ '! Þ!!Ò '! Þ! Ó œ )!! The loss elimination ratio with a deductible of 20 is Answer: E ) &!Þ'( œ Þ$&& 2brove@rogerscom sam@utstattorontoedu

18 30 We are given that 1 * œ T &!Þ!!, and we are asked to find ;' ; ', where ;' is the mortality probability at age 64 for Smith and is the standard mortality probability at age 64 ; ' We are told that for every 5, 5Z for this policy is the same as 5 Z& (whole life policy reserve based on standard mortality) The recursive relationship for reserves for this policy in the 20th L year is Ð* Z 1* ÑÐÞ!$Ñ Ð! ZÑ ; ' œ! Z and the recursive relationship for reserves for the policy based on standard mortality in the 20th year is Ð Z T ÑÐÞ!$Ñ Ð Z Ñ ; œ Z * & &! &! & ' We are told that the policy reserves on the high mortality policy are the same as those on the standard mortality policy, so that * Z œ * Z& and! Z œ! Z& Then, subtracting the standard mortality recursive equation from the high mortality recursive equation we have Ð1* T& ÑÐÞ!$Ñ Ð! Z& Ñ Ð; & L ; & Ñœ! Then since 1* T& œ!þ! and Z œ!þ(, we get ; L ; œ!þ!) Answer: D! & & & L L 31 The loss for one policy is P œ ^ Þ!&], where ^ is the present value random variable for a whole life insurance of 1 for ÐÑ, and ] is the present value random variable for a whole ^ Þ!& Þ!& life annuity-due of 1 for ÐÑ Then ] œ and P œ Ò Ó^ E The mean of P is IÒPÓ œ E Þ!&+ œ E ÐÞ!&ÑÐ Ñ œ Þ!)' Þ!& The variance of Pis Ò Ó Þ!& Z+<Ò^ÓœÒ Ó ÒE ÐE ÑÓœÞ!') Suppose that there are 8 independent policies The total loss is W œ P P â P8, and IÒWÓ œ Þ!)'8, Z +<ÒWÓ œ Þ!')8 The probability of a positive total loss is W Ð Þ!)'8Ñ Þ!')8 TÒW!Ó, and under the normal approximation this Ð Þ!)'8Ñ Þ!')8 becomes TÒW!ÓœTÒ È È Óœ FÐÞ$( È8Ñ In order for this to be less than or equal to 005, we must have FÐÞ$( È8Ñ Ÿ Þ!&, or equivalently, FÐÞ$( È8Ñ Þ*& Since FÐÞ'&Ñ œ Þ*&, follows that we must have Þ$( È 8 Þ'&, or equivalently, 8 ( Answer: 2brove@rogerscom sam@utstattorontoedu

19 32 If the death benefit is,, then the present value random variable of the combined annuity and insurance is [ œ ß!!!],^, Since ] is a whole life annuity-due (of 1 per year) present value random variable and ^ is a discrete whole life insurance of 1 present value random ^ ^ ß!!! ß!!! ß!!! Ñ Z+<Ò^Ó ß!!!, œ œ &!ß!!! variable, we have ] œ Then [ œ ß!!!Ð Ñ,^ œ Ð, Ñ^, with variance Z+<Ò[ÓœÐ, This variance will be 0 if Answer: D 33 The annual cost to the Club W has a compound Poisson distribution with Poisson mean ( prob 5 - œ!!!, and 3-point severity distribution \ œ *! prob 4 (90 of the towing cost) prob 1 Then IÒWÓ œ -IÒ\Ó œ!!!òð(ñðþ&ñ Ð*!ÑÐÞÑ ÐÑÐÞÑÓ œ )'ß!! Z +<ÒWÓ œ - IÒ\ Ó (valid only for compound Poisson) œ (ß *!&ß '!!, and Applying the normal approximation to W, we have W )'ß!! (ß*!&ß'!! *!ß!!! )'ß!! (ß*!&ß'!! T ÒW *!ß!!!Ó œ T Ò È È Ó œ F ÐÞ)Ñ œ Þ! Answer: 34 Suppose that the exponential mean is IÒ\Ó œ ) in all years Î) From the distribution table we have IÒ\ Ó œ ) Ò / Ó With deductible this year, the loss elimination ratio this year is Î) IÒ\ Ó ) Ò / Ó IÒ\Ó ) Î) Î) Þ(! œ œ œ / It follows that / œ Þ$ In the coming year the deductible is Î$ but the distribution of \ remains the same, so that the loss elimination ratio in the coming year is IÒ\ ÐÎ$ÑÓ IÒ\Ó Answer: C ) Ò / ) Î$) Ó Î$ ) Î ) Î$ Î$ œ œ / œ Ð/ Ñ œ ÐÞ$Ñ œþ)! 2brove@rogerscom sam@utstattorontoedu

20 35 We first note that IÒX >Óœ/ We then use the relationship / œ/ : /!À>l!À8l 8 8 Then 'œiòxóœ/ œ/ : / œiòx!ó ÐÞ'ÑÐ/ Ñ!!À!l!!!! From (iv) we have IÒX!Ó œ! Þ!!&Ð! Ñ œ $ Then solving for / results in / œ &! Answer: E!! *! ' *!Àl! > *! 36 From the Illustrative Life Table, ; œ Þ))(( / œ : > Under the uniform distribution assumption, / œ ' Ð ; Ñ > œ ' Ð > ; Ñ > œ ' Ð Þ))((>Ñ > œ Þ*!&'& œ Y *!Àl! > *!! *!! Under the constant force assumption, > : œ Ð: Ñ for! > In this case, :*! œ Þ)$, > so that Then > : œ ÐÞ)$Ñ / œ ' Þ)$ ÐÞ)$Ñ > œ œ Þ*!$' œ G > *! *!Àl! > ; > Ð >Ñ; > *! Þ))((> Þ)$ Þ))((> 68ÐÞ)$ Þ))((>Ñ >œ Þ))(( >œ! > 68ÐÞ)$Ñ Þ))((> Þ)$ Þ))((> Under the hyperbolic assumption, ; œ, so that ; œ *!Àl! Þ)$! Þ)$ Þ))((> Then, / œ ' Ð Ñ> œ ' > œ ÐÞ)$ÑÒ ¹ Ó œ Þ)**!$ œ L Y G L Answer: C 37 With discount rate œþ!&, the one-year present value factor We note that the 2-year discrete endowment insurance has actuarial present : Ñ : Ð; : Ñ] : Ñ ased on Kevin's premiums, we : Ñ œ '!) $&!@:, so that!!!òþ*&; Þ*& Ð ; ÑÓ œ '!) $&!ÐÞ*&ÑÐ ; Ñ It follows that ; œ Þ 1 Then, for Kyra's premium, we : Ñ œ Ñ, so that *!(Þ& œ Þ)&& 1, and then 1 œ )* Answer: 2brove@rogerscom sam@utstattorontoedu

21 38 Using the recursive relationship for reserves from time 9 to time 10, we have Ð* Z 1 ÑÐ 3Ñ, ; * œ : *! Z, so that Ð$ß &$&!()ÑÐÞ!&Ñ!!ß!!!ÐÞ!Ñ œ ÐÞ*)*Ñ! Z Solving for Z results in Z œ $&ß '$'!! Since this is a 10-payment policy, it is paid up as of time 10 so that the prospective form of the reserve is! Z œ!!ß!!!e! ut we have just found that this is 35,636 Therefore, E œ Þ$&'$'! We now use the recursive relationship for discrete insurance, EC CEC Therefore, E, which becomes Þ! Þ*)) Þ$&'$' œ E, and then E œ Þ$''&) Answer: E Þ!& Þ!& 39 / &À'& œ /& / '& / &À'& Under DeMoivre's Law, = / œ, so that / œ $! and / & '& œ!!!! / œ ' : > œ ' : : > œ ' > > Ð ÑÐ Ñ > &À'&! > &À'&! > & > '&!! œ ' > >! Ò!! Ó > œ &Þ&' '!! Then, / œ $!! &Þ&' œ $Þ Answer: &À'& 40 Suppose that \ is the outcome of one trial Then TÒ\œ!ÓœÞ) For the non-zero values of \, the density function is 0ÐÑ œ Þ!!! œ Þ!!!!& for!!! &!!!; this is 2 multiplied by the density function for the value of treasure if treasure is found The distribution Þ)! Ÿ Ÿ!!! function for \ is JÐÑ œ Þ) Þ!!!!&Ð!!!Ñ!!! Ÿ &!!! &!!! We are told to apply the inverse transformation method with large random numbers corresponding to large outcomes This is the same as saying that small random numbers correspond to small outcomes For random numbers less than or equal to 8, the simulated value of \ is 0, and for a random number < between 8 and 1 the simulated value of \ is found by solving JÐÑ œ < The first random number is <œþ(& so the simulated value of \ is 0 The second random number is <œþ)& so the simulated value is, where J ÐÑ œ Þ) Þ!!!!&Ð!!!Ñ œ Þ)& Solving for results in œ!!! The average of the two trials is 1000 Answer: 2brove@rogerscom sam@utstattorontoedu