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临江仙 · 2020年03月17日

问一道题：NO.PZ2020021205000060

  1. FRM I
  2. Financial Markets And Products

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.

一般题目中，什么时间算短期，什么时间算长期。

个人感觉：短期就利用μ服从正态分布来计算；

长期就利用lnSt来计算。

但题目中什么算短期，什么算长期，这个不知道，求指导

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品职答疑小助手雍 · 2020年03月18日

同学你好，我的印象感觉就是按天的都是短期，到月这种数据跨度大一些的就可以算长期了。一般题目的长短跨度会给的比较明显

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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？

2024-10-17 14:18 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，如题，盼复，谢谢！

2024-09-04 10:41 1 · 回答

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2024-08-06 00:46 1 · 回答

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2023-06-08 22:56 1 · 回答

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2023-03-28 21:13 1 · 回答