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上小学 · 2023年03月04日

请问这个N的倒数0,995是啥意思呢？时间短不是正好fuhBSM 符合这个公式的假设吗？

  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.

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.

应该是多少倍的标准差 为何时间长的倒用BSM？不符合假设啊

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1 个答案

pzqa27 · 2023年03月06日

嗨，从没放弃的小努力你好：


请问这个N的倒数0,995是啥意思呢？

这个是指99.5%累积概率密度下对应的正态分布的分为点的大小

为何时间长的倒用BSM？不符合假设啊

题目原文是： Estimate a 99% confidence interval for the price at the end of one day。这里强调的是估计1天的股价区间，因此不涉及特别长的时间段

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虽然现在很辛苦，但努力过的感觉真的很好，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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