ACT455H1S - TEST 2 - MARCH 25, 2008

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1 ACT455H1S - TEST 2 - MARCH 25, 2008 Write name and student number on each page. Write your solution for each question in the space provided. For the multiple decrement questions, it is always assumed that decrements are independent of one another unless indicated otherwise. 1. A fully discrete whole life insurance policy with level premiums issued at age B has a death benefit of 100,000. The policy expenses are as follows 1st Year Renewal Years Percent of Premium 50% 20% Per Policy The policy is based on a two-decrement model, with decrement 1 being death and decrement 2 being policy cancellation. Cancellation can only occur at the end of a year. Interest is at a rate of 3 œ Þ 0 and mortality probabilities are ; B Ð"Ñ œ Þ!& ß ; Ð"Ñ B " œ Þ!& and the policy cancellation probability is Þ& every year. The insurer wishes to have a expected asset share of 3000 per surviving policy at the end of two years. a) If the insurer charges a contract premium of 14,000 per year, and if the insurer pays a cash value of GZ for withdrawals at the end of the first year, and GZ "!!! for withdrawals as the end of the second year, find the value of GZ. b) If the insurer pays a cash value of 500 at the end of the first year and 1000 at the end of the second year, find the contract premium.

2 2. A homogeneous Markov Chain has state space Ö!ß ". Þ% Þ' The one-step transition probability matrix is U œ Þ" Þ*. Suppose that the chain is currently in state 0. " a) Find the probability that it will be in state 0 at least once in the next 8 transitions. " b) Find the expected number of transitions until it is in state 0 again. c) Find the expected number of transitions until the next transition from state 1 to state 0 counting the transition from 1 to 0; this is defined as follows: since the current state is 0, if the sequence of transitions is , then there are 2 transition until the next 1-0 transition, if the sequence of transitions is , then there are 3 transition until the next 1-0 transition, if the sequence of transitions is , then there are 3 transition until the next 1-0 transition, if the sequence of transitions is , then there are 4 transition until the next 1-0 transition, etc.) d) Find the eigenvalues and eigenvectors of U and find the matrix factorization form of U in the form UœY H Y ". e) Find the limiting probabilities U and U using the matrix factorization in d). Ð3ß!Ñ f) Using the matrix relationship 8 " U œ 8U U, take the limit as 8p of both sides of the matrix equation to get two equations from which the limiting probabilities can be found, and find the limiting probabilities. Ð3ß"Ñ

3 3. Consider the following variation on Gambler's ruin. The gambler can win or lose one on each bet, but the win-lose probabilities depend on the amount of money she has. Amount the gambler has Prob. win Prob. lose 1 2/3 1/3 2 1/2 1/2 3 1/3 2/3 As soon as the gambler reaches 4 or 0, her fortune stays at that point Ð!ß!Ñ Ð%ß%Ñ states 0 and 4 are absorbing states, U œ U œ "). Suppose that the gambler begins the game with 2. a) Find the probability that gambler bets at least 3 times before getting to either 0 or 4. b) Find the expected number of times the gambler will have 1,2 or 3 in the future not counting the current state of 2). c) Find the probability that the gambler will ever have 2 in the future. 1 d) States 1, 2 and 3 are transient. The one-step transition matrix for the transient states is Ô! Î! U X œ "Î! "Î states ordered 1, 2, 3 in the matrix rows and columns). Õ! Î! Ø Show by mathematical induction that if 8 is an odd integer!, then the 8-step transition matrix! Î! for the transient states is UX œ Ñ 8 " Ô "Î! "Î. Õ! Î! Ø

4 4. The Boiler Room Sales company ranks each of its sales people at the end of each month according to the sales made by that sales person. There are three rankings: Excellent, Good, Poor. A salesperson who had poor sales for the month is immediately fired and never again rehired. Some sales people die during the month perhaps because of the high pressure under which they have to work). The BRS company has created a homogeneous Markov Chain model to describe transitions in a salesperson's ranking from one month to the next. The model has three states: I - excellent sales for the month just ended, K - good sales for the month just ended, and T - poor sales or died in the month just ended. ÐIßIÑ ÐIßKÑ ÐIßT Ñ Ô U U U Ô Þ Þ' Þ The one-step transition matrix is Ö ÐKßIÑ ÐKßKÑ ÐKßTÑ U U U Ù œ Þ" Þ& Þ%. Õ ÐT ßIÑ ÐT ßKÑ ÐT ßT Ñ U U U Ø Õ!! " Ø A salesperson has just received a ranking of K. 2 a) Find the expected number of times months) in the future that this salesperson will receive a ranking of K. 3 b) Find the expected number of times in the future that this salesperson will have two consecutive months with a ranking of I for instance, if the salesperson's rankings are KKIIKIIIT , then there would be 3 times that this salesperson had consecutive months with a ranking of I - months 3-4, months 6-7 and months 7-8). c) BRS company pays a salesperson a bonus of 1000 whenever that person has two consecutive months with a ranking of I. At a monthly interest rate of 1%, find the actuarial present values of the bonus payments to be made at the end of the next three months that will be made to each of the following salespeople: 1 i) a salesperson whose current rating is K, and 1 ii) a salesperson whose current rating is I.

5 ACT455H1 S - TEST 1 SOLUTIONS - FEBRUARY 12, For both parts a) and b) we use the relationship Ð"Ñ ÐÑ Ð7Ñ Ð EW K I ÑÐ" 3Ñ,; GZ ; œ: EW 5 5 B 5 5 " B 5 B 5 5 " a) Ò"%ß!!!ÐÞ&Ñ &!!!ÓÐ"ÞÑ "!!ß!!!ÐÞ!&Ñ Þ&GZ œ Þ " EW and Ò" EW "%ß!!!ÐÞ)Ñ "!!!ÓÐ"ÞÑ "!!ß!!!ÐÞ!&Ñ Þ&ÐGZ "!!!Ñ œ ÐÞÑÐ!!!Ñ so that Ò"%ß!!!ÐÞ&Ñ &!!!ÓÐ"ÞÑ "!!ß!!!ÐÞ!&Ñ Þ&GZ Þ Ò "%ß!!!ÐÞ)Ñ "!!!ÓÐ"ÞÑ "!!ß!!!ÐÞ!&Ñ Þ&ÐGZ "!!!Ñ œ ÐÞÑÐ!!!Ñ Solving for GZ results in GZ œ '). b) ÒKÐÞ&Ñ &!!!ÓÐ"ÞÑ "!!ß!!!ÐÞ!&Ñ Þ&Ð&!!Ñ œ Þ " EW and Ò" EW KÐÞ)Ñ "!!!ÓÐ"ÞÑ "!!ß!!!ÐÞ!&Ñ Þ&Ð"!!!Ñ œ ÐÞÑÐ!!!Ñ so that ÒKÐÞ&Ñ &!!!ÓÐ"ÞÑ &!!! "& Ò Þ KÐÞ)Ñ "!!!ÓÐ"ÞÑ &!!! &! œ "!! Solving for K results in K œ "ß )*!. 2. a) The process will not be in state 0 in the next 8 transitions is the probability that the chain will transfer to state 1 in the next transition and stay in state 1 for the following 8 " transitions. Ð!ß"Ñ Ð"ß"Ñ 8 8 This probability is U ÐU Ñ œ ÐÞ'ÑÐÞ*Ñ. b) Denote by I!ß! the expected number of transitions until the chain returns to state 0 from state 0, and I "ß! the expected number of transitions to reach state 0 from state 1. I!ß! œ ÐÞ%ÑÐ"Ñ ÐÞ'ÑÐ" I"ß! Ñ and I"ß! œ ÐÞ"ÑÐ"Ñ ÐÞ*ÑÐ" I"ß! Ñ From the second equation we get I œ "! and then from the first equation we get I œ. "ß! c) Denote by I! to "! the expected number of transitions until the next transition from state 1 to state 0 if currently in state 0, and denote by I " to "! he expected number of transitions until the next transition from state 1 to state 0 if currently in state 1. I! to "! œ ÐÞ%ÑÐ" I! to "! Ñ ÐÞ'ÑÐ" I" to "! Ñ I" to "! œ ÐÞ"ÑÐ"Ñ ÐÞ*ÑÐ" I" to "! Ñ From the second equation we get I" to "! œ "!, and then from the first equation we get & I œ œ! to "! Þ'.!ß!

6 d) The eigenvalues of U are the solutions of the equation./>ðu -MÑ œ!. Þ% - Þ'./> ÐU -MÑ œ./> œ - "Þ- Þ œ!. Þ" Þ* - Solutions are -œ"ßþ; these are the eigenvalues. The eigenvector for eigenvalue 1 are of the form B, where U B œ B. C C C This gives equations Þ%B Þ'C œ B and Þ"B Þ*C œ C, which both result in B œ C. " The basic eigenvector for the eigenvalue 1 is. " B B ÞB The eigenvector for eigenvalue Þ are of the form, where œ. C U C ÞC This gives equations Þ%B Þ'C œ ÞB and Þ"B Þ*C œ ÞC, which both result in B œ 'C. ' The basic eigenvector for the eigenvalue 1 is. " Then Y œ " ', H œ "! and Y œ " ' " ' " " œ. " "! Þ ' " " " " " " ' 8 8 " " ' "! e) 8 U œ U œ Y H Y œ " " 8! ÐÞÑ " " " ' 8 ' ' 8 ÐÞÑ ÐÞÑ œ " " 8 ' " 8 ÐÞÑ ÐÞÑ Ð3ß!Ñ " Ð3ß"Ñ ' The limiting probabilities are U œ for state 0 and U œ for state 1. f) U œ 8 " 8 U U is Ð!ß!Ñ Ð!ß"Ñ Ð!ß!Ñ Ð!ß"Ñ 8 " U 8 " U 8U 8U Þ% Þ' Ð"ß!Ñ Ð"ß"Ñ œ Ð"ß!Ñ Ð"ß"Ñ U U U U Þ" Þ* 8 " 8 " 8 8 The limit as 8p is Ð3ß!Ñ Ð3ß"Ñ Ð3ß!Ñ Ð3ß"Ñ U U U U Þ% Þ' Ð3ß!Ñ Ð3ß"Ñ œ Ð3ß!Ñ Ð3ß"Ñ U U U U Þ" Þ* Ð3ß!Ñ Ð3ß"Ñ Ð3ß!Ñ which results in equations Þ% U Þ" U œ U Ð3ß"Ñ Ð3ß!Ñ Ð3ß!Ñ Ð3ß"Ñ so that U œ ' U. It must also be true that U U œ ", Ð3ß!Ñ " Ð3ß"Ñ ' and then U œ ß U œ.

7 3.a) If the gambler starts with 2, the gambler will bet only 2 times before getting to 0 or 4 is the gambler either loses two in a row, or wins two in a row. This probability is " " " " " œ. The probability that gambler bets at least 3 times before getting to either 0 or 4 is the complement of this, which is. b) I 3 denotes the expected number of times the gambler will have 1, 2 or 3 in future is now in state 3. We wish to find I. We get the following equations" " I" œ Ð!Ñ Ð" I Ñ " " I œ Ð" I" Ñ Ð" I Ñ " I œ Ð" I Ñ Ð!Ñ Solving these equations results in I œ %ß I œ &ß I œ %. " c) T34 is the probability that the gambler will ever have 4 in the future given that the gambler currently has 3. " " " " T œ T" T ß T" œ Ð!Ñ Ð"Ñ œ ß T œ Ð"Ñ Ð!Ñ œ " " Then T œ Ñ Ñ œ. Ô! Î! Ô! Î! d) Since UX œ "Î! "Î, we see that UX 8 8 " œ Ð Ñ "Î! "Î Õ! Î! Ø Õ! Î! Ø is valid for 8œ". We note also that Ô! Î! Ô! Î! Ô "Î! "Î Ô "Î! "Î UX œ "Î! "Î Õ Ø "Î! "Î Õ Ø œ! Î!! "!! Î!! Î! Õ"Î! "Î Ø œ Õ"Î! "Î Ø! Î! Suppose that UX 8 œ Ð Ñ 8 " Ô "Î! "Î is valid for odd integers Õ! Î! Ø "ß ß &ß ÞÞÞß 5, We wish to show that the statement is true for the next odd integer 5.! Î! "Î! "Î " Ô Ô UX œ UX U X œ œ Ð Ñ "Î! "Î! "! Õ! Î! Ø Õ"Î! "Î Ø! Î! Ð5 Ñ " Ô œð Ñ "Î! "Î Õ! Î! Ø

8 4.a) Let \ be the expected number of times in the future that someone currently ranked K will be ranked K in the future, and let ] be the expected number of times in the future that someone currently ranked I will be ranked K in the future. \ œ Þ"] Þ&Ð" \Ñ and ] œ Þ] Þ'Ð" \Ñ! " " Solving these two equations results in \œ and ] œ ). b) Let [ be the expected number of times in the future a salesperson will have consecutive II - ranks if the current rank is K, and let ^be the expected number of times in the future a salesperson will have consecutive II - ranks if the current rank is I. [ œ Þ"^ Þ&[ ß ^ œ ÞÐ" ^Ñ Þ'[. " & " " Solving these two equations results in [œ and ^œ ). c)i) The following sequences of rankings for the next three months result in two consecutive II - ratings including the current ranking of K): Sequence that pay bonus at time 2: K I I, prob. ÐÞ"ÑÐÞÑ œ Þ! 2 Sequences that pay bonus at time 3: K K I I, prob. ÐÞ&ÑÐÞ"ÑÐÞÑ œ Þ!", K I I I, prob. ÐÞ"ÑÐÞÑÐÞÑ œ Þ!!%, total prob. of bonus payment at time 3 is.014. Actuarial present value is "!!!ÒÞ!%@ Þ!"%@ Ó œ &Þ)!. ii) The following sequences of rankings for the next three months result in two consecutive II - ratings including the current ranking of I): Sequence that pays bonus at time 1: I I, prob..2 Sequence that pay bonus at time 2: I I I, prob. ÐÞÑÐÞÑœÞ!% Sequences that pay bonus at time 3: I K I I, prob. ÐÞ'ÑÐÞ"ÑÐÞÑ œ Þ!", I I I I, prob. ÐÞÑÐÞÑÐÞÑ œ Þ!!), total prob. of bonus payment at time 3 is.0. Actuarial present value is "!!!ÒÞ@ Þ!%@ Þ!@ Ó œ &'Þ'%.

SIMULATION - PROBLEM SET 1

SIMULATION - PROBLEM SET 1 " if! Ÿ B Ÿ 1. The random variable X has probability density function 0ÐBÑ œ " $ if Ÿ B Ÿ.! otherwise Using the inverse transform method of simulation, find the random observation