UNSW ACTL1101 Final Exam 2020-11-24 Exam ID 00001 Name: Student ID: Signature: 1. (a) (b) X (c) (d) X (e) 2. (a) (b) (c) (d) X (e) 3. (a) (b) (c) (d) X (e) 4. (a) (b) (c) X (d) (e) 5. (a) (b) (c) X (d) (e) 6. (a) X (b) (c) (d) (e) 7. (a) X (b) (c) (d) (e) X 8. (a) (b) (c) (d) (e) X 9. (a) (b) X (c) (d) (e) X 10. (a) (b) (c) (d) X (e) X 11. (a) (b) (c) (d) X (e) 12. (a) (b) (c) (d) (e) X ACTL1101: 00001 2 13. (a) (b) (c) (d) X (e) 14. (a) (b) (c) (d) (e) X 15. (a) (b) (c) (d) (e) X 16. (a) (b) (c) X (d) (e) 17. (a) (b) (c) X (d) (e) 18. (a) (b) (c) (d) X (e) 19. (a) (b) X (c) (d) (e) 20. (a) X (b) (c) (d) (e) 21. (a) (b) (c) X (d) (e) 22. (a) (b) (c) (d) X (e) 23. (a) (b) (c) (d) X (e) 24. (a) (b) (c) (d) X (e) 25. (a) (b) (c) (d) X (e) 26. (a) X (b) (c) (d) (e) 27. (a) X (b) (c) (d) (e) 28. (a) (b) (c) X (d) (e) 29. (a) X (b) X (c) (d) X (e) 30. (a) X (b) (c) (d) (e) 31. 1 . 0 0 0 32. 1 . 0 0 0 33. 1 . 0 0 0 34. 1 . 0 0 0 ACTL1101: 00001 3 1. Problem Let X and Y be two independent Geometric(p) random variables with the same parameter (0 < p < 1). Which of the following statement(s) is(are) necessarily TRUE (select all that applies) (a) X + Y is a Geometric random variable (b) minimum(X,Y ) is a Geometric random variable (c) maximum(X,Y ) is a Binomial random variable (d) X + Y is a Negative Binomial random variable (e) minimum(X,Y ) is a Binomial random variable Solution (a) False (b) True (c) False (d) True (e) False 2. Problem Among all Americans voters, 6% are Asian-American, 48% voted for Trump, and 2.2% are both Asian-American and voted for Trump. If you pick an American voter at random, what is the probability they are neither Asian-American, nor voted for Trump [The choices are expressed in decimal values, i.e. 0.05 ≡ 5%.] (a) 0.54 (b) 0.59 (c) 0.65 (d) 0.48 (e) 0.73 Solution Denoting the event ‘A: Asian’ and ‘T : Trump’. We seek Pr[(A ∪B)C ] = 1 Pr[A ∪B] = 1 (0.48 + 0.06 0.022) ≈ 0.482. (a) False (b) False (c) False (d) True (e) False 3. Problem After their tutorial, 9 hungry students descend on the Quad Foodcourt. Three of those students grab seats at a table, and the remaining 6 form a line to get chicken schnitzels. How many different queuing lines are possible, in total [Every position in the line is distinct, and you don’t know which students sat down.] (a) 504 (b) 55740 ACTL1101: 00001 4 (c) 84 (d) 60480 (e) 38202 Solution There are n = 9 students, and 6 will get in line (but we don’t know which ones), so the number of possible lines is the number of permutations of 6 objects within 9 objects in total, i.e. n · (n 1) · (. . .) · 4. (a) False (b) False (c) False (d) True (e) False 4. Problem In the course ACTN1001, it is known that 17% of students plagiarise their Python code as- signment. If a student did not plagiarise, the chance they have the same code as someone else is 0.1%, while if a student did plagiarise, that chance is 100%. Given that a student has the same code has someone else, what is the probability he plagiarised [The choices are expressed in decimal values, i.e. 0.9900 ≡ 99.00%.] (a) 0.9860 (b) 1.0000 (c) 0.9951 (d) 0.9615 (e) 0.9990 Solution This can be solved by direct application of Bayes theorem. Call ‘P’ the event ‘the student plagiarised’ and ‘S’ the event ‘the student has the same solution as someone else’. We seek to find Pr[P|S]. For this we need Pr[S] = Pr[S|P ]Pr[P ] + Pr[S|P c]P [P c] = 0.17083 From there, we get Pr[P |S] = Pr[S|P ] · Pr[P ] Pr[S] = 0.17 0.17083 = 0.9951414 (a) False (b) False (c) True (d) False (e) False 5. Problem Find E[X] for X a discrete random variable with probability mass function p(x) = ( 1 2 )x+1 , for x = 0, 1, 2, 3, . . . . (a) 0.50 ACTL1101: 00001 5 (b) 1.41 (c) 1.00 (d) 1.63 (e) 2.00 Solution If Y is a Geometric(1/2), we have X = Y 1, so E[X] = 2 1 = 1. (a) False (b) False (c) True (d) False (e) False 6. Problem The annual effective yield (interest rate) I on your latest investment is random (and could be negative), and follows a Normal distribution with mean μ = 0.06 (i.e. 6%) and standard deviation σ = 0.1. What is the probability that, by the end of the year, you turn a profit on this investment (a) 0.73 (b) 0.82 (c) 0.88 (d) 0.50 (e) 0.27 Solution We want Pr[I > 0] where I is the random yield (interest). Hence: Pr[I > 0] = Pr[ I μ σ > μ σ ] = Pr[Z > 0.06 0.1 ] = 0.7257469. (a) True (b) False (c) False (d) False (e) False 7. Problem Let X1, X2, . . . , Xn be a random sample of size n, where all Xi’s have the same distribution with an unknown parameter θ and corresponding estimator θ . In this setting, which of the following quantities is (are) random variables (select all that applies) (a) θ (b) MSE[θ ] (c) Var[θ ] (d) n (e) Xˉ Solution ACTL1101: 00001 6 (a) True (b) False (c) False (d) False (e) True 8. Problem You have two packs of M&M’s (candies). The first pack contains a number n1 of the delicious candies, and the second pack contains n2 of them, with n2 6= n1. Every M&M can be red, or not, so that the total number of red M&M’s in any given pack of size n is modelled by a Binomial(n,p) random variable, where p is an unknown parameter. Denote by X the (random) number of red M&M’s in the first pack, and Y the (random) number of red M&M’s in the second pack. Out of the following list of estimators for p, which one is the best (a) Yn2 (b) Y Xn1+n2 (c) Xn1 (d) 12 ( X n1 + Yn2 ) (e) X+Yn1+n2 Solution (a) False (b) False (c) False (d) False (e) True 9. Problem Consider an unknown parameter θ. Which of the following properties is (are) desirable for any estimator θ (select all that applies) (a) High Var[θ ] (b) Low E[(θ E[θ ])2] (c) High MSE[θ ] (d) High { Var[θ ] + ( E[θ θ] )2} (e) Low E[θ θ] Solution (a) False (b) True (c) False (d) False (e) True ACTL1101: 00001 7 10. Problem A lecturer needs to decide if a student is guilty of plagiarism. We can picture this situation as a hypothesis test with: H0 : the student did not plagiarise versus H1 : the student plagiarised. In this setting, the lecturer needs to decide whether to use a level of significance α1 = 0.007 or α2 = 0.02. What statement(s) best describes the implications of choosing α1 instead of α2 (select all that applies) (a) It increases the chances a student is justly found guilty. (b) It increases the probability to make an Error of Type 1. (c) It increases the chances a student is unjustly found guilty. (d) It decreases the chances a student is unjustly found guilty. (e) It decreases the chances a student is justly found guilty. Solution (a) False (b) False (c) False (d) True (e) True 11. Problem A sample {X1, X2, X3, X4} stemming from an Exponential(β) distribution (with expectation 1/β) is used to estimate the parameter β. If x1 = 0.198, x2 = 1.987, x3 = 3.043, x4 = 0.397, what is your best estimate of the value of β (a) 2.21 (b) 3.20 (c) 1.88 (d) 0.71 (e) 1.41 Solution We have β = 1/Xˉ, hence β = 1 xˉ = 4 0.198 + 1.987 + 3.043 + 0.397 = 0.71. (a) False (b) False (c) False (d) True (e) False ACTL1101: 00001 8 12. Problem In a new taxi start-up called ‘Driver-City’, let X1, . . . , Xn represent the salaries of female drivers while Y1, . . . , Ym represent the salaries of male drivers. You wish to prove that male drivers are paid more, on average, than female drivers. You will check this with a hypothesis test where: H0 : male and female drivers have equal average salary. In what situation would you reject H0 (select all that applies) (a) If Xˉ Yˉ is much larger than 0. (b) If Yˉ /Xˉ is much smaller than 1. (c) If Yˉ Xˉ is much larger or much smaller than 0. (d) If Xˉ/Yˉ is much larger or much smaller than 1. (e) If Yˉ /Xˉ is much larger than 1. Solution (a) False (b) False (c) False (d) False (e) True 13. Problem Find the Present Value of a single payment due exactly 14 years from now, given that the amount of the payment is A = 65 and that the compounding frequency is infinite, with a nominal annual interest rate of 3%. (a) 30.35 (b) 17.91 (c) 0.66 (d) 42.71 (e) 63.08 Solution We simply have: PV = (e δT ·A) = e 0.03·14 · 65 = 42.7080433. (a) False (b) False (c) False (d) True (e) False 14. Problem Let n ≥ 3 be an integer. Let a be the present value of a ‘regular’ annuity (in arrears, no mortality, payments of 1) for n periods. Let b be the present value of an annuity due (no mortality, payments of 1) for n 1 periods. Both are computed on the same effective interest rate i > 0. Which of the following statement(s) is(are) necessarily TRUE (select all that applies) ACTL1101: 00001 9 (a) a < b (b) a > b (c) a+ b > n (d) a = b (e) b+ 1 > a Solution (a) False (b) False (c) False (d) False (e) True 15. Problem You just obtained a loan of amount 300, at an nominal interest rate of 10% (annual). What is the amount of your monthly repayments (paid in arrears over a period of four years) (a) 11.76 (b) 29.17 (c) 18.22 (d) 30.31 (e) 7.61 Solution With a monthly effective rate of i = 0.1/12, we simply solve: Loan = PV of payments, L = P · ( 1 vn i ) P = L · i · ( 1 1 vn ) = 300 · 0.0083333 1 (1 + 0.0083333) 48 = 7.608775. (a) False (b) False (c) False (d) False (e) True 16. Problem A loan of amount 230 is to be repaid with 6 level yearly payments, the first one due one year from now. The effective yearly interest rate is 4%. Consider the very first repayment: what proportion of this payment constitutes interest repaid [The choices are expressed in decimal values, i.e. 0.05 ≡ a proportion of 5%.] (a) 0.040 (b) 0.643 (c) 0.210 (d) 0.079 (e) 0.362 ACTL1101: 00001 10 Solution Call i the yearly interest (effective), L the amount of the loan, and P the yearly payments. We need to solve i · L P = i · L (i · L)/(1 vn) = 1 v n = 1 (1 + i) n = 1 (1.04) 6 = 0.2096855. (a) False (b) False (c) True (d) False (e) False 17. Problem The rate of interest effective on a two-years period is 38%. What is the equivalent rate of interest effective per month [The choices are expressed in decimal values, i.e. 0.0500 ≡ 5.00%.] (a) 0.0470 (b) 0.0272 (c) 0.0135 (d) 0.0086 (e) 0.0379 Solution We simply must find i such that (1 + i)24 = 1 + 0.38, i.e. (1.38)1/24 1 = 0.013511. (a) False (b) False (c) True (d) False (e) False 18. Problem An insurer uses the ‘standard deviation Principle’ to fix the Premium P of an insurance loss X ～ Poisson(λ = 10). If α = 0.35 (α being the ‘risk loading’), what is P (a) 33.03 (b) 0.99 (c) 13.50 (d) 11.11 (e) 11.87 Solution We have P = E[X] + α √ V ar[X] = 10 + 0.35 · √10 = 11.1067972. (a) False (b) False (c) False ACTL1101: 00001 11 (d) True (e) False 19. Problem An group of n = 6 independent and identically distributed losses X1, . . . , Xn are pooled together, so that everyone in the pool will pay the actual average of the losses. This yields that the standard deviation of the individual loss (with pooling) is equal to: b ·√Var[Xi], for a certain constant b > 0. What is b (a) 1.66 (b) 0.41 (c) 0.82 (d) 0.17 (e) 2.45 Solution The standard deviation with pooling is equal to: √ V ar[X]/n. Hence, b = 1/ √ n = 0.4082483. (a) False (b) True (c) False (d) False (e) False 20. Problem Let X and Y be two random variables such that the correlation between X and Y is 0. Which of the following statement(s) is(are) necessarily TRUE (select all that applies) (a) Var[X + Y ] = Var[X]+Var[Y ] (b) X and Y are independent (c) X and Y are dependent (d) E[X] = E[Y ] (e) Var[X] = Var[Y ] Solution (a) True (b) False (c) False (d) False (e) False 21. Problem Let X be a Binomial(n = 2, p = 1/2) random variable. Further, let Y = X5. What is Cov[X,Y ] (a) 0.00 (b) 4.00 (c) 8.00 (d) 24.83 ACTL1101: 00001 12 (e) 17.09 Solution We must find E[XY ] E[X]E[Y ] = E[X6] E[X]E[X5] = ( 0 + 1 2 · 1 + 1 4 · 26 ) 1 · ( 0 + 1 2 · 1 + 1 4 · 25 ) = 8 (a) False (b) False (c) True (d) False (e) False 22. Problem Let X ≥ 0 represent the age-at-death of a new-born. The CDF of X is given by FX(x) = 1 e (x/λ)2 , x ≥ 0. Let λ = 90. What is the probability a life aged exactly 25 today lives at least another 16 years (a) 0.993 (b) 0.955 (c) 0.839 (d) 0.878 (e) 0.934 Solution We have the survival function s(x) = e (x/λ) 2 , with which we find Pr[X > 41|X > 25] = s(41) s(25) = 0.8777703. (a) False (b) False (c) False (d) True (e) False 23. Problem Let 0 ≤ X ≤ ω be a continuous random variable with hazard rate μ(x) = 1ω x for 0 ≤ x < ω. What is the survival function s(x) of X (a) ω/x 1 (b) 1 √x/ω (c) 1 (x/ω)2 ACTL1101: 00001 13 (d) 1 x/ω (e) x/ω Solution (a) False (b) False (c) False (d) True (e) False 24. Problem A Life Table features the following numbers: l17 = 1000, l18 = 988, l19 = 972, l20 = 960, l21 = 932, l22 = 916. Dilbert just turned 18. Use the Table to find the probability he dies before his 21th birthday. (a) 0.073 (b) 0.191 (c) 0.028 (d) 0.057 (e) 0.041 Solution We have (a) False (b) False (c) False (d) True (e) False 25. Problem Charlie Mansion is a prisoner sentenced to death. The sentence will be carried out in 3 years. Charlie is exactly 55 now, and for people with his characteristics: q55 = 0.03, q56 = 0.03, q57 = 0.04. If each year he spends in prison costs 10,000 (paid at the beginning of the year, but only if he is still alive), what is the Expected Present Value of the cost of his future stay in prison Use an effective annual interest rate of 5%. (a) 22118 (b) 22783 (c) 17773 (d) 27772 (e) 28594 ACTL1101: 00001 14 Solution We have: EPV = 10, 000(1 + v · p55 + v2p55p56) = 10, 000(1 + v · (1 q55) + v2 · (1 q55)(1 q56)). (a) False (b) False (c) False (d) True (e) False 26. Problem Consider a life aged x = 63, for which you are given the following information: qx = 0.02, qx+1 = 0.03, qx+2 = 0.04 px = 0.98, 2px = 0.9506. Further, the annual effective interest rate is 6%. A Term Insurance contract with death benefit of 100,000 is set on this life, with a coverage period of n = 3 years. What is the Expected Present Value of the death benefit (which is to be paid at the end of the year of death, if death occurs within the coverage period) (a) 7696 (b) 8158 (c) 11447 (d) 9979 (e) 7915 Solution The answer is the sum ∑2 k=0 v k+1 · qx+k · kpx. (a) True (b) False (c) False (d) False (e) False 27. Problem Year 1 Year 2 Year 3 2017 1200 1700 2100 2018 1700 2600 3200 2019 900 1300 1600 Consider the above Run-Off triangle of cumulative claims for accident years 2017, 2018 and 2019 (columns represent development years). It has already been completed for future years (2020 and beyond). From this triangle, what is the amount of the total reserves for outstanding claims liabilities (as at 31/12/2019) [No discounting is used.] ACTL1101: 00001 15 (a) 1300 (b) 6100 (c) 3897 (d) 10200 (e) 5775 Solution We need the amount of incremental future claims, i.e. Reserve = (3200 2600) + (1300 900) + (1600 1300) = 1300. (a) True (b) False (c) False (d) False (e) False 28. Problem An insurance policy covers a risk X ～ Uniform[0, 97]. If a deductible of d = 20 is included in the policy, what is the expected value of the payment made by the insurer (a) 51.1 (b) 46.4 (c) 30.6 (d) 38.5 (e) 75.2 Solution Using the formula from the lecture notes: E[Y ] = ∫ ∞ 0 yfS(y + d)dy = ∫ 97 d 0 y 97 dy = (77)2 2 · 97 = 30.562. (a) False (b) False (c) True (d) False (e) False 29. Problem Which of the following statement(s) is(are) included in the CAS ratemaking principles (select all that applies) (a) A rate provides for the costs associated with an individual risk transfer. (b) A rate is an estimate of the expected value of future costs. (c) A rate provides for the pure insurance costs associated with the transfer of risk. (d) A rate provides for all costs associated with the transfer of risk. (e) A rate is an estimate of the expected value of present costs. Solution ACTL1101: 00001 16 (a) True (b) True (c) False (d) True (e) False 30. Problem Year 1 Year 2 Year 3 2017 1150 1620 2030 2018 1980 3140 2019 840 Consider the above Run-Off triangle of cumulative claims for accident years 2017, 2018 and 2019 (columns represent development years). Use the Chain Ladder method with weighted averages to find the factor f1 (i.e. the factor developing Year 1 to Year 2). No discounting is used. (a) 1.521 (b) 1.497 (c) 1.164 (d) 1.199 (e) 1.225 Solution We divide the total amount experienced in development year 2 (for accident year 2017 and 2018) by the total amount experienced in development year 1 (of those same accident years), i.e. f 1 = C1,2 + C2,2 C1,1 + C2,1 = 1620 + 3140 1150 + 1980 = 1.521. (a) True (b) False (c) False (d) False (e) False 31. Problem Let X1 ～ Poisson(λ) and X2 ～ Poisson(2λ) be two independent random variables, and let Y = X2 X1. Consider the following two estimators for λ: λ = Y, and λ = X1 +X2 3 . (a) (1pt) Calculate Var[Y ]. (b) (2pt) Between the two estimators above, is there one that you consider to be better Provide at least two reasons to justify your answer. (c) (1pt) Explain carefully why: Pr[Y = 0] = ∞∑ k=0 ( e λλk k! ) · ( e 2λ(2λ)k k! ) ACTL1101: 00001 17 Solution (a) By independence, Var[Y ] = Var[X2] + ( 1)2 Var[X1] = 3λ. (b) There is no doubt λ is the better estimator. This is true because, although both estimators are unbiased, λ has a much lower variance than λ , indeed V ar[λ ] = 1 9 (λ+ 2λ) = λ/3. But also: λ could be negative, which does not make sense for the parameter λ of a Poisson. (c) This is because Pr[Y = 0] = Pr[X2 = X1] = ∞∑ k=0 Pr[X1 = X2 = k] which, by independence, is the expression above, where both PMF (evaluated at every k) are multiplied and summed over all k’s. 32. Problem Four friends play a betting game, where each participant contributes the same amount of money (15$ each) to a pot. The game is of type ‘winner takes all’: three people will achieve a net loss of 15$, and one person will end up with a net gain of 3× 15 = 45$. Everyone has the same probability to win (1/4). Denote X and Y the future net positions of Player 1 and Player 2, respectively, upon playing the game. (a) (1pt) Explain why a ‘risk-lover’ individual would be keen to play this game. (b) (1pt) Show that E[X · Y ] = 225. (c) (2pt) Find the correlation between X and Y . Hint: use information from part b). (d) (2pt) Assume two of the four players apply the following strategy: if either of them wins they will split the gains half-half. Those two players both have the same utility function v(w) = 1 e w. Given what you know, explain why their strategy is sensible. (e) (3pt) Assume that a 5th player, Josephine, has the possibility to join this game (con- tributing the same amount 15$). She has a secret strategy that improves her chances of winning to p > 1/5 (i.e. beating random chance). Given that Josephine has an initial wealth w0 = 24 and a utility function v(w) = ln(w), what values of p would make her decide to play Solution (a) A risk lover prefers to hold a random wealth (with given expectation) as opposed to holding a fixed amount equal to that same expectation. Here, the expectation of playing is 0, and the expectation of not playing is also 0. So, a risk-lover would play this game (and any game with a null expected gain). (b) The product X · Y will be equal to ( 15)2 = 225 if both players lose (chance of 1/2). It will be equal to 15 · 45 = 675 if either one of them the wins (chance of 1/2). So, E[X · Y ] = 1 2 ( 675 + 225) = 225. ACTL1101: 00001 18 (c) Because E[X] = E[Y ] = 0, Cov[X,Y ] = E[X · Y ]. Also, Var[X] = E[X2] = ( 15)2 · 0.75 + (45)2 · 0.25 = 675. Hence, ρ(X,Y ) = E[XY ] V ar[X] = 225 675 = 1/3. (d) From their utility function, the two players are risk averse. Hence, they do not like uncertainty. By splitting their possible gains half-half, they reduce the variance of their future wealth (and keep the expected gains equal), so it is a sensible decision for them. This is akin to the pooling of insurance risks (maintain the expectation while reducing the variance), although here we have ‘gambling risks’. (e) Call W the wealth of Josephine in case she plays. She will play if v(w0) < E[v(W )]. She has a probability p of a 60$ gain, and a probability 1 p of a 15$ loss. Hence, she will play if ln(24) < p ln(24 + 60) + (1 p) ln(24 15) ln(24) < p ln(84) + (1 p) ln(9) ln(24/9) < p ln(84/9) ln(24/9)/ ln(84/9) < p 0.4391264 < p 33. Problem A researcher found that the lifetime (in years) of a given breed of turtles is well-modelled by a random variable X ≥ 0 with: fX(x) = 2 63 ( 1 x 63 ) FX(x) = x 63 ( 2 x 63 ) , for 0 ≤ x ≤ 63, where ω = 63 is the theoretical maximum age for that breed of turtles. The force of mortality μ(x) associated with this mortality model is illustrated below. 0 10 20 30 40 50 60 0. 0 0. 2 0. 4 0. 6 0. 8 1. 0 x μ(x ) ACTL1101: 00001 19 (a) (2pt) Koopa is a turtle aged exactly 10. What is the probability Koopa survives at least up to age 35, but not up to age 53 (b) (2pt) Explain why the random variable X would not be a good model for human mor- tality in Australia. Give at least two reasons. (c) (2pt) We can note that f(x) is strictly decreasing in x, while from the graph above, μ(x) is strictly increasing in x. Explain carefully how this is possible. (d) (2pt) Let T (x) represent the future lifetime of a turtle aged x. We can show that E[T (x)] = ∫ w x 0 t · fX(t+ x) s(x) dt. Explain every element in the above equation. (Note: you are not asked to prove the above equation, but rather to explain intuitively why it holds). Solution (a) We must find: Pr[35 ≤ X ≤ 53|X > 10] = s(35) s(53) s(10) = FX(53) FX(35) s(10) = 0.2435 (b) There are many reasons: The maximum age is not appropriate (i.e. for humans it should be above 100, maybe 110 or 120). The shape of the density function (linearly decreasing) does not fit at all human mortality (which has a peak at old ages, around 85). Consequently, the force of mortality does not fit that of humans (in Australia) either (it should be decreasing in the first few years of life, and not increasing at such a crazy rate in the very old years). For humans, mortality is fairly different for males and females, which is not cap- tured here in a unique model for X. (c) This is possible because they represent two very different things: fX(x) is the likeli- hood a new-born dies around age x, while μ(x) represents the likelihood a turtle aged exactly x dies in the next instant. Hence, while on the whole more turtles die young, it is still possible that older turtles have a higher mortality than young one. Said in a different manner: fewer turtles die old because it’s unlikely they reach an older age, given the (relatively) high mortality at younger ages. However, an older turtle still has more chances to die soon than a young one. More mathematically, we can also see that, because μ(x) = f(x)/s(x), it suffices for s(x) to decrease at a faster rate than f(x) for their ratio to increase, and this is indeed the case here. (d) We can explain each element of the equation as follows: The integral starts at 0 because the minimum future lifetime is 0. The integral stops at ω x, because the turtle is already aged x and can then only live an additional ω x years. We are computing an expectation, hence in the integral there is a ‘t’ in front: every possible value for the future lifetime is considered. The density we use (i.e. the weight of each value t) is not fX(t), but rather fX(t+ x)/s(x), which is the conditional density of a lifetime, given the life has already survived x years. ACTL1101: 00001 20 34. Problem On the Facebook page ‘UNSW Love Letters’, new posts are published according to a Pois- son Process with parameter λ = 6, where time is measured in days. Each post is identified with a unique integer number (#1, #2, #3, etc.). (a) (2pt) In a given week, find the probability there is at least one post published each day. (b) (2pt) Call T the time elapsed (in days) between the appearance of two consecutive even-numbered posts (e.g. #2 then #4). What is the standard deviation of T (c) (3pt) The number of ‘reactions’ (like, love, sad, etc.) any given post receives has a Negative Binomial(r = 2, p = 0.05) distribution. If every single reaction generates 0.05$ (5 cents) in ad revenue for Facebook, what is the expected value of the total revenue this page generates for Facebook annually (1 year = 365 days) (d) (1pt) Joseph, a big fan of the page, eagerly awaits for the next post to be published. He reasons like this: ‘I’ve been sitting here refreshing the page periodically for a long while, surely a new post will appear soon’. Do you agree with this line of thinking (e) (1pt) Do you think it is realistic to model the arrival of Facebook posts on a given page with a Poisson Process Explain your answer. Solution (a) The number of posts per day has distribution Poisson(λ = 6), so the probability that on a single day there is at least one post is 1 e 6. To get to this same probability for 7 days we need to take this number to the power 7, i.e. Pr[N1 > 0, N2 > 0, . . . , N7 > 0] = (1 e 6)7 = 0.9827772. (b) We need to recognise that T = T1+T2, where T1 and T2 are two independent Exp(λ = 6). Then, Var[T ] = Var[T1 + T2] = 2/λ2, so that sd(T ) = √ 2/λ = 0.2357023. (c) The number of posts in a year has distribution N ～ Poisson(365 · λ). Call Mi the number of likes generated by post i, and Xi the revenue generated from that same post i. We then have Xi = 0.05Mi, so E[Xi] = 0.05 · r/p = 2. Then, if S is the total revenue generated in a year, i.e. S = N∑ i=1 Xi, we have E[S]= E[N ]· E[Xi] = 365 · λ · 2 = 4380. (d) Joseph’s reasoning is seriously flawed. Indeed, the time until the next post has Expo- nential distribution, hence it is memoryless: the fact one has been waiting for a long time does not influence in any way the likelihood a post is going to appear soon.
