11 COLLECTED PROBLEMS Do the following problems for coursework 1. Problems 11.4 and 11.5 constitute one exercise leading you through the basic ruin arguments. 2. Problems 11.1 through to 11.13 but excluding problem 11.11 constitute one exercise leading you through the age replacement problems. 3. Problem 11.18 to understand the Cramer-Lundberg model The problems below relate to each section of the course and are cross referenced to the notes. 11.1 (The Gamma & Erlang Distribution. See: 4.6). The Erlang distributions arise as the distributions of sums of independent and identically distributed exponential variates. If X 1, X 2,..., X k are k independent and identically distributed exponential variates with density the distribution of the sum f X (x) = λe λx, T k = X 1 + X 2 +... + X k is Erlang k (Gamma with integer parameter k) with density Obtain the distribution by an inductive argument. f(x k, λ) = λ (λx)k 1 e λt, x, k, λ > Γ (k) 1. Show that the sum of two independent random variables, Z = X + Y, X F X (x), Y F Y (y), with densities f X (x) and f Y (y) has density f Z (z) = z f X (z u)f X (u)du = z f X (u)f X (z u)du 2. Write the exponential distribution for a single RV with parameter λ as f 1 (x λ) = λe λx 3. Use the result of (1) to show that the sum of two iid exponential variates with parameter λ is gamma 2, namely 4. Write f 2 (x λ) = λ (λx) e λx Γ (2) = λ 2 xe λx f k (x λ) = λ (λx)k 1 e λx Γ (k) and set up an inductive argument by considering T k = T k 1 + X k. 11.2 (The Gamma & Erlang Distribution Again. See: 1.2). The results above can be obtained using probability generating functions.
88 11 COLLECTED PROBLEMS 1. Show that the moment generating function, E [e zx ], for the exponential density f X (x) = λe λx is M X (z) = λ λ z, z < λ 2. Use the result that the moment generating of T k is M Tk (z) = M k X (z) 3. Calculate directly the moment generating function of f k (x λ assuming z < λ. Hint: use the change of variables u = (λ z)x and recall the definition of the gamma function. 4. Use the uniqueness of generating functions to complete the derivation. 11.3 (A Ruin Problem. See: 3.4). You are in control of a fund whose day to day value is a random amount Y. The contingency is also random of amount X, independent of Y. Given that the random variables have distributions and densities F X, f x, F Y, and f Y respectively. Write down the probability that the claim will exceed the amount in the fund. Given that the densities are, y < a 1 f Y (y) = a, a x 2a, 2a < y < f X (x) = 1 µ e x µ, x < evaluate the probability. 11.4 (A Ruin Problem - continuation. See: 3.6). Suppose instead that the quantities are independently distributed. A long term study has revealed the following data for the fund values and the claims. The analyst has computed the empirical distributions. Use this data to estimate the ruin probability. X ˆF X ˆFY 2.38. 4.228. 6.912.13 8.2525.228 1.5.1587 12.7475.5 14.988.8413 16.9772.9772 18.9962.9987 2.9996 1. 1. 1. Table 11.1. historical data The true value obtained analytically is.2989. 11.5 (A Ruin Problem - continuation. See: 3.7.2). Suppose instead that the quantities are normally distributed, Y N ( µ Y, σ 2 Y )) and X N ( µ X, σ 2 X )), σ Y, µ X, and σ X are fixed, but assume also that the variables are correlate with correlation ρ XY. Write down the ruin probability. By calling on investors you can determine the mean level in the fund. On the basis of past history µ X = 5, σ X = 1.5, σ Y =.5 and ρ XY =.3. Find the value of µ Y that will ensure that the probability of ruin is 5%. 11.6 (Optimal Level of Cover. See: 3.1). An insurer sells insurance against damage. The policy provides cover against damage above an amount d, the actual level of damage is random variable D. If the actual damage exceeds d the costs that fall to the insurer are proportional to the amount by which D exceeds d, namely C (D d). The cost of the policy is proportional to the level chosen, C 1 d. The actual damage has distribution P [D x] = F D (x) with survival function R D (x) = 1 F D (x) and E [D] = µ D. 1. What is the probability of a claim? 2. Explain why the revenue received by the insurer when protection level d is chosen can be written as where 1 {A} = { 1, x A, x / A is the indicator function C (D d)1 {D>d} C 1 d.
11 COLLECTED PROBLEMS 89 3. Derive the expected cost of this choice. 4. Show how to derive an optimum choice of protection d. 5. If the density f D (x) = x 8, x 4, what is the optimum level of protection which maximizes the return to the insurer? 11.7 (Poisson processes. See: 6.2). Claims arrive at an insurance office according to a Poisson process at rate λ = 1 per hour. 1. Given that 7 claims came in between 1 and 11 AM,what is the probability that no claims come in during the 1 minute interval 1 to 1 :1 PM? 2. What is the variance of the number of claims that come in during a 2-hour period? 11.8 (Poisson). Payments arrive according to a Poisson process at a rate of λ 1 = 3 per day. Independently, claims arrive according to a Poisson process at rate λ 2 = 1 per day. 1. Show that the times between the arrivals of claims are exponentially distributed with F X (x λ 2 ) = 1 exp ( λ 2 x) and by extension that the time between payments are exponentially distributed with F Y (y λ 1 )1 exp ( λ 2 y). [Hint: think about the probability of at least one event]. 2. What is the probability that the first to arrive is a claim? Derive this by defining X as the time of arrival of a claim and Y the time of arrival of a payment, derive the probability P [X < Y ]. Hint: see 3.4.2. 3. What is the expected time to the arrival of the first event? Hint: there are two ways, first, if each process is Poisson, what is the superposed process followed by both types of events, payments and claims; secondly, take X and Y as defined in part 2 and derive the distribution of min{x, Y } by calculating P [min{x, Y }] > t. 4. What is the probability that the first 3 to arrive are all payments, but the 4th is a claim? What is the return period of claims (see 3.2). Hint: use the approach of part 2 to derive p = P [Y X] and use the exponential lack of memory to note that each arrival can be regarded as restarting the process. 5. If we do not distinguish between events, what is mean and variance of the time at which the n-th event occurs? Hint: the time of the n-th event is at the sum of n inter-arrival times, the Poisson inter-arrival times are exponential iid. 11.9 (Investment. See 6.2). You put a call to tender out which closes R months in the future. The offers arrive at times {t n n 1} in a Poisson stream with rate µ such that R > 1/µ. Your objective is to accept the last offer before time R. This strategy is problematic because if you were to choose the offer at t n < R another offer might arrive at time t n+1 < R. You choose the following policy: fix a time τ [, R],and accept the first offer (if any) received after time τ but before time R. You succeed in your objective if and only if exactly ONE such offer occurs; otherwise you fail. 1. For the chosen fixed τ what is the probability that you succeed? 2. What value of τ maximizes your probability of success,and what is this maximal probability? 11.1 (Capital Asset Replacement. See: 6.5). You have to replace a piece of capital equipment regularly. The policy you choose is to fix a time τ and replace the equipment after it has been in service for τ units of time. If the equipment fails before τ it is replaced, but you incur extra costs of lost service and delivery costs. your objective is to determine a value of τ that minimizes the average cost of ownership. Assume the lifetime distribution of the equipment is F T (t) with survival function R T (t) = 1 F T (t), the cost of a planned replacement is C and the cost of an unplanned replacement is C + C 1. 1. Explain why this process is a renewal process. 2. Show that the length of a cycle is described by the RV L where L = t1 {t τ} + τ1 {t>τ} τ 3. Show that the expected length of the cycle is E [L] = R T (u)du (Hint: integration by parts). 4. Calculate the expected cost of one cycle. 5. State how the renewal-reward theorems allow you to calculate the average cost of ownership using the two previous results. 6. Explain how to determine an optimal policy, τ, which minimizes the average cost. 11.11 (Replacement continuation). Show that if F T (t) = 1 e µt (e.g.,exponential lifetimes),then τ = is the optimal value of t. In other words, you would never get rid of a used car that is still working. Explain intuitively why this makes sense (HINT:memoryless property).
9 11 COLLECTED PROBLEMS 11.12 (replacement continuation). Now suppose that F T is the uniform distribution over the interval (2, 8) years; { (t 2)/6, t (2, 8) ; F T (t) =, t < 2. Suppose further (in units of ten thousand) that C = 2 and C 1 = 4/1. In view of the domain of F T restrict your attention to the case 2 < τ < 8. Find the optimal value of τ. 11.13 (Replacement continuation). In problem 11.1 the equipment had no residual value. Suppose now that there is a residual value V (t) when the replacement is planned; the residual value is zero if the equipment is replaced on failure. Write down the expected cost taking account of the residual value. What is now the optimum policy τ? Assume that V (t) = C (8 τ)/8, use the information from question 11.12 to determine an optimal policy τ. 11.14 (Compound Processes. See 7..1 & 7.2). Claims arrive in a Poisson stream with rate λ. The size of the claims is X F X. The fund fails if the claims exceed a critical value c. Show that the time to failure is exponentially distributed F T (t) = 1 exp { θt} where θ = (1 F X (c)) λ 11.15 (Compound Processes. See 7.4). Consider a simplified version of the Cramer-Lundberg model in which the claims are all of equal size, µ, thus not random; the premiums are paid at a rate c per unit time. The number of events in [, t] is N t. 1. Write down the value U t of the fund at time t. 2. Assuming that the arrivals are Poisson with rate λ, write down P [N t = n] and P [N t n]. Determine an expression for the probability of ruin at time t. 3. Given µ = 2, λ = 1, determine the probability of ruin at time 1 for a safety loading ρ when ρ = 1 and when ρ = 1.2. Some partial Poisson tables have been provided for your help. 11.16 (Compound Processes. See 7.2). A compound Poisson process is generated by a Poisson process with rate λ and independent random variables X F X (x). N t S t = Show that the process S t when viewed at the sequence of arrival intervals is Markov. 11.17 (Compound Processes. See 7.4). Consider the Cramer-Lundberg model with initial capital u, iid claims X and Poisson arrivals with rate λ. Examine the process only at the arrival instants T k = k Y i. At these instants the fund value is X i { k } U k = u + ct k S k = u + c Y i S k = u + k (cy i X i ) 1. Show that U k is a Markov process. 2. Use the result of part 1 to show that at the time of the first event at T 1 = Y 1 with claim X 1 the future process is a copy of the process starting with initial capital u + cy 1 X 1. 11.18 (Compound Processes. See 7.4). Consider the Cramer-Lundberg model with initial capital u. The claim size X is discrete with values in the range to m ( ) m P [X = x] = θ x (1 θ) m x x and the arrivals are Poisson with rate λ. 1. Derive the probability generating function of the Poisson probability distribution with mean λt.
2. Derive the probability generating function of the binomial distribution of claim sizes. 3. Use lemma 1.1 to obtain the probability generating function of the claims process N t S t = X i. 4. Given m = 6, θ = 1/2, λ = 4, Show that the generating function reduces to 5. Show that and that i= t 4 t+ G St (z) = e 16 (1+z)6 d dz G S t (z) = 3 8 (1 + z)5 t G St (z) G St () = e 63 16 t. 11 COLLECTED PROBLEMS 91 Use these results to find P [S t = ], P [S t = 1], P [S t = 2], P [S t = 3]. 6. If premiums accrue to the fund at the rate c = 13 per unit time and initial capital is zero, indicate how you would use the generating function to calculate the ruin probability at time t. tabulation of P [N = n] n Poisson mean 7 8 9 1 11 1.64.27.11.5.2 2.223.17.5.23.1 3.521.286.15.76.37 4.912.573.337.189.12 5.1277.916.67.378.224 6.149.1221.911.631.411 7.149.1396.1171.91.646 8.134.1396.1318.1126.888 9.114.1241.1318.1251.185 1.71.993.1186.1251.1194 11.452.722.97.1137.1194 12.263.481.728.948.194 13.142.296.54.729.926 14.71.169.324.521.728 15.33.9.194.347.534 16.14.45.19.217.367 17.6.21.58.128.237 18.2.9.29.71.145 19.1.4.14.37.84 2..2.6.19.46 21..1.3.9.24 22...1.4.12 23....2.6 24....1.3 25.....1 26..... 27..... 28..... 29..... 3..... Table 11.2. Poisson table tabulation of P [N n] n Poisson mean 7 8 9 1 11 1.73.3.12.5.2 2.296.138.62.28.12 3.818.424.212.13.49 4.173.996.55.293.151 5.37.1912.1157.671.375 6.4497.3134.268.131.786 7.5987.453.3239.222.1432 8.7291.5925.4557.3328.232 9.835.7166.5874.4579.345 1.915.8159.76.583.4599 11.9467.8881.83.6968.5793 12.973.9362.8758.7916.6887 13.9872.9658.9261.8645.7813 14.9943.9827.9585.9165.854 15.9976.9918.978.9513.974 16.999.9963.9889.973.9441 17.9996.9984.9947.9857.9678 18.9999.9993.9976.9928.9823 19 1..9997.9989.9965.997 2 1..9999.9996.9984.9953 21 1. 1..9998.9993.9977 22 1. 1..9999.9997.999 23 1. 1. 1..9999.9995 24 1. 1. 1. 1..9998 25 1. 1. 1. 1..9999 26 1. 1. 1. 1. 1. 27 1. 1. 1. 1. 1. 28 1. 1. 1. 1. 1. 29 1. 1. 1. 1. 1. 3 1. 1. 1. 1. 1. Table 11.3. Cumulative Poisson Table