开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)|3.2.5(安卓)APP下载

sarahjia · 2020年02月26日

问一道题:NO.PZ2017092702000094

问题如下:

A stock is priced at $100.00 and follows a one-period binomial process with an up move that equals 1.05 and a down move that equals 0.97. If 1 million Bernoulli trials are conducted, and the average terminal stock price is $102.00, the probability of an up move (p) is closest to:

选项:

A.

0.375.

B.

0.500.

C.

0.625.

解释:

C is correct.

The probability of an up move (p) can be found by solving the equation: (p)uS + (1 – p)dS = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so that p = 0.625.

老师您好,我看了其他同学提问下的回答,“二叉树假设下一期只有两种情况,要么股价上涨为Su(即答案解析公式中的uS),概率为P;要么股价下跌为Sd(即答案解析中dS)”对于这一句我有点疑惑,题目中说的是 with up move 1.05 and down move 0.97,所以我理解的是这个down move 0.97是在up move 1.05 下的一个分支,请问为什么不是我这么理解呢 如果按老师说的那题目不应该是Or down move 0.97吗 ,不好意思老师们呜呜呜我老有这种奇奇怪怪的问题

1 个答案

星星_品职助教 · 2020年02月26日

同学你好,

如果把题目中这句话的并列关系补全,应该是“ with an up move that equals 1.05 and WITH a down move that equals 0.97.”这样应该就能理解了,这个并列关系就是二叉树的两叉。

其实S0=100,下一个阶段画二叉树肯定是一上(105)和一下(97)。如果是100变到105再变到97,首先就只是一支的变化,不是“二叉”了,其次股价就会变成两阶段,也就不是“one-period ”二叉树了。

可以做完题目以后对问题做总结,例如这道题,应该做到再遇到就知道这种描述指的是两叉的画法

 

 

sarahjia · 2020年02月27日

谢谢您!!

  • 1

    回答
  • 0

    关注
  • 497

    浏览
相关问题

NO.PZ2017092702000094 问题如下 A stois price$100.00 anfollows a one-periobinomiprocess with up move thequals 1.05 ana wn move thequals 0.97. If 1 million Bernoulli trials are concte anthe average terminstopriis $102.00, the probability of up move (p) is closest to: A.0.375. B.0.500. C.0.625. C is correct.The probability of up move (p) cfounsolving the equation: (p)uS + (1 – p) = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so thp = 0.625.100到105是move up,概率为p;100到97是move wn,概率为1-p,这是一期二叉树的情况,此时这个一期二叉树的均值就是105×p+97×(1-p)。根据题干,这个均值为102105×p+97×(1-p)=102,可以直接解得p=0.625 没看答案是 我用的二项分布算的,E(X) =np算呢,p= 102/1million 。。。。没答案,,,然后就不知道知识点了这咋能看出来是二叉树呢,我记得课上讲的时候咋的还有个tree的单词。。。咋理解呀老师

2022-12-18 22:27 1 · 回答

NO.PZ2017092702000094问题如下A stois price$100.00 anfollows a one-periobinomiprocess with up move thequals 1.05 ana wn move thequals 0.97. If 1 million Bernoulli trials are concte anthe average terminstopriis $102.00, the probability of up move (p) is closest to:A.0.375.B.0.500.C.0.625. C is correct.The probability of up move (p) cfounsolving the equation: (p)uS + (1 – p) = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so thp = 0.625.100到105是move up,概率为p;100到97是move wn,概率为1-p,这是一期二叉树的情况,此时这个一期二叉树的均值就是105×p+97×(1-p)。根据题干,这个均值为102105×p+97×(1-p)=102,可以直接解得p=0.625 请问为什么上涨是涨到105,下跌是跌到97? 题目不是说的是up move equals 1.05吗?

2022-09-09 21:49 1 · 回答

NO.PZ2017092702000094 问题如下 A stois price$100.00 anfollows a one-periobinomiprocess with up move thequals 1.05 ana wn move thequals 0.97. If 1 million Bernoulli trials are concte anthe average terminstopriis $102.00, the probability of up move (p) is closest to: A.0.375. B.0.500. C.0.625. C is correct.The probability of up move (p) cfounsolving the equation: (p)uS + (1 – p) = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so thp = 0.625.100到105是move up,概率为p;100到97是move wn,概率为1-p,这是一期二叉树的情况,此时这个一期二叉树的均值就是105×p+97×(1-p)。根据题干,这个均值为102105×p+97×(1-p)=102,可以直接解得p=0.625 老师好,看到这道题时尝试用二项分布求概率来解答,解不出来,感觉跟二叉树有些混淆了,请问怎么区分题目是考哪个知识点?

2022-06-17 10:05 1 · 回答

NO.PZ2017092702000094问题如下A stois price$100.00 anfollows a one-periobinomiprocess with up move thequals 1.05 ana wn move thequals 0.97. If 1 million Bernoulli trials are concte anthe average terminstopriis $102.00, the probability of up move (p) is closest to: A.0.375. B.0.500. C.0.625. C is correct.The probability of up move (p) cfounsolving the equation: (p)uS + (1 – p) = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so thp = 0.625.100到105是move up,概率为p;100到97是move wn,概率为1-p,这是一期二叉树的情况,此时这个一期二叉树的均值就是105×p+97×(1-p)。根据题干,这个均值为102105×p+97×(1-p)=102,可以直接解得p=0.625 另外,1 million Bernoulli trials are concte题干里面的这句话如何理解呢,二叉树列跟伯努利试验有啥关联吗

2022-03-26 13:23 1 · 回答

NO.PZ2017092702000094 0.500. 0.625. C is correct. The probability of up move (p) cfounsolving the equation: (p)uS + (1 – p) = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so thp = 0.625. 100到105是move up,概率为p;100到97是move wn,概率为1-p, 这是一期二叉树的情况,此时这个一期二叉树的均值就是105×p+97×(1-p)。 根据题干,这个均值为102 105×p+97×(1-p)=102,可以直接解得p=0.625 我知道要算概率,但是不知道怎样用二叉树计算,麻烦老师讲解一下

2022-03-16 09:38 1 · 回答