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Game Franchises This section of sample problems and solutions is a part of The Actuary's Free Study Guide for Exam 4 / Exam C, authored by Mr. Stolyarov. This is Section 6buy christian louboutin shoes3 of the Study Guide. See an index of all sections by following the link in this paragraph. All information in the preface to the problems in this study guide is found in Section 15.5 of Loss Models: From Data to Decisions (cited below). "A loss function lj(θ^j, θj) describes the penalty louis vuitton bags replicapaid by the investigator when θ^j is the estimate and θj is the true value of the jth parameter." (407) "The Bayes estimate for a given loss function is the one that minimizes the expected loss given the posterior distribution of the parameter in question." Actuaries. Loss Models: From Data to Decisions, (Third Edition), 2008, by Klugman, S.A., Panjer, H.H. and Willmot, G.E., Chapter 15, pp. 407-408. Original Problems and Solutions from The Actuary's Fre buy christian louboutin shoese Study Guide Problem S4C63-1. The mean of an exponential distribution is θ. The median of an exponential distribution is ln(2)*θ. The mode of an exponential distribution is 0. A posterior distribution is exponential with parameter θ = 90. Find the Bayes estimate of the mean for a loss function associated with this posterior distribution under the following ciChristian Louboutin on salercumstances: (a) The loss function is a squared-error loss function. (b) The loss function is an absolute lo buy chi flat iron ukss function. (c) The loss function is a zero-one loss function. Solution S4C63-1. We use Theorem 63.1: "For squared-error loss, the Bayes estimate is the mean of the posterior distribution; for absolute loss, it is a median; for zero-one loss, it is a mode." Thus, for (a), the Bayes estimate is the mean of the posterior distribution, or θ = 90. (q3)(3q*(1-q)2)(3q2*(1-q)) = Model distribution = 9q6(1-q)3. Thus, (Joint distributionbuy lv handbags) = (8q3)(9q6(1-q)3) = (72q9(1-q)3). /(Marginal distribution) = (72q9(1-q)3)/0.0251748252 = 2860q9(1-q)3. Thus, ∫ E(Y│q)* (Posterior distribution)*dq = 01∫3q*2860q9(1-q)3*dq = 01∫8580q10(1-q)3*dq = E(Y│X) = 15/7 = 2.142857143. Thus, we can expect 15/7 elephants to be in the room on Day 3. Problem S4C63-4. Similar to Question 15 of the Exam C Sample Questions from the Society of Actuaries. The probability that a squirrel will eat a nut in a given hour is p. The prior distribution of p is uniform on the interval from 0 to 0.75. During every hour for the past 14 hours, the squirrel has eaten a nut. Estimate the posterior probability that the squirrel will also eat a nut during the 15th hour