Review Test Submission: Practice Test 4 - ECON7070 S1 2024
复习考试提交:模拟考试 4 - ECON7070 S1 2024
Content 内容
User | Fan Fei 范飞 |
---|---|
Course |
[ECON7070] Economic Analysis of Strategy (St Lucia). Semester 1, 2024
[ECON7070]战略经济分析(圣卢西亚)。2024年第一学期 |
Test |
Practice Test 4 - ECON7070 S1 2024
模拟考试 4 - ECON7070 S1 2024 |
Started | 03/06/24 15:21 |
Submitted | 03/06/24 15:28 |
Status | Completed |
Attempt Score 尝试得分 |
0 out of 100 points
0 分(满分 100 分) |
Time Elapsed 经过的时间 |
7 minutes out of 1 hour
1 小时中有 7 分钟 |
Instructions |
The test is a mixture of multiple choice and short calculation questions. There are 10 questions in total, each worth an equal number (10) marks. Answer all questions. Questions will appear one at a time, but you may go back and change your answers once you move on to the next question. Once you begin your test, a timer will count down for 60 minutes. Be sure to submit your test within the 60 minutes. After submitting the Practice Test, you will see your answers, the correct answers, and feedback. These will not be displayed after submitting the real Online Test. Please carefully note the following points about your test:
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Results Displayed 显示的结果 |
All Answers, Submitted Answers, Correct Answers, Feedback, Incorrectly Answered Questions
所有答案、提交的答案、正确答案、反馈、错误回答的问题 |
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Suppose Player 1 and Player 2 plan to go to either a football match, or a concert in the evening. Players 1 and 2 are away from each other during the day and neither know which event the other is going to go to. By the evening, Players 1 and 2 are each either happy or upset independently of each other. Player 1 only knows that Player 2 is happy with probability 0.1 and upset with probability 1-0.1. Player 2 only knows that Player 1 is happy with probability 0.1 and upset with probability 1-0.1. Each player knows with certainty whether they themselves are happy or upset. Denote choosing to go to the football match by F, and choosing to go to the concert by C.
假设玩家 1 和玩家 2 计划在晚上去看一场足球比赛或一场音乐会。 玩家 1 和 2 白天彼此远离,谁也不知道对方要参加哪个活动。 到了晚上,玩家 1 和 2 各自独立地感到高兴或沮丧。 玩家 1 只知道玩家 2 对概率 0.1 感到满意,对概率 1-0.1 感到不安。 玩家 2 只知道玩家 1 对概率 0.1 感到满意,对概率 1-0.1 感到不安。 每个玩家都肯定地知道他们自己是高兴还是沮丧。 F表示选择去看足球比赛,C表示选择去看演唱会。If both players are happy, the utilities of each player given their choice of event are given in the payoff matrix:
如果两个玩家都满意,则每个玩家在选择事件时的效用都会在收益矩阵中给出:F C F 8,2 0,0 C 0,0 2,8 If Player 1 is happy but Player 2 is upset, the utilities of each player given their choice of event are given in the payoff matrix:
如果玩家 1 很高兴,但玩家 2 不高兴,则每个玩家选择的事件的效用将在收益矩阵中给出:F C F 2,0 0,2 C 0,2 2,0 If Player 1 is upset but Player 2 is happy, the utilities of each player given their choice of event are given in the payoff matrix:
如果玩家 1 不高兴,但玩家 2 高兴,则每个玩家选择的事件的效用都会在收益矩阵中给出:F C F 0,2 2,0 C 2,0 0,2 Finally if both players are upset, the utilities of each player given their choice of event are given in the payoff matrix:
最后,如果两个玩家都心烦意乱,则每个玩家在选择事件时的效用都会在收益矩阵中给出:F C F 0,0 8,8 C 2,2 0,0 In the ex-ante normal form game associated with this Bayesian game, what is the expected utility of Player 2 if both players choose the pure strategy CF (i.e., C if happy and F if unhappy)? Round your answer to 2 decimal places.
在与此贝叶斯博弈相关的事前正态博弈中,如果两个玩家都选择纯策略 CF(即 C 如果高兴,则 F 不高兴),玩家 2 的预期效用是什么?将您的答案四舍五入到小数点后 2 位。Selected Answer: 选择答案: [None Given]
[未给出]
Correct Answer: 正确答案: 0.26 ± 0.01
Response Feedback: 响应反馈: Multiply the payoff of Player 2 by the probability of the type profile occuring for each of the 4 possible type profiles. Then sum up the values.
将玩家 2 的收益乘以 4 个可能的类型配置文件中每个配置文件出现类型配置文件的概率。 然后对值进行总结。 -
Consider a second price auction for a single item with two bidders. Suppose bidder 1 has a value uniformly drawn in the interval from 0 to 1, while bidder 2 has a value 0.9.
考虑对一件物品进行第二次价格拍卖,其中有两个投标人。假设投标人 1 的值在 0 到 1 的区间内均匀绘制,而投标人 2 的值为 0.9。In the Bayesian equilibrium in undominated strategies, what is the seller's expected revenue? Give your answer to two decimal places.
在非主导策略的贝叶斯均衡中,卖方的预期收入是多少?把你的答案给出两个小数位。Selected Answer: 选择答案: [None Given]
[未给出]
Correct Answer: 正确答案: 0.50 ± 0.01
Response Feedback: 响应反馈: Because both players' dominant strategy is to bid their value, we know that 90% of the time the second bidder (whose value is 0.9) will win, and 10% of the time the first bidder wins.
因为两个参与者的主导策略都是出价,所以我们知道 90% 的时间第二个出价者(其价值为 0.9)会获胜,而第一个出价者有 10% 的时间会获胜。If the first bidder wins (probability = 0.1), the price is equal to the second bidder's bid, which is 0.9.
如果第一个出价者获胜(概率 = 0.1),则价格等于第二个出价者的出价,即 0.9。If the second bidder wins (probability = 0.9), the price is the expected bid of the first bidder, conditional on it being less than the second bidder's bid. That is, the first bidder's expected bid, which is also the expected price, when the second bidder wins is the average of the range (0, 0.9), which is 0.45.
如果第二个投标人获胜(概率 = 0.9),则价格是第一个投标人的预期出价,条件是它低于第二个投标人的出价。也就是说,当第二个投标人获胜时,第一个投标人的预期出价,也就是预期价格,是范围(0,0.9)的平均值,即0.45。So the expected revenue is equal to:
因此,预期收入等于:(probability first bidder wins)*(expected revenue if first bidder wins) + (probability second bidder wins)*(expected revenue if second bidder wins)
(第一个投标人获胜的概率)*(第一个投标人获胜的预期收入)+(第二个投标人获胜的概率)*(第二个投标人获胜的预期收入)= 0.1*0.9 + 0.9*0.45
= 0.495.
Don't forget to round to two decimal places! You should enter 0.50, which Blackboard will automatically format to "0.5" (don't worry about that).
别忘了四舍五入到小数点后两位!您应该输入 0.50,Blackboard 会自动将其格式化为“0.5”(不用担心)。 -
Two firms simultaneously decide whether to Enter (E) or stay out (O). The state of the market can be strong or weak and it is known by Firm 1 only. Firm 2 believes that a strong and a weak state are equally likely.
两家公司同时决定是进入(E)还是退出(O)。市场状态可强可弱,只有公司 1 知道。公司 2 认为强国和弱国的可能性相同。Letting Firm 1 be the row choser, the payoffs when the market is weak are:
让公司 1 成为行选择者,当市场疲软时的回报是:E O E 1, 1 2, 2 O 2, 1 3, 2 The payoffs when the market is strong are:
市场强劲时的回报是:E O E 5, 3 8, 2 O 2, 3 3, 2 The ex-ante normal form associated with this Bayesian game is:
与此贝叶斯博弈相关的事前正态形式为:Selected Answer: 选择答案: [None Given]
[未给出]
Answers: 答案: A.B.
湾。
C.D.Response Feedback: 响应反馈: See the lecture notes for Bayesian Games.
请参阅贝叶斯游戏的讲义。 -
Consider the extensive form game with imperfect imformation below. In this game, Player 2 is weak with probability
and strong with probability 1-
.
考虑下面信息不完美的广泛形式游戏。在这个游戏中,玩家 2 的概率较弱,概率为 1-
的强。
Suppose
= 3 and
= -2 and suppose Player 1's strategy is to choose E. What is the smallest probability
for which this strategy is part of a subgame perfect equilibrium? Round your answer to two decimal places.
假设= 3 和
= -2,假设玩家 1 的策略是选择 E。该策略是子博弈完美均衡的一部分的最小概率
是多少?将您的答案四舍五入到小数点后两位。
Selected Answer: 选择答案: [None Given]
[未给出]
Correct Answer: 正确答案: 0.40
Response Feedback: 响应反馈: Given Player 2 always chooses C when weak and F when strong, calculate Player 1's expected payoff from choosing E. Find
such that this payoff is greater than 0 (Player 1's payoff of choosing Q).
假设玩家 2 在弱时总是选择 C,在强时总是选择 F,计算玩家 1 选择 E 的预期收益。 -
Select all statements that are true.
选择所有为 true 的语句。Selected Answers: 部分答案: [None Given]
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Answers: 答案: Response Feedback: 响应反馈: All answers are taken straight from the lecture notes and textbook readings.
所有答案都直接来自讲义和教科书阅读材料。 -
Amy (the row player) has just started dating Brad (column player). Each person can choose either to (C)ommit to the relationship, or (D)ump their partner and break up. Amy prefers a partner who will commit. Unfortunately, Amy knows there are two types of men in the world: Gentlemen and Playboys. She doesn’t yet know Brad's type, but from experience, she assigns probability 3/4 to Brad being a Playboy and probability 1/4 that he is a Gentleman. Brad knows his own type. The payoffs for each action and type profile are given as follows:
艾米(排球员)刚刚开始和布拉德(列球员)约会。每个人都可以选择(C)省略这段关系,或者(D)拒绝他们的伴侣并分手。艾米更喜欢一个愿意承诺的合作伙伴。不幸的是,艾米知道世界上有两种类型的男人:绅士和花花公子。她还不知道布拉德的类型,但根据经验,她认为布拉德是花花公子的概率为 3/4,他是绅士的概率为 1/4。布拉德知道自己的类型。每个操作和类型配置文件的收益如下:Brad is a Playboy (probability 3/4):
布拉德是花花公子(概率 3/4):Commit Dump Commit 3 , 0 3 , 3
Dump 0 , 3
0 , 3 Brad is a Gentleman (probability 1/4):
布拉德是个绅士(概率 1/4):Commit Dump Commit 3 , 3 3 , 0
Dump 0 , 3
0 , 0 What is the unique Bayesian-Nash equilibrium of this game?
这个博弈独特的贝叶斯-纳什均衡是什么?Selected Answer: 选择答案: [None Given]
[未给出]
Answers: 答案: Response Feedback: 响应反馈: Notice that to Brad, both CC and CD are strictly dominated by DC. Given that Brad will never play CC or CD, Amy always strictly prefers Dump to Commit. Given that Amy is always Dumping, Brad therefore strictly prefers DD. Thus the only Bayesian-Nash equilibrium of this game is (D, DD).
请注意,对 Brad 来说,CC 和 CD 都严格由 DC 主导。鉴于 Brad 永远不会播放 CC 或 CD,Amy 总是严格更喜欢 Dump 而不是 Commit。鉴于艾米总是倾倒,布拉德因此严格更喜欢 DD。因此,该博弈的唯一贝叶斯-纳什均衡是(D,DD)。 -
Consider the following Bayesian variant of the Bank Run game, where player 1 has one type and player 2 has two types
考虑以下 Bank Run 游戏的贝叶斯变体,其中玩家 1 有一种类型,玩家 2 有两种类型.1 \ 2
W
R
1 \ 2
W
R
W
50, 50
100, 0
W
50, 50
100, 0
R
0, 100
150, 150
R
0, 100
0,0
Type I I型
Type II II型
Probability: p 概率:p
Probability: 1 – p 概率:1 – p
What is the minimum value of p such that Player 1 plays R in a pure strategy Bayesian Nash equilibrium?
玩家 1 在纯策略贝叶斯纳什均衡中扮演 R 的最小值是多少?Selected Answer: 选择答案: [None Given]
[未给出]
Correct Answer: 正确答案: 0.5
Answer range +/- 答案范围 +/- 0.01 (0.49 - 0.51 )Response Feedback: 响应反馈: First, note that R is never a best-response for player 2 if the state is (Type) II. That means we can write out the ex-ante normal form game as a 2x2, where we only consider strategies WW and RW for player 2. In this game, note that R of player 1 can never be part of an equilibrium if player 2 plays W in Type 1. What's left is to consider whether (R, RW) can be an equilibrium in pure strategies. Player 1's payoff from (R, RW) is 150p and player 2's payoff is 100+50p. Checking for deviations:
首先,请注意,如果状态为 (类型) II,则 R 从来都不是玩家 2 的最佳响应。这意味着我们可以将事前正常形式的游戏写成 2x2,其中我们只考虑玩家 2 的策略 WW 和 RW。在这个游戏中,请注意,如果玩家 2 在类型 1 中玩 W,则玩家 1 的 R 永远不能成为均衡的一部分。剩下的就是考虑(R,RW)是否可以成为纯策略中的均衡。玩家 1 从 (R, RW) 获得的收益为 150p,玩家 2 的收益为 100+50p。检查偏差:- player 1 has no profitable deviation so long as 150p >= 50 + 50p
- 玩家 1 只要 150p >= 50 + 50p 就没有盈利偏差- player 2 has no profitable deviation so long as 100 + 50p >= 100 (a non-binding condition)
- 玩家 2 只要 100 + 50p >= 100(非约束条件),就没有盈利偏差Hence, p >= 0.5. 因此,p >= 0.5。
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Garry and Ross are sales representatives for the same company. Their manager informs them that of the two of them, whoever sells more this year wins a new car. Each day, Garry and Ross privately decide how hard they will work for the day.
Garry 和 Ross 是同一家公司的销售代表。他们的经理告诉他们,在他们两个人中,今年卖得更多的人将赢得一辆新车。每天,Garry 和 Ross 都会私下决定他们当天的工作强度。Select all features from the list below that MOST LIKELY apply to this situation. (Note: partial credit is possible for this question.)
从下面的列表中选择最有可能适用于这种情况的所有功能。(注意:这个问题可以部分学分。Selected Answers: 部分答案: [None Given]
[未给出]
Answers: 答案: Response Feedback: 响应反馈: The game is: Simultaneous, because each day the two sales representatives privately decide how hard to work that day; Not zero-sum (one person gets nothing and the other gets a positive payoff); and a game of Imperfect Information, because neither sales representative knows how hard the other will work.
游戏是:同时进行,因为每天两个销售代表私下决定当天的工作强度;不是零和博弈(一个人一无所获,另一个人得到正回报);以及不完全信息的游戏,因为两个销售代表都不知道对方会有多努力。 -
Six thousand players each pay $10,000 to enter the World Series of Poker. Each starts the tournament with $10,000 in chips, and they play No‑ Limit Texas Hold ’Em (a type of poker) until someone wins all the chips. The top 600 players each receive prize money according to the order of finish, with the winner receiving more than $8,000,000.
六千名玩家每人支付 10,000 美元参加世界扑克系列赛。每个人都以 10,000 美元的筹码开始比赛,然后他们玩无限注德州扑克(一种扑克),直到有人赢得所有筹码。前 600 名选手按完赛顺序获得奖金,获胜者将获得超过 8,000,000 美元。Select all features from the list below that MOST LIKELY apply to this situation. (Note: partial credit is possible for this question.)
从下面的列表中选择最有可能适用于这种情况的所有功能。(注意:这个问题可以部分学分。Selected Answers: 部分答案: [None Given]
[未给出]
Answers: 答案: Response Feedback: 响应反馈: Each hand of poker is a distinct simultaneous game, but the tournament requires repeated hands, so it is also sequential. Therefore, although the tournament is not repeated, the individual hands are, so this game may be considered a repeated, simultaneous game. There is imperfect and incomplete (asymmetric) information: no player knows the cards that the other players hold.
扑克的每一手牌都是一个不同的同时游戏,但锦标赛需要重复的手牌,所以它也是连续的。因此,虽然比赛不是重复的,但个人手牌是重复的,所以这个游戏可以被认为是一个重复的、同时进行的游戏。存在不完美和不完整(不对称)的信息:没有玩家知道其他玩家持有的牌。
Monday, 3 June 2024 15:28:46 o'clock AEST
2024 年 6 月 3 日星期一 15:28:46 点(澳大利亚东部标准时间)