第五章效用函数.ppt
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1、Session 5 效用函数,Session Topic,期望货币损益值准则的局限 效用函数的定义和公理 效用函数的构成 风险和效用的关系 损失函数、风险函数和贝叶斯风险,期望货币损益值准则的局限,期望货币损益值准则的局限,以期望货币损益值为标准的决策方法一般只适用于下列几种情况: (1)概率的出现具有明显的客观性值,而且比较稳定; (2)决策不是解决一次性问题,而是解决多次重复的问题; (3)决策的结果不会对决策者带来严重的后果。 如果不符合这些情况,期望货币损益值准则就不适用,需要采用其他标准。 用期望值作为决策准则的根本条件是,决策有不断反复的可能。,所谓决策有不断重复的可能,包括下列三
2、层涵义: 第一,决策本身即为重复性决策。 第二,重复的次数要比较多,尤其是当存在对于决策后果有重 大影响的小概率事件时,只有重复次数相当多时才能用期望值 来作为决策标准,因为只有这样其平均后果才接近于后果的期 望值。 例如,要决定是否按月投保火险,而且需要决定的不是投 保一个月(投保一次),而是决定10年内(即120个月)是否投 保,这就重复120次了。但因为失火损失较大,而失火概率又非 常小,比如说仅万分之一,即120个月也不一定会失一次火,所 以其实际平均后果就和期望值相差很大。,假定投保者投保资产为12万元,而保险费规定为万分之二 (保险 费征收率一定比失火概率大,否则保险公司就无盈利可
3、图了), 那么每月应交保险费24元,这是在投保情况下每月的支出。如果 不去投保,则损失的期望值为120000(1/10000)=12元,比投 保的支出小得多。如按期望值标准,则谁也不会去投保了。可是 实际上决策者还是会去投保的,这是因为实际平均损失与其期望 值大不一样,如果这120个月中没有失火,则1元损失也没有,但 万一失火一次,则等于每月平均损失120000/120=1000元,比计 算的损失期望值大80多倍。所以计算出来的损失期望值对决定是 否投保的决策者来说毫无意义,决策者往往会按“不怕一万,只 怕万一”的心里去投保。因为拥有12万元资产的决策者来说,每 月支出24元同其资产额相比几乎
4、等于零,而万一失火却会遭受惨 重损失。,第三,每次决策后果都不会给决策者造成致命的威胁,否则,如 果有此威胁,一旦真的产生此种致命后果,决策者就不可能再作 下一次决策,从而也失去了重复的可能性。这就像投机者把全部 资本孤注一掷一样,一旦失败,资本赔光,下一次也就无法再投 机了。对于有此致命危险的重复性决策,期望值标准的采用也就 受到了限制。 最后,采用期望值标准时,还得假定在不断重复作出相同决策时 其客观条件不变,这一方面包括了个自然状态的概率不变,另一 方面亦包括决策后果函数不变。,Example 1 St. Petersburg paradox A prime motivator for
5、Bernoullis work on the evaluation or risky ventures was the famous St. Petersburg game. In current terms, a fair coin is tossed until a head appears. If the first head occurs at the nth toss, the payoff is 2n$. Suppose you own title to one play of the game; that is, you can engage in it without cost
6、. What is the least amount you would sell your title for? According to the Bernoullis, this least amount is your equivalent monetary value of the game. He observed that the expected payoff (1/2)2 + (1/4)22 + (1/8)23 + = 1 +1 +1 + is infinite, but most people would sell title for a relatively small s
7、um, and he asked for an explanation of such a flagrant violation of maximum expected return.,Daniel showed how his theory resolves the issue by providing a unique solution s to the equation for any finite w0, where s is the minimum selling price or equivalent monetary value. Moreover, except for the
8、 very rich, a person would gladly sell title for about $25 or $30. The effect of w0 can be seen indirectly by estimating your minimum selling price when the payoff at n is 2n cents instead of 2n dollars and comparing 100 times this estimate to your answer from the preceding paragraph. Unlike Bernoul
9、li, Cramer pays little attention to initial wealth, and for x 0 sets v(x) = . In his terms, the minimum selling price is the value of s that satisfies which is a little under $6.,Example 2 A Game illustrating the St. Petersburg paradox,A casino makes repeated independent tosses of a fair coin until
10、a tail occurs. A gambler, starting with a stake of $1, is offered the following wager. After each toss the gambler will be given two choices. He may either take away his winnings from the previous tosses of the coin. In this case the game will end. Alternatively he may use all his winnings from prev
11、ious tosses plus his original stake money as a stake for the next toss of the coin. This stake will be tripled by the casino if a head is tossed on the next throw. On the other hand, if a tail is thrown the gambler will lose all his winnings from previous tosses of the coin together with his origina
12、l stake money.,Suppose that the gambler is instructed to follow the EMV algorithm when playing this game. Assume r consecutive heads have been thrown and denote the gamblers original stake plus total winnings as Sr. His expected pay-off for withdrawing from the game is clearly Sr. However, his expec
13、ted pay-off for continuing to play is at least (1/2) 3Sr+(1/2)0=(3/2) Sr (the expected pay-off for playing once more). So under the EMV algorithm the gambler should continue to stake his winnings until a tail is thrown. But since a tail will be thrown eventually with probability one, by following th
14、e EMV algorithm the gambler ensures that he will lose his original stake money with certainty!,Clearly, in the simple game given above, very rational people will not want to follow the dictates of the EMV algorithm. It is therefore necessary to modify the EMV algorithm so that optimal decisions can
15、be defined sensibly for situations like the one given above. It will be shown in the next section that such a modification is possible provided that your client is prepared to commit himself to following certain rules (or axioms). It also generalizes the EMV approach to problems when clients objecti
16、ves are not only the maximization of pay-off.,Homework: A medical laboratory has to test N samples of blood to see which have traces of a rare disease. The probability any one patient has the disease is p, and given p, the probability of any group of patients having the disease is uninfluenced by th
17、e existence or otherwise of the disease in any other disjoint group of patients. Because p is believed to be small it is suggested that the laboratory combine the blood of d patients into equal sized pools of n = N/d samples where d is a divisor of N. Each pool of samples would then be tested to see
18、 if it exhibited a trace of the infection. If no trace were found then the individual samples comprising the group would be known to be uninfected. If on the other hand a trace were found in the pooled sample it is then proposed that each of the n samples comprising that pool be tested individually.
19、 If it costs 1 to test any sample of blood, whether pooled or unpooled, find the Bayes decision for the optimal size of the groups of patients for a given value of p.,Answer You are given the space of decisions you are to consider is the set of divisions of N. All the uncertainty in the experiment e
20、xists because you do not know (d) the number of tests your client will need to do if he chooses to pool the samples into groups if d samples. His monetary loss is just L(d)= (d). The uncertain quantity (d) can be broken down into two components, being the sum of the number n of tested pools plus the
21、 number of individual patients that subsequently need to be checked. If d = 1, and he chooses to test samples individually, the second component of this sum is known to be zero. So the corresponding expected loss L(1) = N, the number of patients. Suppose d 1 and you choose to pool the samples in som
22、e way. If denotes the probability that a pool has no trace of diseased blood, then is the probability that no patient in the pool has the infection. Since patients have the disease independently it follows by the laws of probability that = (1-p)d,So, since he will test n = N/d such pooled samples, t
23、he expected number of pooled samples that need retesting is n(1-). If a pool is found to have traces of the disease then all members of the pool will be retested. So the expected number of samples that subsequently need retesting is dn(1-) = N1-(1-p)d Adding this number to the chosen number of pools
24、 (n = N/d ) gives the expected number of tests (or equivalently the expected loss in $) for choosing to use sample pools of size d 1. Combining these results gives that Although is not a linear function of d (as it was in our first example), given p, can be easily calculated for each divisor d of N
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