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Lich · 2024年10月17日

如题

  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.

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

50 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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1 个答案

pzqa27 · 2024年10月17日

嗨，爱思考的PZer你好：


可以这么算，解析用的是离散的形式，你用连续的算法算出来结果也差不多。

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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. 老师，如题，盼复，谢谢！

2024-09-04 10:41 1 · 回答

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. 老师好，请问这题为啥不用469页的公式求解？反而去了一种较为简单的公式。看其他同学的提问，助教老师说是因为时间长短的原因？一天的用简便公式？考试时候如何去区分该用哪个公式？另外关于一年天数的选择，是否在FRM里固定取252天？谢谢。

2024-08-06 00:46 1 · 回答

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. 我算出来的 99% confinintervfor the percentage return is between:0.0595%+2.33*1.2599%=2.995%0.0595%-2.33*1.2599%=-2.876%为什么跟解析不一样？

2023-06-08 22:56 1 · 回答

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.这是哪个知识点。。。

2023-03-28 21:13 1 · 回答