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知道分子 · 2020年04月04日

问一道题:NO.PZ2017092702000094 [ CFA I ]

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.

if 1 million Bernoulli trials are conducted这句话怎么理解,是进行了100万次伯努利实验么?那是不是要算次方?
1 个答案

杨正昆 Danny · 2020年04月06日

这个题一开始我也没懂,后来发现是这个一百万次理解的问题。

这其实是伯努利分布和二项分布的差别

即这是,1百万次伯努利实验的结果分布,而不是1次 n=1million 的二项分布

所以1百万的作用只是形成了这个分布,本质上就是相当于样本数量,所以不需要乘次方

这个题求的是伯努利分布的期望,所以~~

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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

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2022-09-09 21:49 1 · 回答

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