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常馨予 · 2022年06月12日

为什么解题方法与将以471页例题不同呢

  1. FRM I
  2. Financial Markets And Products

NO.PZ2020021205000060

问题如下：

A stock has an expected return of 15% and a volatility of 20%. The current price of the stock is USD 50. Estimate a 99% confidence interval for the price at the end of one day.

选项：

解释：

Here, we are dealing with a short time period, and so it is reasonable to assume that the return is normally distributed. The return has a mean of 15% X (1 /252) = 0.0595%, and a standard deviation of 20% X 1/252\sqrt{1/252} = 1.2599%. The 99% confidence interval for the percentage return is between:

0.0595 - 1.2599 X N−1N^{-1}(0.995) = -3.186%

and

0.0595 + 1.2599 X N−1N^{-1}(0.995)= +3.305%

The confidence interval for the stock price is therefore between 50 X 0.96814 = 48.4 and

50 X 1.03305 = 51.7.

讲义471页例题中使用的是价格服从lognormal，计算出lnS的区间后再推算出S的区间，这里为什么不使用这种方法呢？怎么区别解题时应该使用哪种方法？

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DD仔_品职助教 · 2022年06月12日

嗨，从没放弃的小努力你好：


这俩题条件给的是一样的，但是问题问的时间区间不一样。

讲义的题目要求是6个月的股票价格，对于长期的股票价格变动我们用的是log，做法和例题一样。

而这道题问的是一天的情况，对于短期一天两天五天这种，我们使用正态分布来计算，像答案解释的一样，收益率在均值左右开出对应置信区间Z值倍的标准差，根据收益率即可计算出一天结束时的价格区间。

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努力的时光都是限量版，加油！

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NO.PZ2020021205000060 问题如下 A stohexpectereturn of 15% ana volatility of 20%. The current priof the stois US50. Estimate a 99% confinintervfor the prithe enof one y. Here, we are aling with a short time perio anso it is reasonable to assume ththe return is normally stribute The return ha meof 15% X (1 /252) = 0.0595%, ana stanrviation of 20% X 1/252\sqrt{1/252}1/252​ = 1.2599%. The 99% confinintervfor the percentage return is between:0.0595 - 1.2599 X N−1N^{-1}N−1(0.995) = -3.186%an.0595 + 1.2599 X N−1N^{-1}N−1(0.995)= +3.305%The confinintervfor the stopriis therefore between 50 X 0.96814 = 48.4 an0 X 1.03305 = 51.7. The confinintervfor the stopriis therefore between 50 X 0.96814 = 48.4 an0 X 1.03305 = 51.7.老师，上面这个一步想确认一下，是分别用50*exp(-3.186%) 、50*exp(+3.305%)吗？

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