MATH590统计数学理论算法

你要去参加比赛，想决定是否把赌注押在霹雳马上。你想应用决策理论来做决定。你使用的信息来自两个独立赛马专家。数据X表示建议的专家数量你可以在霹雳上下注（当然，因为他们相信这匹马会赢得比赛）。如果您决定不下注，而Thunderbolt没有获胜，或者当您下注时，Thunderbol特赢得比赛，什么都没输。如果Thunderbolt没有获胜，并且您决定下注他，你的主观判断是，你的损失将是不这样做的成本的四倍
下注，但霹雳确实赢了（因为你会错过其他投资机会你的钱）。您已经为两位赛马专家调查了正确中奖的历史如下所示。当霹雳成为赢家时，两位专家都正确地预测了他以5/6的概率获胜（以及以1/6的概率失利）。当霹雳没有赢得了一场比赛，两位专家都预测他会赢3/5。你听取两位专家的意见。MATH590MATH5905 Term One 202 Assignment One Statistical InferenceUniversity of New South WalesSchool of Mathematics and StatisticsMATH5905 Statistical InferenceTerm One 2023Assignment OneGiven: Friday 24 February 2023 Due date: Sunday 12 March 2023

1. Instructions

This assignment is to be completed collaboratively by a group of at most 3students. (If you are unable to join a group you can still present it as an individual assignment.)The same mark will be awarded to each student within the group, unless I have good reasonsto believe that a group member did not contribute appropriately. This assignment must besubmitted no later than 11:59 pm on Sunday, 12 March 2023. The first page of the submittedPDF should be this page. Only one of the group members should submit the PDF file onMoodle, with the names, student numbers and signatures of the other students in the groupclearly indicated on this page.I/We declare that this assessment item is my/our own work, except where acknowledged, andhas not been submitted for academic credit elsewhere. I/We acknowledge that the assessor ofthis item may, for the purpose of assessing this item reproduce this assessment item and providea copy to another member of the University; and/or communicate a copy of this assessmentitem to a plagiarism checking service (which may then retain a copy of the assessment item onits database for the purpose of future plagiarism checking). I/We certify that I/We have readand understood the University Rules in respect of Student Academic Misconduct.Name Student No. Signature Date1MATH590

2.Term One 202

3 Assignment One Statistical InferenceProblem Onei) It is known that for independent Poisson distributed random variables X2 (and zero else). Here c is a normalizing constant.a) Show that c = 516 .b) Find the marginal density fX(x) and FX(x).c) Find the marginal density fY (y) and FY (y).d) Find the conditional density fY |X(y|x).e) Find the conditional expected value a(x) = E(Y |X = x).Make sure that you show your working and do not forget to always specify the support ofthe respective distribution.Problem TwoLet X and Y be independent uniformly distributed in (0, 1) random variables. Further, letU(X,Y ) = X + Y, V (X,Y ) = Y ?X be a transformation.a) Sketch the support S(X,Y ) of the random vector X,Y in R2.b) Sketch the support S(U,V ) of the random vector (U, V ) in R2.c) Determine the Jacobian of the transformationd) Determine the density of the random vector (U, V )Justify completely each step.Problem ThreeYou are going to the races and want to decide whether or not to bet on the horse Thunderbolt.You want to apply decision theory to make a decision. You use the information from twoindependent horse-racing experts. Data X represents the number of experts recommendingyou to bet on Thunderbolt (due, of course, to their belief that this horse will win the race).If you decide not to bet and Thunderbolt does not win, or when you bet and Thunderboltwins the race, nothing is lost. If Thunderbolt does not win and you have decided to bet onhim, your subjective judgment is that your loss would be four times higher than the cost of notbetting but the Thunderbolt does win (as you will have missed other opportunities to investyour money).You have investigated the history of correct winning bets for the two horse-racing experts andit is as follows. When Thunderbolt had been a winner, both experts have correctly predictedhis win with probability 5/6 (and a loss with a probability 1/6). When Thunderbolt has notwon a race, both experts had a prediction of 3/5 for him to win. You listen to both expertsand make your decision based on the data X.a) There are two possible actions in the action space A = {a0, a1} where action a1 represents ※Thunderbolt not winning§.Define the appropriate loss function L(牟, a) for this problem.b) Compute the probability mass function (pmf) for X under both states of nature.c) The complete list of all the non-randomized decisions rules D based on x is given by:d1 a1For the set of non-randomized decision rules D compute the corresponding risk points.d) Find the minimax rule(s) among the non-randomized rules in D.e) Sketch the risk set of all randomized rules D generated by the set of rules in D. Youmight want to use R (or your favorite programming language) to make this sketch moreprecise.f) Suppose there are two decisions rules d and d∩. The decision d strictly Hence,given a choice between d and d∩ we would always prefer to use d. Any decision ruleswhich is strictly dominated by another decisions rule (as d∩ is in the above) is said to beinadmissible.Correspondingly, if a decision rule d is not strictly dominated by any otherdecision rule then it is admissible. Show on the risk plot the set of randomized decisionsrules that correspond to the admissible decision rules.g) Find the risk point of the minimax rule in the set of randomized decision rules D anddetermine its minimax risk. Compare the two minimax risks of the minimax decisionrule in D and in D. Comment.h) Define the minimax rule in the set D in terms of rules in D.i) For which prior on is the minimax rule in the set D also a Bayes rule?j) Prior to listening to the two experts, you believe that Thunderbolt will win the race withprobability 1/2. Find the Bayes rule and the Bayes risk with respect to your prior.k) For a small positive ? = 0.1, illustrate on the risk set the risk points of all rules whichare ?-minimax.Problem FourIn a Bayesian estimation problem, we sample n i.i.d. observations X = (X1, X2, . . . , Xn)from a population with conditional distribution of each single observation being thegeometricdistributionfX1, x = 0, 1, 2, . . . ; is considered as random in the interval is interpreted as a probability of success in a single trial in a success-failure scheme, givean interpretation of the conditional distribution of the random variable X1(You mayuse the integrate function in R or another numerical integration routine from your favouriteprogramming package to answer the question.)
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