ST4020 1. A random variable X has probability density function fX(x) = ↵e x ↵ (+ ex)↵+1 for 1 < x <1 where ↵, and are positive real numbers. (a) Find the corresponding cumulative distribution function FX(x). [3] (b) Suggest how to generate random variates from the density fX(x). [2] [Total 5] 2. A probability distribution of a continuous random variable Y is said to belong to the expo- nential family if its density has the form f(y; ✓,) = exp  y✓ b(✓) a() + c(y,) where a, b and c are functions. (a) Sketch the proof that E  @` @✓ (Y ) = E " @2` @✓2 (Y ) + ✓ @` @✓ (Y ) ◆2# = 0 where ` is the log-likelihood function. In your proof you do not need to justify the exchange of derivatives and expectation symbols. [4] (b) Show that the mean of Y is b0(✓). [2] (c) Derive the variance of Y in terms of ✓, , a and b. [2] [Total 8] Continued. . . 2 ST4020 3. Consider the collective risk model S = PN i=1Xi where the claim amounts Xi > 0 are independent and identically distributed, and are independent of the number of claims N . (a) Define the cumulant generating function KS(t) in terms of the moment generating function MS(t) of S. Show that d2 dt2 KS(t) t=0 = Var(S) . [2] (b) Assuming that N ⇠ Bin(n, p), for some n 1 and 0 < p < 1, compute the third cumulant and the coecient of skewness of S. Is S always positively skewed? HINT: if N ⇠ Bin(n, p), then MN (t) = (pet + q)n, where q = 1 p. [4] [Total 6] 4. Let ⇤ be a random variable having the Pareto distribution with parameters ↵ = 4 and = 10. The probability density function of the distribution of ⇤ is given by f⇤(y) = ↵↵ (+ y)↵+1 , y > 0, with mean E[⇤] = /(↵ 1). Assume that the claim amount on an insurance policy is X = ⇤+ 12. Excess of loss reinsurance is purchased, with retention level 16. (a) Find the probability that the reinsurer is involved in the claim. [2] (b) Find the survivor function of the amount Y paid by the reinsurer given that it is involved in the claim. [3] (c) Find the mean amount paid out by the reinsurance company given that it is involved in the claim. [2] (d) Find the mean claim amount paid by the insurance company. [3] [Total 10] Continued. . . 3 ST4020 5. Consider the AR(2) process defined by Xt + 1 6 Xt1 1 3 Xt2 = 1 8 + et , where {et : t 2 Z} is a white noise with E[e21] = 1. (a) By considering the characteristic polynomial of this autoregression, or otherwise, show that X is stationary. [2] (b) State the Yule Walker equations for the above AR(2) model. By solving them find Var(Xt) for t 0. [6] [Total 8] 6. Based on the proposal form, an applicant for life insurance is classified as standard life (1), an impaired life (2) or uninsurable (3). The proposed form is not a perfect classifier and may place the applicant into the wrong category. The decision to place the applicant in state i is denoted by di, and the correct state for the applicant is ✓i. The loss function for this decision is given below. Correct state Decision of applicant d1 d2 d3 ✓1 0 6 9 ✓2 13 0 4 ✓3 21 16 0 (a) Determine the minimax solution when assigning an applicant to a category. [1] (b) Based on the application form, the correct category for a new applicant appears to be as impaired life. However, of applicants which appear to be impaired lives, 16% are in fact standard lives and 24% are uninsurable. Determine the Bayes solution for this applicant. [4] [Total 5] Continued. . . 4 ST4020 7. A portfolio of an insurance company consists of one risk. Assume that the aggregate claims process follows a compound Poisson process where the Poisson parameter equals while the claims amounts X1, X2, ... are independent and exponentially distributed with parameter µ. Furthermore, we assume that the rate of premium per unit of time is c and the initial surplus is U units where one unit of money represents one pound. (a) State the definition of the probability of ultimate ruin. [2] (b) Under which conditions does there exist an adjustment coecient? Compute the ad- justment coecient for the portfolio when this exists. [4] (c) Assume that = 20, µ = 3, c = 50 and U = 60. Find an upper bound for the probability of ultimate ruin. [3] [Total 9] 8. The delay triangles given below relate to a certain portfolio of insurance policies, for which it may be assumed that the claims are fully run o↵ by the end of development year 2. The number of reported claims is shown in the table below: Development year Accident year 0 1 2 2015 442 151 50 2016 623 111 2017 681 The cost of settled claims during each year (in 0000s) is given in the table below: Development year Accident year 0 1 2 2015 6321 1901 701 2016 7012 2237 2017 7278 Estimate the outstanding claims reserve using the average cost per claim method with grossing up factors. Write down the details of your calculations including the accumulated claim numbers, the accumulated claim costs, the grossing-up factors for the claim numbers and the accumulated claim costs, and the average costs per claim tables. [11] Continued. . . 5 ST4020 9. The table below shows aggregate annual claim statistics for three risks over a period of seven years. Annual aggregate claims for risk i in year j is Xij . Risk i Xi S2i i=1 127.9 335.1 i=2 88.9 65.1 i=3 149.7 33.9 where Xi = 1 7 7X j=1 Xij and S2i = 1 6 7X j=1 (Xij Xi)2. (a) Calculate the credibility premium of each risk under the assumptions of EBCT model 1. [6] (b) Explain why the credibility factor is relatively high in this case. [2] [Total 8] TOTAL: [70] END. 6 ST402 Risk Theory 2019 Exam: Outline solutions 1. (a) [unseen with this distribution] We have FX(x) = Z x 1 ↵ey ↵ (+ ey)↵+1 dy = Z ex 0 ↵ ↵ (+ z)↵+1 dz (putting z = ey, dz = eydy) = 1 ↵ (+ ex)↵ . [3] (b) [unseen with this distribution] We can generate variates by inversion. The inverse of the cumulative function is 1 log (1 x)1/↵ 1. Hence, given a Uniform(0,1) random variate U , we set X = 1 log U1/↵ 1. [2] [Total 5] 2. (a) [seen] We have E  @ @✓ l(Y ) = Z @ @✓ f(y; ✓,)dy = @ @✓ Z f(y; ✓,)dy = 0. [1] We also have @2 @✓2 l + ✓ @` @✓ ◆2 = @2 @✓2 f f . Hence, E " @2` @✓2 (Y ) + ✓ @` @✓ ◆2 (Y ) # = Z @2 @✓2 f(y; ✓,)dy = @2 @✓2 1 = 0. [3] (b) [seen] Using Q2(a), we have E[ @ @✓ l(✓,;Y )] = E  Y b0(✓) a() = E[Y ] b0(✓) a() = 0. This gives E[Y ] = b0(✓). [2] (c) [seen] Using again Q2(a), we obtain E h b00(✓)a() + (Yb 0(✓)2) a2() i = 0 which gives Var(Y ) = a()b00(✓). [2] [Total 8] 3. (a) [first part bookwork proof of second part unseen] KS(t) = logMS(t), where MS(t) is the moment generating function of S. d2 dt2 KS(t) = d2 dt2 logMS(t) = MS(t)M 00S(t) [M 0S(t)]2 [MS(t)]2 . 1 Since MS(0) = 1, we therefore have d2 dt2 KS(t) t=0 = M 00S(0) [M 0S(0)]2 = Var(S) . [2] (b) [proof unseen] Using the hint, and the fact that MS(t) = MN (logMX(t)): d3 dt3 KS(t) = n d3 dt3 log(pMX(t) + q) = npM 000 X (t)(pMX(t) + q) 1 3np2M 00X(t)(pMX(t) + q)2M 0 X(t) + 2n(pM 0 X(t)) 3(pMX(t) + q) 3. [2] Hence, E ⇥ (S E[S])3⇤ = d3 dt3 KS(t)|t=0 = npm3 3np2m1m2 + 2np3m31. The coecient of skewness is (npm3 3np2m1m2 + 2np3m31)/(npm2 np2m21)3/2. S can be positive or negative skewed. [2] [Total 6] 2 ST402 Risk Theory 2020 Exam: Outline solutions 4. (a) [seen for di↵erent distribution] p = Pr(X > M) = Pr(⇤ > M 12) = ✓ +M 12 ◆↵ = ✓ + 4 ◆↵ = ✓ 10 14 ◆4 = (0.714)4 = 0.2599. [2] (b) [seen for di↵erent distribution] Survivor function of payment Y by reinsurance com- pany, conditional on Y > 0, is Pr(X > M + y |X > M) = Pr(⇤ > M + y 12|⇤ > M 12) = ✓ +M + y 12 ◆↵✓ +M 12 ◆↵ = ✓ +M 12 +M + y 12 ◆↵ . Hence Y ⇠ Pareto(↵, 14). [3] (c) Y |Y > 0 ⇠ Pareto(↵, +M 12) = Pareto(4, 14) which implies that E[Y |Y > 0] = (+M 12)/(↵ 1) = 14/3 = 4.66. [2] (d) [seen for di↵erent distribution] EX = E[⇤+12] = 12+/(↵1) = 12+10/3 = 15.3333 It follows that, if µ is mean claim paid by insurer, then µ = EX pEY = 15.3333 0.2599⇥ 4.6666 = 14.12. [3] [Total 10] 3 ST402 Risk Theory 2020 Exam: Outline solutions 5. (a) [relevant theorem given in lectures, but example of application completely unseen] The characteristic polynomial is given by 1 + z 6 z 2 3 = 1 3 (2 z) ✓ 3 2 + z ◆ . The roots of this equation (-3/2 and 2) both have absolute value greater than 1, and so by Result 3.1 of chapter 9 in the lectures, X is stationary. [2] (b) [unseen for this autoregression] Define k = Cov(Xt, Xtk). The Yule-Walker equations are 0 = 1 6 1 + 1 3 2 + 2 1 = 1 6 0 + 1 3 1 2 = 1 6 1 + 1 3 0 k = 0 for k 3. [3] The second and third of these equations yield 1 = 1 4 0 and 2 = 9 24 0 . Putting these values into the first equation shows that 0 = 1 6 0 + 2 , and so Var(Xt) = 0 = 6 5 2 = 6 5 . [3] [Total 8] 4 ST402 Risk Theory 2020 Exam: Outline solutions 7. (a) [bookwork ]The probability of ultimate ruin is give by (u) = P (U(t) < 0 for some t < 1|U(0) = u) where (U(t), t 0) is the surplus process and u is the initial cashflow, i.e., the probability that the surplus goes negative. [2] (b) [first part is bookwork, details of second part unseen] The adjustment coecient is the unique positive solution to µµr cr = 0 . This is found to be R = µ /c. [2] The condition for the existence of an adjustment coecient is cµ > . [2] (c) [unseen] We use Lundberg’s inequality which states that (U)  exp{RU} where (U), R and U are, respectively, the probability of ultimate ruin, the adjustment coecient and the initial surplus. We get (u)  eRU = e60⇥(3 2050 ) = e60⇥2.6 = e156. [3] [Total 9] 6 ST402 Risk Theory 2020 Exam: Outline solutions 9. (a) [similar to seen exercise] The overall mean is X = (127.9 + 88.9 + 149.7)/3 = 122.167. We also have E[s2(✓)] = 1 3 3X i=1 0@1 6 7X j=1 (Xij Xi)2 1A = 1 3 (335.1 + 65.1 + 33.9) = 144.7. We also have V ar(m(✓)) = 1 2 X i=12 (Xi X)2 = 1 7 E[S2(✓)] = (127.9 122.1)2/2 (88.9 122.1)2/2 + (149.7 122.1)2/2 144.7/7 = 928.14. [3] So the credibility factor Z is 7/ (7 + 144.7/928.14) = 0.978213. The credibility premia for the risks are: For risk 1: 0.978213⇥ 127.9 + (1 0.9782213)⇥ 122.167 = 127.8; For risk 2: 0.978213⇥ 88.9 + (1 0.9782213)⇥ 122.67 = 89.6; For risk 3: 0.978213⇥ 149.7 + (1 0.9782213)⇥ 122.67 = 149.1. [3] (b) [unseen] The date show that the variation within risks is relatively low as S2i are low particularly for risks 2 and 3. Hoever, there seems to be a high variation between the average claims on the risks. The variation cannot be explained just by the variability in the claims and must be due to variability in the underlying parameter. So we cannot put relatively little weight on the information provided by the data set as a whole. We must put more weight on the data from the individual risks leading to a relatively high credibility factor. [2] [Total 8] TOTAL: [70] 9

























































































































































































































































































































































































































































































































































学霸联盟