老师,这道题是不是考察求二项分布的均值和方差那一部分的内容,可是为什么公式看不懂呢?
问题如下图:
选项:
A.
B.
C.
解释:
菲菲_品职助教 · 2018年10月01日
同学你好,这道题考察的是二项分布。我们一句一句来看题目。首先,你知道当前股票的价格为100,然后有个关键词“one-period binomial”,就说明可以通过二叉树的图解形式来做这道题。up move1.05和down move 0.97,我们要马上提取信息知道股票上升之后的价格为105,下降之后的价格为97。下一句我们提取的有效信息是在经过了1 million次试验之后得到的“average terminal stock price is 102”,知道多次试验后的平均价格为102。最后是问题:题目求的是股价上升的概率。下一步就是画图了:
通过图我们可以很直观的看懂这道题目的本质,并且求出p,即股价上升的概率。
加油~~
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
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?
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?
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
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