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 generated by the (uniform decimal) random number r œ.716. A).358 B).608 C).716 D).858 E) 1.16 ". A binomial distribution with n œ 3 and p œ.4 is simulated by the inverse transform method with the uniform random numbers.31,.71,.66,.48,.19. How many of the generated random variables are equal to? A) 1 B) C) 3 D) 4 E) 5 3. A mixed exponential" distribution has P[ X œ 0 ] œ p (0 p 1) and density function -x f(x) œ (1 p) - e for x 0. The distribution can be simulated by the inverse transform method as follows: if U is a random uniform (0,1) value and U p, then X œ 0 ; if p Ÿ U 1 then X œ 1 1 p 1 1 p 1 U A) - ln(1 U) B) - ln(1 U) C) - ln Š 1 U D) - ln Š 1 U E) - ln Š 1p sin x 4. X has density function f(x) =, 0 Ÿ x Ÿ 1, and f(x) = 0 elsewhere. Use the inverse transform method to simulate three values of X based on uniform [0, 1] values È 1 + È3 u 1 = È, u =.5, u 3 =. Find the mean of the simulated values x 1, x, x 3. 4 191 191 171 171 51 A) 36 B) 4 C) 36 D) 4 E) 1 5. You are given a probability distribution with the following probability density function: The inverse transformation method was used to generate a random observation this distribution by using <, a uniform random number between 0 and 1. Determine the value of <. " " $ & & A) ' B) $ C) ) D) " E) ' B œ $Þ& " from

6. You are to use the inverse transform method to generate two random observations from the $ B " Ÿ B Ÿ " distribution with probability density function 0ÐBÑ œ œ.! elsewhere You are to use the following random numbers from the uniform distribution on Ò!ß "Ó:!Þ&!!& ß!Þ! Calculate the sum of the resulting random observations. A)!Þ( B)!Þ' C)!Þ& D)!Þ E)!Þ$ 7. The random variable \ has distribution function JÐBÑ. You are given: w " " w " JÐ!Ñœ!ßJÐBÑœ ) for!b, TÒ\œÓœ, JÐBÑœ for B$, JÐ$Ñœ" You simulate \ by using a function of Y, the uniform distribution on the interval Ò!ß"Ó, and $ " " ( obtain the following sequence of values from Y: ß ß ß ). Determine \, the sample mean. "* "( A) ' B) 1 C) D) ) E) 3 8. A company is insured against liability suits. The number of suits for a given year is distributed as follows: Number of suits: 0 1 3 Probability: 0.5 0. 0. 0.1 For each of three years, you simulate the number of suits per year by generating one uniform random number from the interval Ò!ß "Ó and then applying the inverse transformation method. Your three random numbers are 0.83, 0.59, 0.19. The liability loss from an individual suit is a random variable with the following cumulative! for B! distribution function JÐBÑ œ " for Bœ!. B "! for!bÿ"! You generate uniform random numbers from the interval Ò!ß "Ó and apply the inverse transformation method to simulate the amounts of the individual losses. The first nine uniform random numbers, in order, are: 0.51, 0.01, 0.78, 0.74, 0.03, 0.69, 0.17, 0.86, 0.8 Use, in order, as many of these random numbers as needed to simulate the company's total loss due to liability suits over the three year period. A) 0.0 B) 0. C) 5.6 D) 5.8 E) 14.4

9. Your company insures a risk that is modeled as a surplus process as follows: (i) Interarrival times for claims are independent and exponentially distributed with mean 1/3. (ii) Claim size equals "! >, where t equals the time the claim occurs. (iii) Initial surplus equals 5. (iv) Premium is collected continuously at rate ->. (v) 3œ! You simulate the interarrival times for the first three claims by using 0.5, 0., and 0.1, respectively, from the uniform distribution on Ò!ß "Ó, where small random numbers correspond to long interarrival times. Of the following, which is the smallest - such that your company does not become insolvent from any of these three claims? A) B) 35 C) 49 D) 113 E) 141 10. When generating random variables, it is important to consider how much time it takes to complete the process. Consider a discrete random variable \ with the following distribution: 5 " $ & T <9,Ð\ œ 5Ñ Þ"& Þ"! Þ& Þ! Þ$! Of the following algorithms, which is the most efficient way to simulate \? A) If YÞ"& set \œ" and stop. If YÞ& set \œ and stop. If YÞ&! set \œ$ and stop. If YÞ(! set \œ and stop. Otherwise set \œ& and stop. B) If YÞ$! set \œ& and stop. If YÞ&! set \œ and stop. If YÞ(& set \œ$ and stop. If YÞ)& set \œ and stop. Otherwise set \œ" and stop. C) If YÞ"! set \œ and stop. If YÞ& set \œ" and stop. If YÞ& set \œ and stop. If YÞ(! set \œ$ and stop. Otherwise set \œ& and stop. D) If YÞ$! set \œ& and stop. If YÞ&& set \œ$ and stop. If YÞ(& set \œ and stop. If YÞ*! set \œ" and stop. Otherwise set \œ and stop. E) If YÞ! set \œ and stop. If YÞ$& set \œ" and stop. If YÞ& set \œ and stop. If YÞ(& set \œ& and stop. Otherwise set \œ$ and stop.

Problems 11 and 1 relate to the following situation. A random sample _ of size 30 is taken from the distribution of the random variable \. The sample mean is \ $! œ Þ)!*, and the sample variance is W$! œ "Þ'$ Þ 10 additional random values are drawn in the following order: \ $" œ $Þ) ß \ $ œ Þ ß &Þ& ß &Þ" ß Þ& ß Þ! ß $Þ! ß Þ ß $Þ* ß &Þ* Þ 11. It is decided that sampling will stop when the estimated standard deviation of \ 5 is first less than.. How many samples values after the 30th are needed to satisfy this criterion? A) B) 4 C) 6 D) 8 E) 10 1. It is decided that sampling will stop when the width of the 95 confidence interval for. is first less than Þ(& (the 97.5 percentile of the standard normal distribution is 1.96). Ho many sample values after the 30th are needed to satisfy this criterion? A) B) 4 C) 6 D) 8 E) 10 13. An estimate is being made of the probability : of flipping a head with a particular coin. The coin will be flipped until the estimated standard deviation of the estimated value of : is less than.075. In the first 40 coin flips there are 5 heads and 15 tails. The next 10 coin flips are T H H H T H H T H H. With which flip is the stopping criterion reached? A) 41 B) 4 C) 43 D) 44 E) 45 14. To estimate IÒ\Ó, you have simulated \ß\ß\ß\ " $ and \& with the following results: 1 3 4 5 You want the standard deviation of the estimator of IÒ\Óto be less than 0.05. Estimate the total number of simulations needed. A) Less than 150 B) At least 150, but less than 400 C) At least 400, but less than 650 D) At least 650, but less than 900 E) At least 900 15. A random sample of 18 data points has a sample mean of 8 and an unbiased sample variance of 4. \"* œ & and \! œ * are added to the sample. Find the updated unbiased sample variance based on all 0 data points. A) 4.0 B) 4.1 C) 4. D) 4.3 E) 4.4 16. (CAS) A scientist perform experiments, each with a 60 success rate. Let \ represent the number of trials until the first success. Use the inverse transform method to simulate the random variable, \, and the following random numbers (where low numbers correspond to a high number of trials): 0.15, 0.6, 0.37, 0.78. Generate the total number of trials until three successes result. A) 3 B) 4 C) 5 D) 6 E) 7

17. (CAS) Two actuaries are simulating the number of automobile claims for a book of business. For the population that are studying: i) The claim frequency for each individual driver has a Poisson distribution. ii) The means of the Poisson distributions are distributed as a random variable, A. iii) A has a gamma distribution. In the first actuary's simulation, a driver is selected and one year's experience is generated. This process of selecting a driver and simulating one year is repeated R times. In the second actuary's simulation, a driver is selected an R years of experience are generated for that driver. Which of the following is/are true? I. The ratio of the number of claims the first actuary simulates to the number of claims the second actuary simulates should tend towards 1 as R tends to infinity. II. The ratio of the number of claims the first actuary simulates to the number of claims the second actuary simulates will equal 1, provided that the same uniform random numbers are used. III. When the variances of the two sequences of claim counts are compared the first actuary's sequence will have a smaller variance because more random numbers are used in computing it. A) I only B) I and II only C) I and III only D) II and II only E) None of I, II, or III is true 18. (SOA) Insurance for a city s snow removal costs covers four winter months. (i) There is a deductible of 10,000 per month. (ii) The insurer assumes that the city s monthly costs are independent and normally distributed with mean 15,000 and standard deviation,000. (iii) To simulate four months of claim costs, the insurer uses the Inverse Transform Method (where small random numbers correspond to low costs). (iv) The four numbers drawn from the uniform distribution on [0,1] are: 0.5398 0.1151 0.006 0.7881 Calculate the insurer s simulated claim cost. (A) 13,400 (B) 14,400 (C) 17,800 (D) 0,000 (E) 6,600 19. (SOA) You are simulating a continuous surplus process, where claims occur according to a Poisson process with frequency, and severity is given by a Pareto distribution with parameters α œ and ) œ "!!!. The initial surplus is 000, and the relative security loading is 0.1. Premium is collected continuously, and the process terminates if surplus is ever negative. You simulate the time between claims using the inverse transform method (where small numbers correspond to small times between claims) using the following values from the uniform distribution on [0,1]: 0.83, 0.54, 0.48, 0.14. You simulate the severities of the claims using the inverse transform method (where small numbers correspond to small claim sizes) using the following values from the uniform distribution on [0,1]: 0.89, 0.36, 0.70, 0.61. Calculate the simulated surplus at time 1. (A) 1109 (B) 1935 (C) 185 (D) 400 (E) Surplus becomes negative at some time in [0,1].

(SOA) Use the following information for Questions 0 and 1. Lucky Tom finds coins on his 60 minute walk to work at a Poisson rate of per minute. 60 of the coins are worth 1 each; 0 are worth 5 each; 0 are worth 10 each. The denominations of the coins found are independent. Two actuaries are simulating Tom s 60 minute walk to work. (i) The first actuary begins by simulating the number of coins found, using the procedure of repeatedly simulating the time until the next coin is found, until the length of the walk has been exceeded. For each coin found, he simulates its denomination, using the inverse transform algorithm. The expected number of random numbers he needs for one simulation of the walk is J. (ii) The second actuary uses the same algorithm for simulating the times between events. However, she first simulates finding coins worth 1, then simulates finding coins worth 5, then simulates finding coins worth 10, in each case simulating until the 60 minutes are exceeded. The expected number of random numbers she needs for one complete simulation of Tom s walk is KÞ 0. Which of the following statements is true? (A) Neither is a valid method for simulating the process. (B) The first actuary s method is valid; the second actuary s is not. (C) The second actuary s method is valid; the first actuary s is not. (D) Both methods are valid, but they may produce different results from the same sequence of random numbers. (E) Both methods are valid, and will produce identical results from the same sequence of random numbers. 1. Determine which of the following ranges contains the ratio JK /. (A) 0.0 J/ KŸ0.4 (B) 0.4 J/ KŸ0.8 (C) 0.8 J/ KŸ1. (D) 1. J/ KŸ1.6 (E) 1.6 J/ K. (SOA) You wish to simulate a value, ], from a two point mixture. With probability 0.3, ] is exponentially distributed with mean 0.5. With probability 0.7, ] is uniformly distributed on Ò$ß $Ó. You simulate the mixing variable where low values correspond to the exponential distribution. Then you simulate the value of ], where low random numbers correspond to low values of ]. Your uniform random numbers from Ò! ß "Ó are 0.5 and 0.69 in that order. Calculate the simulated value of Y. (A) 0.19 (B) 0.38 (C) 0.59 (D) 0.77 (E) 0.95

3. (SOA) An actuary is evaluating two methods for simulating XÐBCÑ, the future lifetime of the joint-life status of independent lives ÐBÑ and ÐCÑ: (i) Mortality for ÐBÑ and ÐCÑ follows De Moivre s law with = œ "!!., which states that > JXÐDÑÐ>Ñ œ " "!!D for! D "!! and! > "!! D. (ii)! B Ÿ C "!! (iii) Both methods select random numbers V", and V independently from the uniform distribution on Ò!ß "Ó. (iv) Method 1 sets: (a) X ÐBÑ œ Ð"!!BÑV" ß (b) X ÐCÑ œ Ð"!!CÑV ß (c) XÐBCÑ œ smaller of XÐBÑand XÐCÑ (v) Method first determines which lifetime is shorter: (a) If V" Ÿ!Þ&!, it chooses that ÐBÑ is the first to die, and sets XÐBCÑ œ XÐBÑ œ Ð"!!BÑV. (b) If R "!Þ&!, it chooses that ÐCÑis the first to die, and sets XÐBCÑ œ XÐCÑ œ Ð"!!CÑV. Which of the following is correct? (A) Method 1 is valid for BœCbut not for BC; Method is never valid. (B) Method 1 is valid for BœCbut not for BC; Method is valid for BœCbut not for BC. (C) Method 1 is valid for BœCbut not for BC; Method is valid for all Band C. (D) Method 1 is valid for all B and C; Method is never valid. (E) Method 1 is valid for all B and C; Method is valid for B œ C but not for B C. 4. (SOA) You are simulating a compound claims distribution. (i) The number of claims, R, is binomial with 7 œ $ and mean 1.8. (ii) Claim amounts are uniformly distributed on Ö"ß ß $ß ß &. (iii) Claim amounts are independent and are independent of the number of claims. (iv) You simulate the number of claims, R, and the amounts of each of those claims, \" ß \ ß ÞÞÞß \ R. Then you repeat another R, its claim amounts, and so on until you have performed the desired number of simulations. (v) When the simulated number of claims is!, you do not simulate any claims amounts. (vi) All simulations use the inverse transform method, with low random numbers corresponding to few claims or small claim amounts. (vii) Your random numbers from Ð!ß "Ñ are 0.7, 0.1, 0.3, 0.1, 0.9, 0.5, 0.5, 0.7, 0.3, and 0.1. Calculate the aggregate claim amount associated with your third simulated value of R. (A) 3 (B) 5 (C) 7 (D) 9 (E) 11 5. \ is a mixture of two exponential distributions. Distribution 1 has a mean of 1 and a mixing weight of.5 and distribution has a mean of and a mixing weight of.75. \ is simulated using the inverse transformation method with a uniform Ð!ß "Ñ value of.7. Find the simulated value of \.

SIMULATION - PROBLEM SET 1 SOLUTIONS 3 x if 0 Ÿ x Ÿ 1. F(x) = 0 if x Ÿ 0, F(x) = 1 if x 4, and F(x) = š. 1 1 3 x if Ÿ x Ÿ 4 1 r =.716 œjðbñ= x p x =.608. Answer: B. 1. s : 0 1 3 0(s).16.43.88.064 J (s).16.648.936 1.00 A binomial value of will be simulated by a uniform (0,1) value that is both greater than or equal to.648 and less than.936. Thus, the uniform numbers.71 and.66 result in simulated binomial values of. Answer: B. 3. For x 0, F(x) = P[ X Ÿ x ] = P[ X = 0 ] + P[ 0 X Ÿ x] = p + ' -t (1 p) - e dt! if B! -x = p + (1 p)(1 e ). Thus, F(x) = : if B œ! -B : Ð" :ÑÐ" / Ñ if B! Thus, for U p, the inversion method implies that -x 1 1p U = F(x) = p + (1 p)(1 e ) p x = - ln Š 1U. Answer: D.! x B 4. The c.d.f. is F(x) = ' sin t 1 cos x! dt = for 0 Ÿ x Ÿ 1. 1 1 cos x By the inversion method, for uniform u we have x = F (u), or equivalently, u =. È 1 1 cos x1 1 1 u 1= = cos x 1 = x 1 =, È Ä È Ä 4 1 u =.5 Ä cos x = 0 Ä x =, + È3 È3 51 x 1 + x + x3 191 u 3 = 4 Ä cos x 3 = Ä x 3 = 6. 3 = 36. Answer: A. 5. From the diagram of the density function we must have " " " ÐÑ ÐÑ œ " p œ ', since the total probability Ÿ\Ÿ) is 1. From the inverse transformation method, " " " & < œ J\ ÐBÑ œ J\ Ð$Þ&Ñ œ ÐÑÐ ' Ñ Ð"Þ&ÑÐ ' Ñ œ " (this is the integral of 0ÐBÑ from to 3.5). Answer: D $ B 6. J\ ÐBÑœ' B" " 0Ð>Ñ.>œ "ŸBŸ". We solve?œj\ ÐBÑ for B. $ $ B" " B" Þ&!!& œ p B" œ Þ", Þ! œ p B œ Þ). B" B œ Þ(. Answer: A

! B Ÿ!!B B 7. \ has a mixed distribution with cdf ) $ JÐBÑ œ Bœ. B" B$ " B $ According to the inversion method of simulation, given a uniform random number? from Ò!ß "Ó, the simulated value of \ is " " $ $ Bœ)? if!ÿ?, Bœ if Ÿ?Ÿ, Bœ?" if?ÿ". $ " " ( & Then,?" œ p B" œ ß? œ p B œ ß? $ œ p B$ œ ß? œ ) p B œ. The sample mean is Þ& "( œ ). Answer: D 8. The cdf of R (number of suits per year) is 8 À! " $ JÐ8Ñ À Þ& Þ( Þ* " The simulated numbers of suits in the three years are?" œ Þ)$ p 8" œ ß? œ Þ&* p 8 œ " ß? $ œ Þ"* p 8$ œ!. The liability loss \ from a suit has a mixed distribution. If?Þ& then Bœ!, and if B " Þ&Ÿ?Ÿ" then?œjðbñœ! Þ The simulated liability losses from the 3 suits are B" " B$ "?" œþ&"œ! pb" œþ ß? œþ!"pb œ! ß? $ œþ()œ! pb" œ&þ'. Total liability losses in 3 years œþ!&þ)œ&þ). Answer: D 9. The interarrival time for claims is exponential. We must simulate the three claim times. We are told that where small random numbers correspond to long interarrival times. This is the opposite of the usual inverse transform method. Instead of solving for interarrival claim time > from?œjð>ñ(the simulation method that has small random number? corresponding to small interarrival time > ), we solve for > from the equation? œ "JÐ>Ñ. Since interarrival time for " claims is exponential with mean $, "JÐ>Ñœ/$>. $> The simulated interarrival times are > ß> and >, where Þ&œ/ " " $ p > " œþ$", $> $> $ Þœ/ p> œþ&$', and Þ"œ/ p> $ œþ('). These are interarrival times, so the actual successive claim times are at times.31,.767, and Þ$" Þ('( "Þ&$& 1.535. The claim amounts are "! œ "Þ(!, "! œ &Þ)&, and "! œ $Þ). Þ$" & The total premium received up to time.31 is ' Þ$"! ->.> œ - & œ Þ!!!"$-. Þ('( & The total premium received up to time.767 is ' Þ('(! ->.> œ - & œ Þ!&$-. "Þ&$& & The total premium received up to time 1.535 is ' "Þ&$&! ->.> œ - & œ "Þ(-. Surplus at time each of the claim times is: - at time.31 surplus is & Þ!!!"$- "Þ(!!, - at time.767 surplus is & Þ!&$- "Þ( &Þ)& œ Þ!&$- Þ&&, and - at time 1.535 surplus is & "Þ(- "Þ(! &Þ)& $Þ) œ "Þ(- $'Þ)$. In order for the company to be solvent at each claim time it must be true that Þ!&$- Þ&&, and "Þ(- $'Þ)$ p - )Þ" and - "Þ(. Answer: C

10. In order to generate \ values as quickly as possible, we want the \-value with the largest probability to be the first possible one generated, and second largest probability to be the second possible one generated, etc. We place the \-values in order of largest to smallest probability: \ œ 5 & $ " : 5 Þ$! Þ& Þ! Þ"& Þ"! Cumulative Þ$! Þ&& Þ(& Þ*! " The simulation procedure with generate \œ& if YÞ$!, it would generate \œ$ if Þ$!ŸY Þ&&, it would generate \œ if Þ&&ŸY Þ(&, it would generate \œ" if Þ(&ŸY Þ*!, and it would generate \œ if Þ*!ŸY. The reason this method results in the least time required for simulation is that the probability that the stopping time occurs on the first step is higher than any of the other simulation methods. For instance, with YœÞ&$ the stopping points for each of the 5 methods given are A. Þ&! Ÿ Þ&$ Þ(! so set \ œ (stop on the 4th step) B. Þ&! Ÿ Þ&$ Þ(& so set \ œ $ (stop on the 3rd step) C. Þ& Ÿ Þ&$ Þ(! so set \ œ $ (stop on the 4th step) D. Þ$! Ÿ Þ&$ Þ&& so set \ œ $ (stop on the nd step) E. Þ& Ÿ Þ&$ Þ(& so set \ œ & (stop on the 4th step) D stops sooner than the other methods. In general, method D stops at the same time or sooner than any of the other four methods, but never later. Note that all five answers result in a valid simulation method since they reproduce the probabilities correctly. Answer: D W5 W$! "Þ'$ 11. The stopping rule is È Þ Þ 5 È œé $! $! œþ!( ÞÞ \ \ Successive sample means can be found from the relationship \ 5" œ \ 5 5" 5 5" : \ \ \ œ \ $" $! $Þ)Þ)!* ÞÞ((' $" $! $" œ Þ)!* $" œ Þ((' ß \ $ œ Þ((' $ œ Þ'*' ß &Þ&Þ'*' \ $$ œ Þ'*' $$ œ Þ(! ß \ $ œ Þ($" ß \ $& œ Þ( ß \ $' œ Þ(! ß \ $( œ Þ'&) ß \ $) œ Þ'&" ß \ $* œ Þ'$ ß \! œ Þ'' Þ Successive sample variances can be found from the relationship 5" W5" œ Ð 5 ÑW5 Ð5 "ÑÐ\ 5" \ 5Ñ : * W$" W$" œ Ð $! ÑW$! $"Ð\ $" \ $! Ñ œ "Þ& p È œ Þ!" Þ Þ $" $! W$ W$ œ Ð $" ÑW$" $Ð\ $ \ $" Ñ œ "Þ"* p È œ Þ"" Þ Þ $ W$$ W$ W$$ œ "Þ$* p È œ Þ!' Þ Þ W$ œ "Þ$&' p œ Þ"**( Þ Þ $$ È$ We see that the stopping rule is reached at the 34-th data point. Answer: B

1. The width of the 95 confidence interval is 5 È 5. From the calculations in Problem 9, we see that 5 œ $! p interval width is Ð$Þ*ÑÉ $! œ Þ)! ß 5 œ $" p interval width is Ð$Þ*ÑÉ $" œ Þ()* ß 5 œ $ p interval width is Ð$Þ*ÑÉ $ œ Þ)& ß 5 œ $$ p interval width is Ð$Þ*ÑÉ $$ œ Þ)!' ß 5 œ $ p interval width is Ð$Þ*ÑÉ $ œ Þ()$ Þ Continuing with the calculations in Problem 9, W$& œ "Þ$") p interval width is Ð$Þ*ÑÉ $& œ Þ('", W$' œ "Þ*& p interval width is Ð$Þ*ÑÉ $' œ Þ($ Þ We see that the stopping rule is reached at the 36-th data point. Answer: C 13. \ œ " if head.. if tail \ 3 š 5! œ number of heads in 5 flips 5 The stopping criterion is É \ Ð"\ 5 5 Ñ 5 Þ!(&. & After the first 40 flips, we have \! œ œ Þ'& ß so that É \! Ð"\! Ñ!! œ Þ!('& Þ & 5 œ " p \ " œ œ Þ'"! p É \" Ð"\ " Ñ " " œ Þ!(' ß ' 5 œ p \ œ œ Þ'"* p É \" Ð"\ " Ñ " œ Þ!(* Þ Answer: B 14. The estimator of IÒ\Ó is the sample mean \, which has variance Z+<Ò\Ó Z+<Ò\Óœ 8, if there are 8simulated observations. ÈZ+<Ò\Ó The standard deviation of \ is È. From the data given, the estimated variance of \ is 8 = œ " ÐB BÑ œ " & \ D 3 ÒÐÑ Ð"Ñ! " Óœ. È&Î In order for the standard deviation of \ to be less than.05 we must have È Þ!&, 8 which translates to 8 "!!!. Answer: E ")\") \ "* "* 15. The sample mean based on the first 19 points is \ "* œ "* œ "*, "*\"* \! "&) and the sample mean based on all 0 points is \! œ! œ! Þ We use the relationship 5" " W œ 5" 5" 5 Ð\ 3 \ 5" Ñ œ Ð 5 Ñ W5 Ð5 "ÑÐ\ 5" \ 5Ñ to get 3œ" "( W"* œ Ð") ÑW") Ð"*ÑÐ\ "* \") Ñ, and then ") W! œ Ð"* ÑW"* Ð!ÑÐ\! \"* Ñ "( œ Ð"* ÑW") Ð")ÑÐ\ "* \") Ñ Ð!ÑÐ\! \"* Ñ "( "* "&) "* œ Ð"* ÑÐÑ Ð")ÑÐ "* )Ñ Ð!ÑÐ! "* Ñ œ Þ!*. Answer: B

16. To say that low numbers correspond to a low number of trials is the standard form of the inverse transform method: given a uniform random number < we find the integer 8 such that J\ Ð8"ÑŸ<J\ Ð8Ñ, and the simulated value of \ is 8. If we apply the inverse transform method in the form where low numbers correspond to a high number of trials, then given a uniform random number <, we find the integer 7such that J\ Ð7"ÑŸ"<J\ Ð7Ñ. The random variable \ is the number of success until the first trial. This is a geometric distribution with probability function :ÐBÑ and distribution function J ÐBÑ as follows: \ œ 8 " $ ÞÞÞ $ :Ð8Ñ Þ' ÐÞÑÐÞ'Ñ œ Þ ÐÞÑ ÐÞ'Ñ œ Þ!*' ÐÞÑ ÐÞ'Ñ œ Þ!$) ÞÞÞ J Ð8Ñ Þ' Þ) Þ*$' Þ*( Each uniform number < simulates the number of trials until the next success. We must simulate \ three times to get \" (the simulated number of trials until the first success, \ (the additional number of trials until the second success) and \ $ (the additional number of trials until the third success). Then the total number of trials until the third success is \" \ \ $. The first random number is < " œ!þ"&, so that " < " œ!þ)&. We see that J ÐÑ œ Þ) Ÿ Þ)& Þ*$' œ J Ð$Ñ so that the simulated value of \ is 3. This is \", the simulated number of trials until the first success. The second random number is < œ!þ', with " < œ!þ$). We see that!þ$) Þ' œ J Ð"Ñ, so that the simulated value of \ is 1. < $ œ!þ$(, so that " < $ œ!þ'$. We see that J Ð"Ñ œ Þ' Ÿ Þ'$ Þ) œ J ÐÑ, so that the simulated value of \ is. Then \ \ \ œ$"œ'. Answer: D $ " $ 17. I. The combination of a Poisson claim count with mean A and a gamma distribution for A results in a negative binomial distribution being simulated by actuary 1. The average number of claims simulated by actuary 1 in R trials is R IÒAÓ. The second actuary selects a driver with Poisson parameter - and the average number of claims in R years for that driver will be R-. RIÒAÓ The ratio R- tends to 1 only if the second actuary's driver's - is equal to IÒAÓ. False II. This is false for the same reason as I. False III. For actuary 1, the variance of the sequence generated is the variance of a negative binomial distribution. For actuary, the variance of the sequence generated is the variance of the Poisson distribution with parameter - (the -for the driver chosen by actuary ). Either variance could be larger than the other depending on the value of - for actuary 's driver. False Answer: E 18. According to the inverse transform method, given uniform [0,1] number?, the simulated standard normal ^ value is D, where TÒ^ DÓ œ?. From the given uniform values, we get? Þ&$*) Þ""&" Þ!!' Þ())" D!Þ" "Þ Þ&!Þ) The loss random variable \ has a mean of 15,000 and standard deviation of 000. The simulated values of \ are!!!d "&ß!!! À "&ß!! "ß '!! "!ß!!! "'ß '!!. After monthly deductibles, the insurer pays &ß!! ß '!!! 'ß '!! in the four winter months, for total claim cost of 14,400. Answer: B

" 19. The time between successive claims has an exponential distribution with mean. > The simulated inter-claim times are > where? œ JÐ> Ñ œ "/ 3 3 3 3, or equivalently, " > 3 œ 68Ð"? 3Ñ. We use the relationship? œ JÐ>Ñ because we are told that "small numbers correspond to small times between claims"; this is the standard form of the inverse transform method. The simulated inter-claim times are " " > " œ 68Ð" Þ)$Ñ œ Þ)*, > œ 68Ð" Þ&Ñ œ Þ$*. The second simulated claim occurs after time 1, so it is irrelevant. The simulated claim amount of the first claim is B, where ) α? œ J ÐBÑ œ " Ð Ñ œ " Ð "!!! B) B"!!! Ñ, or equivalently, B œ "!!! É " "? "!!!. The simulated claim amount is B œ "!!! É " "Þ)* "!!! œ!"&. ) "!!! The expected claim amount per claim is α" œ " œ "!!!, so that the expected aggregate claim per period is ÐÑÐ"!!!Ñ œ!!!. With relative security loading of.1, the premium rate is Ð"Þ"ÑÐ!!!Ñ œ!!. With initial surplus of 000, the simulated surplus at time 1 is!!!!!!"& œ ")&. Answer: C 0. D. 1. Actuary 1: Average number of times that must be simulated in one hour is Ð'!ÑÐÑ œ "!. For each one of those times, a simulation of coin denomination must be made. Average number of simulations in total is! œ J. Actuary : Denomination 1 coins are found at the rate of ÐÞ'ÑÐÑ œ "Þ per minute, so that in one hour, an average of Ð'!ÑÐ"ÞÑ œ ( times must be simulated. For denomination 5, coins are found at rate ÐÞÑÐÑ œ Þ per minute, and an average of Ð'!ÑÐÞÑ œ must be simulated. Similarly, for denomination 10, an average of 4 simulations are needed per hour. Average total is ( œ "! œ K. J ÎK œ. Answer: E. The mixing probability for the exponential distribution is.3. We are told that low values correspond to the exponential distribution when simulating the mixing variable. Therefore the exponential distribution is used if the uniform random number for simulating the mixing variable is ŸÞ$. Since the first uniform random number is Þ&ŸÞ$, we use the exponential distribution for ]. We are also told that low random numbers correspond to low values of ]. This just means that we are applying the usual inverse transformation method. The cdf of the exponential distribution with mean.5 is JÐCÑ œ "/ C. The second uniform random number is.69, so that the simulated value of ] is the value of C C which satisfies " / œ Þ'* p C œ Þ&)'. Answer: C

3. Under DeMovire's Law, the future lifetime of someone at age D is uniformly distributed on the interval = D. Therefore Method 1 is a valid simulation of both times of death. The two individuals are independent, so if BœCthen there is an equal chance that either is first to die. However, given that B is first to die, the conditional distribution of B's time of death is no longer uniform (same for C). Therefore Method is never valid. Answer: D 4. The distribution of R binomial with mean $; œ "Þ), so that ; œ Þ' is the probability of a claim. Then the distribution of number of claims is T ÒR œ 5Ó œ Š 5 $5 5 ÐÞ'Ñ ÐÞÑ R! " $ : Þ!' Þ)) Þ$ Þ"' J Þ!' Þ$& Þ() "Þ! The first uniform number.7 simulates claims, since Þ$& Þ( Ÿ Þ(). We then use.1 and.3 to simulate claim amounts. The next uniform number is.1 and is used to simulated the second R. This will be 1, since Þ!' Þ" Ÿ Þ$&. We use.9 to simulate the one claim amount. The next uniform number is.5, which simulates claims. The uniform numbers.5 and.7 simulate the claim amounts. The claim amount distribution is \ " $ & : Þ Þ Þ Þ Þ J Þ Þ Þ' Þ)Þ "Þ! The uniform number.5 simulates a claim amount of 3, and the uniform number.7 simulates a claim amount of 4. Aggregate claim amount simulated associated with the 3rd simulated value of R is 7. Answer: C B 5. J ÐBÑ œ ÐÞ&ÑÐ" / Ñ ÐÞ(&ÑÐ" / Ñ. B BÎ We must solve for B from the equation ÐÞ&ÑÐ" / Ñ ÐÞ(&ÑÐ" / Ñ œ Þ(. B BÎ The equation can be written as Þ&/ Þ(&/ Þ$ œ!. BÎ Substituting Cœ/ results in the quadratic equation C $C"Þœ!. $ È $ Ð"ÞÑ Solving for C we get C œ œ Þ$&( ß $Þ$&(. BÎ We ignore the negative root, since Cœ/ must be!. Solving for B, we get B œ 68ÐÞ$&(Ñ œ Þ!'. BÎ